First order differential equation. Types of differential equations, solution methods

The first order, having the standard form $y"+P\left(x\right)\cdot y=0$, where $P\left(x\right)$ is a continuous function, is called linear homogeneous. The name "linear" is explained the fact that the unknown function $y$ and its first derivative $y"$ are included in the equation linearly, that is, to the first degree. The name "homogeneous" comes from the fact that there is a zero on the right side of the equation.

Such a differential equation can be solved using the separation of variables method. Let's imagine it in standard form method: $y"=-P\left(x\right)\cdot y$, where $f_(1) \left(x\right)=-P\left(x\right)$ and $f_(2) \left(y\right)=y$.

Let's calculate the integral $I_(1) =\int f_(1) \left(x\right)\cdot dx =-\int P\left(x\right)\cdot dx $.

Let's calculate the integral $I_(2) =\int \frac(dy)(f_(2) \left(y\right)) =\int \frac(dy)(y) =\ln \left|y\right|$ .

Let's write it down common decision in the form $\ln \left|y\right|+\int P\left(x\right)\cdot dx =\ln \left|C_(1) \right|$, where $\ln \left|C_( 1) \right|$ is an arbitrary constant, taken in a form convenient for further transformations.

Let's perform the transformations:

\[\ln \left|y\right|-\ln \left|C_(1) \right|=-\int P\left(x\right)\cdot dx ; \ln \frac(\left|y\right|)(\left|C_(1) \right|) =-\int P\left(x\right)\cdot dx .\]

Using the definition of a logarithm, we get: $\left|y\right|=\left|C_(1) \right|\cdot e^(-\int P\left(x\right)\cdot dx ) $. This equality, in turn, is equivalent to the equality $y=\pm C_(1) \cdot e^(-\int P\left(x\right)\cdot dx ) $.

Replacing the arbitrary constant $C=\pm C_(1) $, we obtain the general solution of the linear homogeneous differential equation: $y=C\cdot e^(-\int P\left(x\right)\cdot dx ) $.

Having solved the equation $f_(2) \left(y\right)=y=0$, we find special solutions. By a usual check we are convinced that the function $y=0$ is a special solution of this differential equation.

However, the same solution can be obtained from the general solution $y=C\cdot e^(-\int P\left(x\right)\cdot dx ) $, putting $C=0$ in it.

So the final result is: $y=C\cdot e^(-\int P\left(x\right)\cdot dx ) $.

The general method for solving a first-order linear homogeneous differential equation can be represented as the following algorithm:

1. To solve this equation, it must first be presented in the standard form of the $y"+P\left(x\right)\cdot y=0$ method. If this was not achieved, then this differential equation must be solved by a different method.
2. We calculate the integral $I=\int P\left(x\right)\cdot dx $.
3. We write the general solution in the form $y=C\cdot e^(-I) $ and, if necessary, perform simplifying transformations.

Problem 1

Find the general solution to the differential equation $y"+3\cdot x^(2) \cdot y=0$.

We have a linear homogeneous equation of the first order in standard form, for which $P\left(x\right)=3\cdot x^(2) $.

We calculate the integral $I=\int 3\cdot x^(2) \cdot dx =x^(3) $.

The general solution has the form: $y=C\cdot e^(-x^(3) ) $.

Linear inhomogeneous differential equations of the first order

Definition

A first order differential equation that can be represented in the standard form $y"+P\left(x\right)\cdot y=Q\left(x\right)$, where $P\left(x\right)$ and $ Q\left(x\right)$ -- known continuous functions, is called a linear inhomogeneous differential equation. The name "inhomogeneous" is explained by the fact that the right side of the differential equation is nonzero.

The solution of one complex linear inhomogeneous differential equation can be reduced to the solution of two simpler ones differential equations. To do this, the required function $y$ should be replaced by the product of two auxiliary functions $u$ and $v$, that is, put $y=u\cdot v$.

We differentiate the accepted replacement: $\frac(dy)(dx) =\frac(du)(dx) \cdot v+u\cdot \frac(dv)(dx) $. We substitute the resulting expression into this differential equation: $\frac(du)(dx) \cdot v+u\cdot \frac(dv)(dx) +P\left(x\right)\cdot u\cdot v=Q\ left(x\right)$ or $\frac(du)(dx) \cdot v+u\cdot \left[\frac(dv)(dx) +P\left(x\right)\cdot v\right] =Q\left(x\right)$.

Note that if $y=u\cdot v$ is accepted, then one of the auxiliary functions can be chosen arbitrarily as part of the product $u\cdot v$. Let us choose the auxiliary function $v$ so that the expression in square brackets becomes zero. To do this, it is enough to solve the differential equation $\frac(dv)(dx) +P\left(x\right)\cdot v=0$ for the function $v$ and choose the simplest particular solution for it $v=v\left(x \right)$, nonzero. This differential equation is linear homogeneous and is solved by the method discussed above.

We substitute the resulting solution $v=v\left(x\right)$ into this differential equation, taking into account the fact that now the expression in square brackets is equal to zero, and we obtain another differential equation, but now with respect to the auxiliary function $u$: $\ frac(du)(dx) \cdot v\left(x\right)=Q\left(x\right)$. This differential equation can be represented as $\frac(du)(dx) =\frac(Q\left(x\right))(v\left(x\right)) $, after which it becomes obvious that it allows immediate integration. For this differential equation it is necessary to find a general solution in the form $u=u\left(x,\; C\right)$.

Now we can find a general solution to this first-order linear inhomogeneous differential equation in the form $y=u\left(x,C\right)\cdot v\left(x\right)$.

The general method for solving a first-order linear inhomogeneous differential equation can be represented as the following algorithm:

1. To solve this equation, it must first be presented in the standard form of the method $y"+P\left(x\right)\cdot y=Q\left(x\right)$. If this was not achieved, then this differential equation must be solved by another method.
2. We calculate the integral $I_(1) =\int P\left(x\right)\cdot dx $, write a particular solution in the form $v\left(x\right)=e^(-I_(1) ) $, execute simplifying transformations and choose the simplest non-zero option for $v\left(x\right)$.
3. We calculate the integral $I_(2) =\int \frac(Q\left(x\right))(v\left(x\right)) \cdot dx $, after which we write the expression in the form $u\left(x, C\right)=I_(2) +C$.
4. We write the general solution of this linear inhomogeneous differential equation in the form $y=u\left(x,C\right)\cdot v\left(x\right)$ and, if necessary, perform simplifying transformations.

Problem 2

Find the general solution to the differential equation $y"-\frac(y)(x) =3\cdot x$.

We have a first-order linear inhomogeneous equation in standard form, for which $P\left(x\right)=-\frac(1)(x) $ and $Q\left(x\right)=3\cdot x$.

We calculate the integral $I_(1) =\int P\left(x\right)\cdot dx =-\int \frac(1)(x) \cdot dx=-\ln \left|x\right| $.

We write a particular solution in the form $v\left(x\right)=e^(-I_(1) ) $ and perform simplifying transformations: $v\left(x\right)=e^(\ln \left|x\ right|) $; $\ln v\left(x\right)=\ln \left|x\right|$; $v\left(x\right)=\left|x\right|$. For $v\left(x\right)$ we choose the simplest non-zero option: $v\left(x\right)=x$.

We calculate the integral $I_(2) =\int \frac(Q\left(x\right))(v\left(x\right)) \cdot dx =\int \frac(3\cdot x)(x) \ cdot dx=3\cdot x $.

We write the expression $u\left(x,C\right)=I_(2) +C=3\cdot x+C$.

We finally write down the general solution of this linear inhomogeneous differential equation in the form $y=u\left(x,C\right)\cdot v\left(x\right)$, that is, $y=\left(3\cdot x+C \right)\cdot x$.

First order differential equations resolved with respect to the derivative

How to solve first order differential equations

Let us have a first order differential equation resolved with respect to the derivative:
.
Dividing this equation by , at , we get equation of the form:
,
Where .

Next, we look to see if these equations belong to one of the types listed below. If not, then rewrite the equation in the form of differentials. To do this, we write and multiply the equation by .
.

We obtain an equation in the form of differentials: If this equation is not an equation in full differentials
.
, then we consider that in this equation is an independent variable, and is a function of .

Divide the equation by:
,
Next, we look to see if this equation belongs to one of the types listed below, taking into account that we have swapped places.
.

If a type has not been found for this equation, then we see if it is possible to simplify the equation by simple substitution. For example, if the equation is:

then we notice that .

;
.
Then we make a substitution.
.

After this, the equation will take a simpler form:

If this does not help, then we try to find the integrating factor.

Separable equations
,
Divide by and integrate. When we get:
;
.
Equations reducing to separable equations

Homogeneous equations

We solve by substitution:
;
.
where is a function of .
;
.
Then

We separate the variables and integrate.

Equations reducing to homogeneous

Enter the variables and:

There are three methods for solving linear equations.

2) Bernoulli's method.
We are looking for a solution in the form of a product of two functions and a variable:
.
;
.
We can choose one of these functions arbitrarily. Therefore, we choose any non-zero solution of the equation as:
.

3) Method of variation of constant (Lagrange).
Here we first solve the homogeneous equation:

The general solution of the homogeneous equation has the form:
,
where is a constant. Next, we replace the constant with a function that depends on the variable:
.
Substitute into the original equation. As a result, we obtain an equation from which we determine .

Bernoulli's equations

By substitution, Bernoulli's equation is reduced to a linear equation.

This equation can also be solved using the Bernoulli method. That is, we are looking for a solution in the form of a product of two functions depending on the variable:
.
Substitute into the original equation:
;
.
We choose any non-zero solution of the equation as:
.
Having determined , we obtain an equation with separable variables for .

Riccati equations

It is not resolved in general view. Substitution

The Riccati equation is reduced to the form:
,
where is constant; ;
.

Next, by substitution:
,
Where .

it is reduced to the form:
Properties of the Riccati equation and some special cases of its solution are presented on the page

Riccati differential equation >>>

Jacobi equations
.

Solved by substitution:

Equations in total differentials
.
Given that
.
If this condition is met, the expression on the left side of the equality is the differential of some function:
.
Then
.

From here we obtain the integral of the differential equation:
;
;
;
.

To find the function, the most convenient way is the method of sequential differential extraction. To do this, use the formulas:

Integrating factor If a first-order differential equation cannot be reduced to any of the listed types, then you can try to find the integrating factor. An integrating factor is a function, when multiplied by which, a differential equation becomes an equation in total differentials. The first order differential equation has

infinite number

integrating factors. However, there are no general methods for finding the integrating factor.

Equations not solved for the derivative y"

Equations that can be solved with respect to the derivative y"

First you need to try to solve the equation with respect to the derivative.
,
If possible, the equation can be reduced to one of the types listed above. consistent solution simpler equations:
;
;

;
.
We believe.
Then
;
.
or .

Next we integrate the equation: As a result, we obtain the expression of the second variable through the parameter.:
More
general equations or are also solved in parametric form. To do this, you need to choose a function such that
original equation
;
.

could be expressed either through the parameter.

To express the second variable through the parameter, we integrate the equation:

Equations resolved for y

Clairaut equations

This equation has a general solution

Lagrange equations

We are looking for a solution in parametric form. We assume where is a parameter.

Equations leading to Bernoulli's equation
These equations are reduced to the Bernoulli equation if we look for their solutions in parametric form by introducing a parameter and making the substitution .
References:

V.V. Stepanov, Course of differential equations, "LKI", 2015. N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, “Lan”, 2003. A differential equation is an equation that involves a function and one or more of its derivatives. In most practical problems, functions are

physical quantities , the derivatives correspond to the rates of change of these quantities, and the equation determines the relationship between them. This article discusses methods for solving certain types of ordinary differential equations, the solutions of which can be written in the form elementary functions, that is, polynomial, exponential, logarithmic and trigonometric, as well as their inverse functions. Many of these equations appear in real life, although most other differential equations cannot be solved by these methods, and for them the answer is written in the form of special functions or

power series

, or is found by numerical methods.

• To understand this article, you must be proficient in differential and integral calculus, as well as have some understanding of partial derivatives. It is also recommended to know the basics of linear algebra as applied to differential equations, especially second-order differential equations, although knowledge of differential and integral calculus is sufficient to solve them. Preliminary information Differential equations have an extensive classification. This article talks about partial differential equations, which include functions of several variables. This article does not discuss partial differential equations, since the methods for solving these equations are usually determined by their particular form.
  • Below are some examples of ordinary differential equations.
    • d y d x = k y (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=ky)
    • d 2 x d t 2 + k x = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)x)((\mathrm (d) )t^(2)))+kx=0)
  • Below are some examples of partial differential equations.
    • ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2 = 0 (\displaystyle (\frac (\partial ^(2)f)(\partial x^(2)))+(\frac (\partial ^(2 )f)(\partial y^(2)))=0)
    • ∂ u ∂ t − α ∂ 2 u ∂ x 2 = 0 (\displaystyle (\frac (\partial u)(\partial t))-\alpha (\frac (\partial ^(2)u)(\partial x ^(2)))=0)
• Order of a differential equation is determined by the order of the highest derivative included in this equation. The first of the above ordinary differential equations is of first order, while the second is a second order equation. Degree of a differential equation is the highest power to which one of the terms of this equation is raised.
  • For example, the equation below is third order and second degree.
    • (d 3 y d x 3) 2 + d y d x = 0 (\displaystyle \left((\frac ((\mathrm (d) )^(3)y)((\mathrm (d) )x^(3)))\ right)^(2)+(\frac ((\mathrm (d) )y)((\mathrm (d) )x))=0)
• The differential equation is linear differential equation in the event that the function and all its derivatives are in the first degree. Otherwise the equation is nonlinear differential equation. Linear differential equations are remarkable in that their solutions can be used to form linear combinations that will also be solutions to the given equation.
  • Below are some examples of linear differential equations.
  • Below are some examples of nonlinear differential equations. The first equation is nonlinear due to the sine term.
    • d 2 θ d t 2 + g l sin ⁡ θ = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)\theta )((\mathrm (d) )t^(2)))+( \frac (g)(l))\sin \theta =0)
    • d 2 x d t 2 + (d x d t) 2 + t x 2 = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)x)((\mathrm (d) )t^(2)))+ \left((\frac ((\mathrm (d) )x)((\mathrm (d) )t))\right)^(2)+tx^(2)=0)
• Common decision ordinary differential equation is not unique, it includes arbitrary integration constants. In most cases, the number of arbitrary constants is equal to the order of the equation. In practice, the values ​​of these constants are determined based on the given initial conditions, that is, according to the values ​​of the function and its derivatives at x = 0. (\displaystyle x=0.) The number of initial conditions that are necessary to find private solution differential equation, in most cases is also equal to the order of the given equation.
  • For example, this article will look at solving the equation below. This is a second order linear differential equation. Its general solution contains two arbitrary constants. To find these constants it is necessary to know the initial conditions at x (0) (\displaystyle x(0)) And x ′ (0) . (\displaystyle x"(0).) Usually the initial conditions are specified at the point x = 0 , (\displaystyle x=0,), although this is not necessary. This article will also discuss how to find particular solutions for given initial conditions.
    • d 2 x d t 2 + k 2 x = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)x)((\mathrm (d) )t^(2)))+k^(2 )x=0)
    • x (t) = c 1 cos ⁡ k x + c 2 sin ⁡ k x (\displaystyle x(t)=c_(1)\cos kx+c_(2)\sin kx)

Steps

Part 1

First order equations

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1. Linear equations of the first order. This section discusses methods for solving first-order linear differential equations in general and special cases when some terms are equal to zero. Let's pretend that y = y (x) , (\displaystyle y=y(x),) p (x) (\displaystyle p(x)) And q (x) (\displaystyle q(x)) are functions x. (\displaystyle x.)

   D y d x + p (x) y = q (x) (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+p(x)y=q(x ))
   

   P (x) = 0. (\displaystyle p(x)=0.) According to one of the main theorems mathematical analysis, the integral of the derivative of a function is also a function. Thus, it is enough to simply integrate the equation to find its solution. It should be taken into account that when calculating indefinite integral an arbitrary constant appears.

   • y (x) = ∫ q (x) d x (\displaystyle y(x)=\int q(x)(\mathrm (d) )x)

   Q (x) = 0. (\displaystyle q(x)=0.) We use the method separation of variables. This moves different variables to different sides of the equation. For example, you can move all members from y (\displaystyle y) into one, and all members with x (\displaystyle x) to the other side of the equation. Members can also be transferred d x (\displaystyle (\mathrm (d) )x) And d y (\displaystyle (\mathrm (d) )y), which are included in the expressions of derivatives, however, it should be remembered that this is just a symbol that is convenient when differentiating a complex function. Discussion of these members, which are called differentials, is beyond the scope of this article.

   • First, you need to move the variables to opposite sides of the equal sign.
     • 1 y d y = − p (x) d x (\displaystyle (\frac (1)(y))(\mathrm (d) )y=-p(x)(\mathrm (d) )x)
   • Let's integrate both sides of the equation. After integration, arbitrary constants will appear on both sides, which can be transferred to the right side of the equation.
     • ln ⁡ y = ∫ − p (x) d x (\displaystyle \ln y=\int -p(x)(\mathrm (d) )x)
     • y (x) = e − ∫ p (x) d x (\displaystyle y(x)=e^(-\int p(x)(\mathrm (d) )x))
   • Example 1.1. In the last step we used the rule e a + b = e a e b (\displaystyle e^(a+b)=e^(a)e^(b)) and replaced e C (\displaystyle e^(C)) on C (\displaystyle C), since this is also an arbitrary integration constant.
     • d y d x − 2 y sin ⁡ x = 0 (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))-2y\sin x=0)
     • 1 2 y d y = sin ⁡ x d x 1 2 ln ⁡ y = − cos ⁡ x + C ln ⁡ y = − 2 cos ⁡ x + C y (x) = C e − 2 cos ⁡ x (\displaystyle (\begin(aligned )(\frac (1)(2y))(\mathrm (d) )y&=\sin x(\mathrm (d) )x\\(\frac (1)(2))\ln y&=-\cos x+C\\\ln y&=-2\cos x+C\\y(x)&=Ce^(-2\cos x)\end(aligned)))

   P (x) ≠ 0 , q (x) ≠ 0. (\displaystyle p(x)\neq 0,\ q(x)\neq 0.) To find a general solution we introduced integrating factor as a function of x (\displaystyle x) to reduce left side to the common derivative and thus solve the equation.

   • Multiply both sides by μ (x) (\displaystyle \mu (x))
     • μ d y d x + μ p y = μ q (\displaystyle \mu (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+\mu py=\mu q)
   • To reduce the left-hand side to the general derivative, the following transformations must be made:
     • d d x (μ y) = d μ d x y + μ d y d x = μ d y d x + μ p y (\displaystyle (\frac (\mathrm (d) )((\mathrm (d) )x))(\mu y)=(\ frac ((\mathrm (d) )\mu )((\mathrm (d) )x))y+\mu (\frac ((\mathrm (d) )y)((\mathrm (d) )x)) =\mu (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+\mu py)
   • The last equality means that d μ d x = μ p (\displaystyle (\frac ((\mathrm (d) )\mu )((\mathrm (d) )x))=\mu p). This is an integrating factor that is sufficient to solve any first-order linear equation. Now we can derive the formula for solving this equation with respect to μ , (\displaystyle \mu ,) although it is useful for training to do all the intermediate calculations.
     • μ (x) = e ∫ p (x) d x (\displaystyle \mu (x)=e^(\int p(x)(\mathrm (d) )x))
   • Example 1.2. This example shows how to find a particular solution to a differential equation with given initial conditions.
     • t d y d t + 2 y = t 2 , y (2) = 3 (\displaystyle t(\frac ((\mathrm (d) )y)((\mathrm (d) )t))+2y=t^(2) ,\quad y(2)=3)
     • d y d t + 2 t y = t (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )t))+(\frac (2)(t))y=t)
     • μ (x) = e ∫ p (t) d t = e 2 ln ⁡ t = t 2 (\displaystyle \mu (x)=e^(\int p(t)(\mathrm (d) )t)=e ^(2\ln t)=t^(2))
     • d d t (t 2 y) = t 3 t 2 y = 1 4 t 4 + C y (t) = 1 4 t 2 + C t 2 (\displaystyle (\begin(aligned)(\frac (\mathrm (d) )((\mathrm (d) )t))(t^(2)y)&=t^(3)\\t^(2)y&=(\frac (1)(4))t^(4 )+C\\y(t)&=(\frac (1)(4))t^(2)+(\frac (C)(t^(2)))\end(aligned)))
     • 3 = y (2) = 1 + C 4 , C = 8 (\displaystyle 3=y(2)=1+(\frac (C)(4)),\quad C=8)
     • y (t) = 1 4 t 2 + 8 t 2 (\displaystyle y(t)=(\frac (1)(4))t^(2)+(\frac (8)(t^(2)) ))
   
   Solving linear equations of the first order (recorded by Intuit - National Open University).

2. Nonlinear first order equations. This section discusses methods for solving some first-order nonlinear differential equations. Although there is no general method for solving such equations, some of them can be solved using the methods below.

   D y d x = f (x , y) (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=f(x,y))
   d y d x = h (x) g (y) . (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=h(x)g(y).) If the function f (x , y) = h (x) g (y) (\displaystyle f(x,y)=h(x)g(y)) can be divided into functions of one variable, such an equation is called differential equation with separable variables. In this case, you can use the above method:

   • ∫ d y h (y) = ∫ g (x) d x (\displaystyle \int (\frac ((\mathrm (d) )y)(h(y)))=\int g(x)(\mathrm (d) )x)
   • Example 1.3.
     • d y d x = x 3 y (1 + x 4) (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (x^(3))( y(1+x^(4)))))
     • ∫ y d y = ∫ x 3 1 + x 4 d x 1 2 y 2 = 1 4 ln ⁡ (1 + x 4) + C y (x) = 1 2 ln ⁡ (1 + x 4) + C (\displaystyle (\ begin(aligned)\int y(\mathrm (d) )y&=\int (\frac (x^(3))(1+x^(4)))(\mathrm (d) )x\\(\ frac (1)(2))y^(2)&=(\frac (1)(4))\ln(1+x^(4))+C\\y(x)&=(\frac ( 1)(2))\ln(1+x^(4))+C\end(aligned)))

   D y d x = g (x , y) h (x , y) . (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (g(x,y))(h(x,y))).) Let's pretend that g (x , y) (\displaystyle g(x,y)) And h (x , y) (\displaystyle h(x,y)) are functions x (\displaystyle x) And y. (\displaystyle y.) Then homogeneous differential equation is an equation in which g (\displaystyle g) And h (\displaystyle h) are homogeneous functions to the same degree. That is, the functions must satisfy the condition g (α x , α y) = α k g (x , y) , (\displaystyle g(\alpha x,\alpha y)=\alpha ^(k)g(x,y),) Where k (\displaystyle k) is called the degree of homogeneity. Any homogeneous differential equation can be used by suitable substitutions of variables (v = y / x (\displaystyle v=y/x) More v = x / y (\displaystyle v=x/y)) convert to an equation with separable variables.

   • Example 1.4. The above description of homogeneity may seem unclear. Let's look at this concept with an example.
     • d y d x = y 3 − x 3 y 2 x (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (y^(3)-x^ (3))(y^(2)x)))
     • To begin with, it should be noted that this equation is nonlinear with respect to y. (\displaystyle y.) We also see that in in this case You can't separate variables. At the same time, this differential equation is homogeneous, since both the numerator and the denominator are homogeneous with a power of 3. Therefore, we can make a change of variables v = y/x. (\displaystyle v=y/x.)
     • d y d x = y x − x 2 y 2 = v − 1 v 2 (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (y)(x ))-(\frac (x^(2))(y^(2)))=v-(\frac (1)(v^(2))))
     • y = v x , d y d x = d v d x x + v (\displaystyle y=vx,\quad (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac ((\mathrm (d) )v)((\mathrm (d) )x))x+v)
     • d v d x x = − 1 v 2 . (\displaystyle (\frac ((\mathrm (d) )v)((\mathrm (d) )x))x=-(\frac (1)(v^(2))).) As a result, we have the equation for v (\displaystyle v) with separable variables.
     • v (x) = − 3 ln ⁡ x + C 3 (\displaystyle v(x)=(\sqrt[(3)](-3\ln x+C)))
     • y (x) = x − 3 ln ⁡ x + C 3 (\displaystyle y(x)=x(\sqrt[(3)](-3\ln x+C)))

   D y d x = p (x) y + q (x) y n . (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=p(x)y+q(x)y^(n).) This Bernoulli differential equation- a special type of nonlinear equation of the first degree, the solution of which can be written using elementary functions.

   • Multiply both sides of the equation by (1 − n) y − n (\displaystyle (1-n)y^(-n)):
     • (1 − n) y − n d y d x = p (x) (1 − n) y 1 − n + (1 − n) q (x) (\displaystyle (1-n)y^(-n)(\frac ( (\mathrm (d) )y)((\mathrm (d) )x))=p(x)(1-n)y^(1-n)+(1-n)q(x))
   • We use the rule for differentiating a complex function on the left side and transform the equation into linear equation relatively y 1 − n , (\displaystyle y^(1-n),) which can be solved using the above methods.
     • d y 1 − n d x = p (x) (1 − n) y 1 − n + (1 − n) q (x) (\displaystyle (\frac ((\mathrm (d) )y^(1-n)) ((\mathrm (d) )x))=p(x)(1-n)y^(1-n)+(1-n)q(x))

   M (x , y) + N (x , y) d y d x = 0. (\displaystyle M(x,y)+N(x,y)(\frac ((\mathrm (d) )y)((\mathrm (d) )x))=0.) This equation in total differentials. It is necessary to find the so-called potential function φ (x , y) , (\displaystyle \varphi (x,y),), which satisfies the condition d φ d x = 0. (\displaystyle (\frac ((\mathrm (d) )\varphi )((\mathrm (d) )x))=0.)

   • For execution this condition must have total derivative. The total derivative takes into account the dependence on other variables. To calculate the total derivative φ (\displaystyle \varphi ) By x , (\displaystyle x,) we assume that y (\displaystyle y) may also depend on x. (\displaystyle x.)
     • d φ d x = ∂ φ ∂ x + ∂ φ ∂ y d y d x (\displaystyle (\frac ((\mathrm (d) )\varphi )((\mathrm (d) )x))=(\frac (\partial \varphi )(\partial x))+(\frac (\partial \varphi )(\partial y))(\frac ((\mathrm (d) )y)((\mathrm (d) )x)))
   • Comparing the terms gives us M (x , y) = ∂ φ ∂ x (\displaystyle M(x,y)=(\frac (\partial \varphi )(\partial x))) And N (x, y) = ∂ φ ∂ y. (\displaystyle N(x,y)=(\frac (\partial \varphi )(\partial y)).) This is a typical result for equations in several variables, in which the mixed derivatives of smooth functions are equal to each other. Sometimes this case is called Clairaut's theorem. In this case, the differential equation is a total differential equation if next condition:
     • ∂ M ∂ y = ∂ N ∂ x (\displaystyle (\frac (\partial M)(\partial y))=(\frac (\partial N)(\partial x)))
   • The method for solving equations in total differentials is similar to finding potential functions in the presence of several derivatives, which we will briefly discuss. First let's integrate M (\displaystyle M) By x. (\displaystyle x.) Because the M (\displaystyle M) is a function and x (\displaystyle x), And y , (\displaystyle y,) upon integration we get an incomplete function φ , (\displaystyle \varphi ,) designated as φ ~ (\displaystyle (\tilde (\varphi ))). The result also depends on y (\displaystyle y) integration constant.
     • φ (x , y) = ∫ M (x , y) d x = φ ~ (x , y) + c (y) (\displaystyle \varphi (x,y)=\int M(x,y)(\mathrm (d) )x=(\tilde (\varphi ))(x,y)+c(y))
   • After this, to get c (y) (\displaystyle c(y)) we can take the partial derivative of the resulting function with respect to y , (\displaystyle y,) equate the result N (x , y) (\displaystyle N(x,y)) and integrate. You can also first integrate N (\displaystyle N), and then take the partial derivative with respect to x (\displaystyle x), which will allow you to find an arbitrary function d(x). (\displaystyle d(x).) Both methods are suitable, and usually the simpler function is chosen for integration.
     • N (x , y) = ∂ φ ∂ y = ∂ φ ~ ∂ y + d c d y (\displaystyle N(x,y)=(\frac (\partial \varphi )(\partial y))=(\frac (\ partial (\tilde (\varphi )))(\partial y))+(\frac ((\mathrm (d) )c)((\mathrm (d) )y)))
   • Example 1.5. You can take partial derivatives and see that the equation below is a total differential equation.
     • 3 x 2 + y 2 + 2 x y d y d x = 0 (\displaystyle 3x^(2)+y^(2)+2xy(\frac ((\mathrm (d) )y)((\mathrm (d) )x) )=0)
     • φ = ∫ (3 x 2 + y 2) d x = x 3 + x y 2 + c (y) ∂ φ ∂ y = N (x , y) = 2 x y + d c d y (\displaystyle (\begin(aligned)\varphi &=\int (3x^(2)+y^(2))(\mathrm (d) )x=x^(3)+xy^(2)+c(y)\\(\frac (\partial \varphi )(\partial y))&=N(x,y)=2xy+(\frac ((\mathrm (d) )c)((\mathrm (d) )y))\end(aligned)))
     • d c d y = 0 , c (y) = C (\displaystyle (\frac ((\mathrm (d) )c)((\mathrm (d) )y))=0,\quad c(y)=C)
     • x 3 + x y 2 = C (\displaystyle x^(3)+xy^(2)=C)
   • If the differential equation is not a total differential equation, in some cases you can find an integrating factor that allows you to convert it into a total differential equation. However, such equations are rarely used in practice, and although the integrating factor exists, it happens to find it not easy, therefore these equations are not considered in this article.

Part 2

Second order equations

1. Homogeneous linear differential equations with constant coefficients. These equations are widely used in practice, so their solution is of primary importance. In this case, we are not talking about homogeneous functions, but about the fact that there is 0 on the right side of the equation. The next section will show how to solve the corresponding heterogeneous differential equations. Below a (\displaystyle a) And b (\displaystyle b) are constants.

   D 2 y d x 2 + a d y d x + b y = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2)))+a(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+by=0)
   

   Characteristic equation. This differential equation is remarkable in that it can be solved very easily if you pay attention to what properties its solutions should have. From the equation it is clear that y (\displaystyle y) and its derivatives are proportional to each other. From previous examples, which were discussed in the section on first-order equations, we know that only exponential function. Therefore, it is possible to put forward ansatz(an educated guess) about what the solution to a given equation will be.

   • The solution will have the form of an exponential function e r x , (\displaystyle e^(rx),) Where r (\displaystyle r) is a constant whose value should be found. Substitute this function into the equation and get the following expression
     • e r x (r 2 + a r + b) = 0 (\displaystyle e^(rx)(r^(2)+ar+b)=0)
   • This equation indicates that the product of an exponential function and a polynomial must equal zero. It is known that the exponent cannot be equal to zero for any values ​​of the degree. From this we conclude that the polynomial is equal to zero. Thus, we have reduced the problem of solving a differential equation to the much simpler problem of solving an algebraic equation, which is called the characteristic equation for a given differential equation.
     • r 2 + a r + b = 0 (\displaystyle r^(2)+ar+b=0)
     • r ± = − a ± a 2 − 4 b 2 (\displaystyle r_(\pm )=(\frac (-a\pm (\sqrt (a^(2)-4b)))(2)))
   • We got two roots. Since this differential equation is linear, its general solution is a linear combination of partial solutions. Since this is a second order equation, we know that it is really general solution, and there are no others. A more rigorous justification for this lies in theorems on the existence and uniqueness of a solution, which can be found in textbooks.
   • A useful way to check whether two solutions are linearly independent is to calculate Wronskiana. Vronskian W (\displaystyle W) is the determinant of a matrix whose columns contain functions and their successive derivatives. The linear algebra theorem states that the functions included in the Wronskian are linearly dependent if the Wronskian is equal to zero. In this section we can check whether two solutions are linearly independent - to do this we need to make sure that the Wronskian is not zero. The Wronskian is important in solving inhomogeneous differential equations with constant coefficients by the method of varying parameters.
     • W = | y 1 y 2 y 1 ′ y 2 ′ | (\displaystyle W=(\begin(vmatrix)y_(1)&y_(2)\\y_(1)"&y_(2)"\end(vmatrix)))
   • In terms of linear algebra, the set of all solutions to a given differential equation forms a vector space whose dimension is equal to the order of the differential equation. In this space one can choose a basis from linearly independent decisions from each other. This is possible due to the fact that the function y (x) (\displaystyle y(x)) valid linear operator. Derivative is linear operator, since it transforms the space of differentiable functions into the space of all functions. Equations are called homogeneous in those cases when, for some linear operator L (\displaystyle L) we need to find a solution to the equation L [ y ] = 0. (\displaystyle L[y]=0.)

   Let us now move on to consider several specific examples. We will consider the case of multiple roots of the characteristic equation a little later, in the section on reducing the order.

   If the roots r ± (\displaystyle r_(\pm )) are different real numbers, the differential equation has the following solution

   • y (x) = c 1 e r + x + c 2 e r − x (\displaystyle y(x)=c_(1)e^(r_(+)x)+c_(2)e^(r_(-)x ))

   Two complex roots. From the fundamental theorem of algebra it follows that solutions to polynomial equations with real coefficients have roots that are real or form conjugate pairs. Therefore, if complex number r = α + i β (\displaystyle r=\alpha +i\beta ) is the root of the characteristic equation, then r ∗ = α − i β (\displaystyle r^(*)=\alpha -i\beta ) is also the root of this equation. Thus, we can write the solution in the form c 1 e (α + i β) x + c 2 e (α − i β) x , (\displaystyle c_(1)e^((\alpha +i\beta)x)+c_(2)e^( (\alpha -i\beta)x),) however, it is a complex number and is not desirable for solving practical problems.

   • Instead you can use Euler's formula e i x = cos ⁡ x + i sin ⁡ x (\displaystyle e^(ix)=\cos x+i\sin x), which allows us to write the solution in the form trigonometric functions:
     • e α x (c 1 cos ⁡ β x + i c 1 sin ⁡ β x + c 2 cos ⁡ β x − i c 2 sin ⁡ β x) (\displaystyle e^(\alpha x)(c_(1)\cos \ beta x+ic_(1)\sin \beta x+c_(2)\cos \beta x-ic_(2)\sin \beta x))
   • Now you can instead of a constant c 1 + c 2 (\displaystyle c_(1)+c_(2)) write down c 1 (\displaystyle c_(1)), and the expression i (c 1 − c 2) (\displaystyle i(c_(1)-c_(2))) replaced by c 2 . (\displaystyle c_(2).) After this we get the following solution:
     • y (x) = e α x (c 1 cos ⁡ β x + c 2 sin ⁡ β x) (\displaystyle y(x)=e^(\alpha x)(c_(1)\cos \beta x+c_ (2)\sin\beta x))
   • There is another way to write the solution in terms of amplitude and phase, which is better suited for physics problems.
   • Example 2.1. Let us find a solution to the differential equation given below with the given initial conditions. To do this, you need to take the resulting solution, as well as its derivative, and substitute them into the initial conditions, which will allow us to determine arbitrary constants.
     • d 2 x d t 2 + 3 d x d t + 10 x = 0 , x (0) = 1 , x ′ (0) = − 1 (\displaystyle (\frac ((\mathrm (d) )^(2)x)(( \mathrm (d) )t^(2)))+3(\frac ((\mathrm (d) )x)((\mathrm (d) )t))+10x=0,\quad x(0) =1,\x"(0)=-1)
     • r 2 + 3 r + 10 = 0 , r ± = − 3 ± 9 − 40 2 = − 3 2 ± 31 2 i (\displaystyle r^(2)+3r+10=0,\quad r_(\pm ) =(\frac (-3\pm (\sqrt (9-40)))(2))=-(\frac (3)(2))\pm (\frac (\sqrt (31))(2) )i)
     • x (t) = e − 3 t / 2 (c 1 cos ⁡ 31 2 t + c 2 sin ⁡ 31 2 t) (\displaystyle x(t)=e^(-3t/2)\left(c_(1 )\cos (\frac (\sqrt (31))(2))t+c_(2)\sin (\frac (\sqrt (31))(2))t\right))
     • x (0) = 1 = c 1 (\displaystyle x(0)=1=c_(1))
     • x ′ (t) = − 3 2 e − 3 t / 2 (c 1 cos ⁡ 31 2 t + c 2 sin ⁡ 31 2 t) + e − 3 t / 2 (− 31 2 c 1 sin ⁡ 31 2 t + 31 2 c 2 cos ⁡ 31 2 t) (\displaystyle (\begin(aligned)x"(t)&=-(\frac (3)(2))e^(-3t/2)\left(c_ (1)\cos (\frac (\sqrt (31))(2))t+c_(2)\sin (\frac (\sqrt (31))(2))t\right)\\&+e ^(-3t/2)\left(-(\frac (\sqrt (31))(2))c_(1)\sin (\frac (\sqrt (31))(2))t+(\frac ( \sqrt (31))(2))c_(2)\cos (\frac (\sqrt (31))(2))t\right)\end(aligned)))
     • x ′ (0) = − 1 = − 3 2 c 1 + 31 2 c 2 , c 2 = 1 31 (\displaystyle x"(0)=-1=-(\frac (3)(2))c_( 1)+(\frac (\sqrt (31))(2))c_(2),\quad c_(2)=(\frac (1)(\sqrt (31))))
     • x (t) = e − 3 t / 2 (cos ⁡ 31 2 t + 1 31 sin ⁡ 31 2 t) (\displaystyle x(t)=e^(-3t/2)\left(\cos (\frac (\sqrt (31))(2))t+(\frac (1)(\sqrt (31)))\sin (\frac (\sqrt (31))(2))t\right))
   
   Solving nth order differential equations with constant coefficients (recorded by Intuit - National Open University).

2. Decreasing order. Order reduction is a method for solving differential equations when one linearly independent solution is known. This method consists of lowering the order of the equation by one, which allows you to solve the equation using the methods described in the previous section. Let the solution be known. The main idea of ​​order reduction is to find a solution in the form below, where it is necessary to define the function v (x) (\displaystyle v(x)), substituting it into the differential equation and finding v(x). (\displaystyle v(x).) Let's look at how order reduction can be used to solve a differential equation with constant coefficients and multiple roots.

   
   

   Multiple roots homogeneous differential equation with constant coefficients. Recall that a second-order equation must have two linearly independent solutions. If characteristic equation has multiple roots, many solutions Not forms a space since these solutions are linearly dependent. In this case, it is necessary to use order reduction to find a second linearly independent solution.

   • Let the characteristic equation have multiple roots r (\displaystyle r). Let us assume that the second solution can be written in the form y (x) = e r x v (x) (\displaystyle y(x)=e^(rx)v(x)), and substitute it into the differential equation. In this case, most terms, with the exception of the term with the second derivative of the function v , (\displaystyle v,) will be reduced.
     • v ″ (x) e r x = 0 (\displaystyle v""(x)e^(rx)=0)
   • Example 2.2. Let the following equation be given which has multiple roots r = − 4. (\displaystyle r=-4.) When substituting, most terms are reduced.
     • d 2 y d x 2 + 8 d y d x + 16 y = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2)))+8( \frac ((\mathrm (d) )y)((\mathrm (d) )x))+16y=0)
     • y = v (x) e − 4 x y ′ = v ′ (x) e − 4 x − 4 v (x) e − 4 x y ″ = v ″ (x) e − 4 x − 8 v ′ (x) e − 4 x + 16 v (x) e − 4 x (\displaystyle (\begin(aligned)y&=v(x)e^(-4x)\\y"&=v"(x)e^(-4x )-4v(x)e^(-4x)\\y""&=v""(x)e^(-4x)-8v"(x)e^(-4x)+16v(x)e^ (-4x)\end(aligned)))
     • v ″ e − 4 x − 8 v ′ e − 4 x + 16 v e − 4 x + 8 v ′ e − 4 x − 32 v e − 4 x + 16 v e − 4 x = 0 (\displaystyle (\begin(aligned )v""e^(-4x)&-(\cancel (8v"e^(-4x)))+(\cancel (16ve^(-4x)))\\&+(\cancel (8v"e ^(-4x)))-(\cancel (32ve^(-4x)))+(\cancel (16ve^(-4x)))=0\end(aligned)))
   • Similar to our ansatz for a differential equation with constant coefficients, in this case only the second derivative can be equal to zero. We integrate twice and obtain the desired expression for v (\displaystyle v):
     • v (x) = c 1 + c 2 x (\displaystyle v(x)=c_(1)+c_(2)x)
   • Then the general solution of a differential equation with constant coefficients in the case where the characteristic equation has multiple roots can be written in the following form. For convenience, you can remember that to get linear independence just multiply the second term by x (\displaystyle x). This set of solutions is linearly independent, and thus we have found all the solutions to this equation.
     • y (x) = (c 1 + c 2 x) e r x (\displaystyle y(x)=(c_(1)+c_(2)x)e^(rx))

   D 2 y d x 2 + p (x) d y d x + q (x) y = 0. (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^( 2)))+p(x)(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+q(x)y=0.) Order reduction is applicable if the solution is known y 1 (x) (\displaystyle y_(1)(x)), which can be found or given in the problem statement.

   • We are looking for a solution in the form y (x) = v (x) y 1 (x) (\displaystyle y(x)=v(x)y_(1)(x)) and substitute it into this equation:
     • v ″ y 1 + 2 v ′ y 1 ′ + p (x) v ′ y 1 + v (y 1 ″ + p (x) y 1 ′ + q (x)) = 0 (\displaystyle v""y_( 1)+2v"y_(1)"+p(x)v"y_(1)+v(y_(1)""+p(x)y_(1)"+q(x))=0)
   • Because the y 1 (\displaystyle y_(1)) is a solution to a differential equation, all terms with v (\displaystyle v) are being reduced. In the end it remains first order linear equation. To see this more clearly, let's make a change of variables w (x) = v ′ (x) (\displaystyle w(x)=v"(x)):
     • y 1 w ′ + (2 y 1 ′ + p (x) y 1) w = 0 (\displaystyle y_(1)w"+(2y_(1)"+p(x)y_(1))w=0 )
     • w (x) = exp ⁡ (∫ (2 y 1 ′ (x) y 1 (x) + p (x)) d x) (\displaystyle w(x)=\exp \left(\int \left((\ frac (2y_(1)"(x))(y_(1)(x)))+p(x)\right)(\mathrm (d) )x\right))
     • v (x) = ∫ w (x) d x (\displaystyle v(x)=\int w(x)(\mathrm (d) )x)
   • If the integrals can be calculated, we obtain the general solution as a combination of elementary functions. Otherwise, the solution can be left in integral form.

3. Cauchy-Euler equation. The Cauchy-Euler equation is an example of a second order differential equation with variables coefficients, which has exact solutions. This equation is used in practice, for example, to solve the Laplace equation in spherical coordinates.

   X 2 d 2 y d x 2 + a x d y d x + b y = 0 (\displaystyle x^(2)(\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2) ))+ax(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+by=0)
   

   Characteristic equation. As you can see, in this differential equation, each term contains a power factor, the degree of which is equal to the order of the corresponding derivative.

   • Thus, you can try to look for a solution in the form y (x) = x n , (\displaystyle y(x)=x^(n),) where it is necessary to determine n (\displaystyle n), just as we were looking for a solution in the form of an exponential function for a linear differential equation with constant coefficients. After differentiation and substitution we get
     • x n (n 2 + (a − 1) n + b) = 0 (\displaystyle x^(n)(n^(2)+(a-1)n+b)=0)
   • To use the characteristic equation, we must assume that x ≠ 0 (\displaystyle x\neq 0). Dot x = 0 (\displaystyle x=0) called regular singular point differential equation. Such points are important when solving differential equations using power series. This equation has two roots, which can be different and real, multiple or complex conjugate.
     • n ± = 1 − a ± (a − 1) 2 − 4 b 2 (\displaystyle n_(\pm )=(\frac (1-a\pm (\sqrt ((a-1)^(2)-4b )))(2)))

   Two different real roots. If the roots n ± (\displaystyle n_(\pm )) are real and different, then the solution to the differential equation has the following form:

   • y (x) = c 1 x n + + c 2 x n − (\displaystyle y(x)=c_(1)x^(n_(+))+c_(2)x^(n_(-)))

   Two complex roots. If the characteristic equation has roots n ± = α ± β i (\displaystyle n_(\pm )=\alpha \pm \beta i), the solution is a complex function.

   • To transform the solution into a real function, we make a change of variables x = e t , (\displaystyle x=e^(t),) that is t = ln ⁡ x , (\displaystyle t=\ln x,) and use Euler's formula. Similar actions were performed previously when determining arbitrary constants.
     • y (t) = e α t (c 1 e β i t + c 2 e − β i t) (\displaystyle y(t)=e^(\alpha t)(c_(1)e^(\beta it)+ c_(2)e^(-\beta it)))
   • Then the general solution can be written as
     • y (x) = x α (c 1 cos ⁡ (β ln ⁡ x) + c 2 sin ⁡ (β ln ⁡ x)) (\displaystyle y(x)=x^(\alpha )(c_(1)\ cos(\beta \ln x)+c_(2)\sin(\beta \ln x)))

   Multiple roots. To obtain a second linearly independent solution, it is necessary to reduce the order again.

   • It takes quite a lot of calculations, but the principle remains the same: we substitute y = v (x) y 1 (\displaystyle y=v(x)y_(1)) into an equation whose first solution is y 1 (\displaystyle y_(1)). After reductions, the following equation is obtained:
     • v ″ + 1 x v ′ = 0 (\displaystyle v""+(\frac (1)(x))v"=0)
   • This is a first order linear equation with respect to v ′ (x) . (\displaystyle v"(x).) His solution is v (x) = c 1 + c 2 ln ⁡ x . (\displaystyle v(x)=c_(1)+c_(2)\ln x.) Thus, the solution can be written in the following form. This is quite easy to remember - to obtain the second linearly independent solution simply requires an additional term with ln ⁡ x (\displaystyle \ln x).
     • y (x) = x n (c 1 + c 2 ln ⁡ x) (\displaystyle y(x)=x^(n)(c_(1)+c_(2)\ln x))

4. Inhomogeneous linear differential equations with constant coefficients. Inhomogeneous equations look like L [ y (x) ] = f (x) , (\displaystyle L=f(x),) Where f (x) (\displaystyle f(x))- so-called free member. According to the theory of differential equations, the general solution of this equation is a superposition private solution y p (x) (\displaystyle y_(p)(x)) And additional solution y c (x) . (\displaystyle y_(c)(x).) However, in this case, a particular solution does not mean a solution given by the initial conditions, but rather a solution that is determined by the presence of heterogeneity (a free term). An additional solution is a solution to the corresponding homogeneous equation in which f (x) = 0. (\displaystyle f(x)=0.) The overall solution is a superposition of these two solutions, since L [ y p + y c ] = L [ y p ] + L [ y c ] = f (x) (\displaystyle L=L+L=f(x)), and since L [ y c ] = 0 , (\displaystyle L=0,) such a superposition is indeed a general solution.

   D 2 y d x 2 + a d y d x + b y = f (x) (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2)))+a (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+by=f(x))
   

   Method uncertain coefficients. The method of indefinite coefficients is used in cases where the dummy term is a combination of exponential, trigonometric, hyperbolic or power functions. Only these functions are guaranteed to have a finite number of linearly independent derivatives. In this section we will find a particular solution to the equation.

   • Let's compare the terms in f (x) (\displaystyle f(x)) with terms in without paying attention to constant factors. There are three possible cases.
     • No two members are the same. In this case, a particular solution y p (\displaystyle y_(p)) will be a linear combination of terms from y p (\displaystyle y_(p))
     • f (x) (\displaystyle f(x)) contains member x n (\displaystyle x^(n)) and member from y c , (\displaystyle y_(c),) Where n (\displaystyle n) is zero or a positive integer, and this term corresponds to a separate root of the characteristic equation. In this case y p (\displaystyle y_(p)) will consist of a combination of the function x n + 1 h (x) , (\displaystyle x^(n+1)h(x),) its linearly independent derivatives, as well as other terms f (x) (\displaystyle f(x)) and their linearly independent derivatives.
     • f (x) (\displaystyle f(x)) contains member h (x) , (\displaystyle h(x),) which is a work x n (\displaystyle x^(n)) and member from y c , (\displaystyle y_(c),) Where n (\displaystyle n) equals 0 or a positive integer, and this term corresponds to multiple root of the characteristic equation. In this case y p (\displaystyle y_(p)) is a linear combination of the function x n + s h (x) (\displaystyle x^(n+s)h(x))(Where s (\displaystyle s)- multiplicity of the root) and its linearly independent derivatives, as well as other members of the function f (x) (\displaystyle f(x)) and its linearly independent derivatives.
   • Let's write it down y p (\displaystyle y_(p)) as a linear combination of the terms listed above. Thanks to these coefficients in a linear combination this method called the "method of undetermined coefficients". When contained in y c (\displaystyle y_(c)) terms can be discarded due to the presence of arbitrary constants in y c . (\displaystyle y_(c).) After this we substitute y p (\displaystyle y_(p)) into the equation and equate similar terms.
   • We determine the coefficients. At this stage the system is obtained algebraic equations, which can usually be solved without special problems. The solution of this system allows us to obtain y p (\displaystyle y_(p)) and thereby solve the equation.
   • Example 2.3. Let us consider an inhomogeneous differential equation whose free term contains a finite number of linearly independent derivatives. A particular solution to such an equation can be found by the method of indefinite coefficients.
     • d 2 y d t 2 + 6 y = 2 e 3 t − cos ⁡ 5 t (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )t^(2) ))+6y=2e^(3t)-\cos 5t)
     • y c (t) = c 1 cos ⁡ 6 t + c 2 sin ⁡ 6 t (\displaystyle y_(c)(t)=c_(1)\cos (\sqrt (6))t+c_(2)\sin (\sqrt (6))t)
     • y p (t) = A e 3 t + B cos ⁡ 5 t + C sin ⁡ 5 t (\displaystyle y_(p)(t)=Ae^(3t)+B\cos 5t+C\sin 5t)
     • 9 A e 3 t − 25 B cos ⁡ 5 t − 25 C sin ⁡ 5 t + 6 A e 3 t + 6 B cos ⁡ 5 t + 6 C sin ⁡ 5 t = 2 e 3 t − cos ⁡ 5 t ( \displaystyle (\begin(aligned)9Ae^(3t)-25B\cos 5t&-25C\sin 5t+6Ae^(3t)\\&+6B\cos 5t+6C\sin 5t=2e^(3t)-\ cos 5t\end(aligned)))
     • ( 9 A + 6 A = 2 , A = 2 15 − 25 B + 6 B = − 1 , B = 1 19 − 25 C + 6 C = 0 , C = 0 (\displaystyle (\begin(cases)9A+ 6A=2,&A=(\dfrac (2)(15))\\-25B+6B=-1,&B=(\dfrac (1)(19))\\-25C+6C=0,&C=0 \end(cases)))
     • y (t) = c 1 cos ⁡ 6 t + c 2 sin ⁡ 6 t + 2 15 e 3 t + 1 19 cos ⁡ 5 t (\displaystyle y(t)=c_(1)\cos (\sqrt (6 ))t+c_(2)\sin (\sqrt (6))t+(\frac (2)(15))e^(3t)+(\frac (1)(19))\cos 5t)

   Lagrange method. The Lagrange method, or the method of variation of arbitrary constants, is a more general method solving inhomogeneous differential equations, especially in cases where the free term does not contain a finite number of linearly independent derivatives. For example, with free terms tan ⁡ x (\displaystyle \tan x) More x − n (\displaystyle x^(-n)) to find a particular solution it is necessary to use the Lagrange method. The Lagrange method can even be used to solve differential equations with variable coefficients, although in this case, with the exception of the Cauchy-Euler equation, it is used less frequently, since the additional solution is usually not expressed in terms of elementary functions.

   • Let's assume that the solution has the following form. Its derivative is given in the second line.
     • y (x) = v 1 (x) y 1 (x) + v 2 (x) y 2 (x) (\displaystyle y(x)=v_(1)(x)y_(1)(x)+v_ (2)(x)y_(2)(x))
     • y ′ = v 1 ′ y 1 + v 1 y 1 ′ + v 2 ′ y 2 + v 2 y 2 ′ (\displaystyle y"=v_(1)"y_(1)+v_(1)y_(1) "+v_(2)"y_(2)+v_(2)y_(2)")
   • Since the proposed solution contains two unknown quantities, it is necessary to impose additional condition. Let's choose this additional condition in the following form:
     • v 1 ′ y 1 + v 2 ′ y 2 = 0 (\displaystyle v_(1)"y_(1)+v_(2)"y_(2)=0)
     • y ′ = v 1 y 1 ′ + v 2 y 2 ′ (\displaystyle y"=v_(1)y_(1)"+v_(2)y_(2)")
     • y ″ = v 1 ′ y 1 ′ + v 1 y 1 ″ + v 2 ′ y 2 ′ + v 2 y 2 ″ (\displaystyle y""=v_(1)"y_(1)"+v_(1) y_(1)""+v_(2)"y_(2)"+v_(2)y_(2)"")
   • Now we can get the second equation. After substitution and redistribution of members, you can group together members with v 1 (\displaystyle v_(1)) and members with v 2 (\displaystyle v_(2)). These terms are reduced because y 1 (\displaystyle y_(1)) And y 2 (\displaystyle y_(2)) are solutions of the corresponding homogeneous equation. As a result we get the following system equations
     • v 1 ′ y 1 + v 2 ′ y 2 = 0 v 1 ′ y 1 ′ + v 2 ′ y 2 ′ = f (x) (\displaystyle (\begin(aligned)v_(1)"y_(1)+ v_(2)"y_(2)&=0\\v_(1)"y_(1)"+v_(2)"y_(2)"&=f(x)\\\end(aligned)))
   • This system can be converted to matrix equation kind A x = b , (\displaystyle A(\mathbf (x) )=(\mathbf (b) ),) whose solution is x = A − 1 b . (\displaystyle (\mathbf (x) )=A^(-1)(\mathbf (b) ).) For matrix 2 × 2 (\displaystyle 2\times 2) inverse matrix is found by dividing by the determinant, rearranging the diagonal elements, and changing the sign of the non-diagonal elements. In fact, the determinant of this matrix is ​​a Wronskian.
     • (v 1 ′ v 2 ′) = 1 W (y 2 ′ − y 2 − y 1 ′ y 1) (0 f (x)) (\displaystyle (\begin(pmatrix)v_(1)"\\v_( 2)"\end(pmatrix))=(\frac (1)(W))(\begin(pmatrix)y_(2)"&-y_(2)\\-y_(1)"&y_(1)\ end(pmatrix))(\begin(pmatrix)0\\f(x)\end(pmatrix)))
   • Expressions for v 1 (\displaystyle v_(1)) And v 2 (\displaystyle v_(2)) are given below. As in the order reduction method, in this case, during integration, an arbitrary constant appears, which includes an additional solution in the general solution of the differential equation.
     • v 1 (x) = − ∫ 1 W f (x) y 2 (x) d x (\displaystyle v_(1)(x)=-\int (\frac (1)(W))f(x)y_( 2)(x)(\mathrm (d) )x)
     • v 2 (x) = ∫ 1 W f (x) y 1 (x) d x (\displaystyle v_(2)(x)=\int (\frac (1)(W))f(x)y_(1) (x)(\mathrm (d) )x)
   
   Lecture from the National Open University Intuit entitled "Linear differential equations of nth order with constant coefficients."

Practical use

Differential equations establish a relationship between a function and one or more of its derivatives. Since such connections are extremely common, differential equations have found wide application in the most different areas, and since we live in four dimensions, these equations are often differential equations in private derivatives. This section covers some of the most important equations of this type.

• Exponential growth and decay. Radioactive decay. Compound interest. Speed chemical reactions. Concentration of drugs in the blood. Unlimited population growth. Newton-Richmann law. In the real world, there are many systems in which the rate of growth or decay at any given time is proportional to the amount in this moment time or can be well approximated by the model. This is because the solution to this differential equation, the exponential function, is one of the most important functions in mathematics and other sciences. In more general case with controlled population growth, the system may include additional members that limit growth. In the equation below, the constant k (\displaystyle k) can be either greater or less than zero.
  • d y d x = k x (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=kx)
• Harmonic vibrations. In both classical and quantum mechanics, the harmonic oscillator is one of the most important physical systems thanks to its simplicity and wide application to approximate more complex systems, such as a simple pendulum. In classical mechanics, harmonic vibrations are described by an equation that relates the position of a material point to its acceleration through Hooke's law. In this case, damping and driving forces can also be taken into account. In the expression below x ˙ (\displaystyle (\dot (x)))- time derivative of x , (\displaystyle x,) β (\displaystyle \beta )- parameter that describes the damping force, ω 0 (\displaystyle \omega _(0))- angular frequency of the system, F (t) (\displaystyle F(t))- time dependent driving force. The harmonic oscillator is also present in electromagnetic oscillatory circuits, where it can be implemented with greater accuracy than in mechanical systems.
  • x ¨ + 2 β x ˙ + ω 0 2 x = F (t) (\displaystyle (\ddot (x))+2\beta (\dot (x))+\omega _(0)^(2)x =F(t))
• Bessel's equation. The Bessel differential equation is used in many areas of physics, including solving the wave equation, Laplace's equation, and Schrödinger's equation, especially in the presence of cylindrical or spherical symmetry. This second-order differential equation with variable coefficients is not a Cauchy-Euler equation, so its solutions cannot be written as elementary functions. The solutions to the Bessel equation are the Bessel functions, which are well studied due to their application in many fields. In the expression below α (\displaystyle \alpha )- a constant that corresponds in order Bessel functions.
  • x 2 d 2 y d x 2 + x d y d x + (x 2 − α 2) y = 0 (\displaystyle x^(2)(\frac ((\mathrm (d) )^(2)y)((\mathrm (d ) )x^(2)))+x(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+(x^(2)-\alpha ^(2)) y=0)
• Maxwell's equations. Along with the Lorentz force, Maxwell's equations form the basis of classical electrodynamics. These are the four partial differential equations for electrical E (r , t) (\displaystyle (\mathbf (E) )((\mathbf (r) ),t)) and magnetic B (r , t) (\displaystyle (\mathbf (B) )((\mathbf (r) ),t)) fields. In the expressions below ρ = ρ (r , t) (\displaystyle \rho =\rho ((\mathbf (r) ),t))- charge density, J = J (r , t) (\displaystyle (\mathbf (J) )=(\mathbf (J) )((\mathbf (r) ),t))- current density, and ϵ 0 (\displaystyle \epsilon _(0)) And μ 0 (\displaystyle \mu _(0))- electric and magnetic constants, respectively.
  • ∇ ⋅ E = ρ ϵ 0 ∇ ⋅ B = 0 ∇ × E = − ∂ B ∂ t ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t (\displaystyle (\begin(aligned)\nabla \cdot (\mathbf (E) )&=(\frac (\rho )(\epsilon _(0)))\\\nabla \cdot (\mathbf (B) )&=0\\\nabla \times (\mathbf (E) )&=-(\frac (\partial (\mathbf (B) ))(\partial t))\\\nabla \times (\mathbf (B) )&=\mu _(0)(\ mathbf (J) )+\mu _(0)\epsilon _(0)(\frac (\partial (\mathbf (E) ))(\partial t))\end(aligned)))
• Schrödinger equation. In quantum mechanics, the Schrödinger equation is the fundamental equation of motion, which describes the movement of particles according to a change in the wave function Ψ = Ψ (r , t) (\displaystyle \Psi =\Psi ((\mathbf (r) ),t)) with time. The equation of motion is described by the behavior Hamiltonian H^(\displaystyle (\hat (H))) - operator, which describes the energy of the system. One of the widely famous examples The Schrödinger equation in physics is an equation for a single non-relativistic particle that is acted upon by a potential V (r , t) (\displaystyle V((\mathbf (r) ),t)). Many systems are described by the time-dependent Schrödinger equation, and on the left side of the equation is E Ψ , (\displaystyle E\Psi ,) Where E (\displaystyle E)- particle energy. In the expressions below ℏ (\displaystyle \hbar )- reduced Planck constant.
  • i ℏ ∂ Ψ ∂ t = H ^ Ψ (\displaystyle i\hbar (\frac (\partial \Psi )(\partial t))=(\hat (H))\Psi )
  • i ℏ ∂ Ψ ∂ t = (− ℏ 2 2 m ∇ 2 + V (r , t)) Ψ (\displaystyle i\hbar (\frac (\partial \Psi )(\partial t))=\left(- (\frac (\hbar ^(2))(2m))\nabla ^(2)+V((\mathbf (r) ),t)\right)\Psi )
• Wave equation. Physics and technology cannot be imagined without waves; they are present in all types of systems. In general, waves are described by the equation below, in which u = u (r , t) (\displaystyle u=u((\mathbf (r) ),t)) is the desired function, and c (\displaystyle c)- experimentally determined constant. d'Alembert was the first to discover that for the one-dimensional case the solution to the wave equation is any function with argument x − c t (\displaystyle x-ct), which describes a wave of arbitrary shape propagating to the right. The general solution for the one-dimensional case is a linear combination of this function with a second function with argument x + c t (\displaystyle x+ct), which describes a wave propagating to the left. This solution is presented in the second line.
  • ∂ 2 u ∂ t 2 = c 2 ∇ 2 u (\displaystyle (\frac (\partial ^(2)u)(\partial t^(2)))=c^(2)\nabla ^(2)u )
  • u (x , t) = f (x − c t) + g (x + c t) (\displaystyle u(x,t)=f(x-ct)+g(x+ct))
• Navier-Stokes equations. The Navier-Stokes equations describe the movement of fluids. Because fluids are present in virtually every field of science and technology, these equations are extremely important for predicting weather, designing aircraft, studying ocean currents, and solving many other applied problems. The Navier-Stokes equations are nonlinear partial differential equations, and in most cases they are very difficult to solve, since the nonlinearity leads to turbulence, and obtaining a stable solution by numerical methods requires partitioning into very small cells, which requires significant computing power. For practical purposes in hydrodynamics, methods such as time averaging are used to simulate turbulent flows. Even more basic questions such as the existence and uniqueness of solutions for nonlinear equations in partial derivatives, and proving the existence and uniqueness of a solution for the Navier-Stokes equations in three dimensions is one of the mathematical problems of the millennium. Below are the incompressible fluid flow equation and the continuity equation.
  • ∂ u ∂ t + (u ⋅ ∇) u − ν ∇ 2 u = − ∇ h , ∂ ρ ∂ t + ∇ ⋅ (ρ u) = 0 (\displaystyle (\frac (\partial (\mathbf (u) ) )(\partial t))+((\mathbf (u) )\cdot \nabla)(\mathbf (u) )-\nu \nabla ^(2)(\mathbf (u) )=-\nabla h, \quad (\frac (\partial \rho )(\partial t))+\nabla \cdot (\rho (\mathbf (u) ))=0)

• Many differential equations simply cannot be solved using the above methods, especially those mentioned in the last section. This applies when the equation contains variable coefficients and is not a Cauchy-Euler equation, or when the equation is nonlinear, except in a few very rare cases. However, the above methods can solve many important differential equations that are often encountered in various fields of science.
• Unlike differentiation, which allows you to find the derivative of any function, the integral of many expressions cannot be expressed in elementary functions. So don't waste time trying to calculate an integral where it is impossible. Look at the table of integrals. If the solution to a differential equation cannot be expressed in terms of elementary functions, sometimes it can be represented in integral form, and in this case it does not matter whether this integral can be calculated analytically.

Warnings

• Appearance differential equation can be misleading. For example, below are two first order differential equations. The first equation can be easily solved using the methods described in this article. At first glance, a minor change y (\displaystyle y) on y 2 (\displaystyle y^(2)) in the second equation makes it non-linear and becomes very difficult to solve.
  • d y d x = x 2 + y (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=x^(2)+y)
  • d y d x = x 2 + y 2 (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=x^(2)+y^(2))

Educational institution "Belarusian State

agricultural Academy"

Department of Higher Mathematics

DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

Lecture notes for accounting students

correspondence form of education (NISPO)

Gorki, 2013

First order differential equations

The concept of a differential equation. General and particular solutions

When studying various phenomena, it is often not possible to find a law that directly connects the independent variable and the desired function, but it is possible to establish a connection between the desired function and its derivatives.

The relationship connecting the independent variable, the desired function and its derivatives is called differential equation :

Here x– independent variable, y– the required function,
- derivatives of the desired function. In this case, relation (1) must have at least one derivative.

The order of the differential equation is called the order of the highest derivative included in the equation.

Consider the differential equation

. (2)

Since this equation includes only a first-order derivative, it is called is a first order differential equation.

If equation (2) can be resolved with respect to the derivative and written in the form

, (3)

then such an equation is called a first order differential equation in normal form.

In many cases it is advisable to consider an equation of the form

which is called a first order differential equation written in differential form.

Because
, then equation (3) can be written in the form
or
, where we can count
And
. This means that equation (3) is converted to equation (4).

Let us write equation (4) in the form
. Then
,
,
, where we can count
, i.e. an equation of the form (3) is obtained. Thus, equations (3) and (4) are equivalent.

Solving a differential equation (2) or (3) is called any function
, which, when substituting it into equation (2) or (3), turns it into an identity:

or
.

The process of finding all solutions to a differential equation is called its integration , and the solution graph
differential equation is called integral curve this equation.

If the solution to the differential equation is obtained in implicit form
, then it is called integral of this differential equation.

General solution of a first order differential equation is a family of functions of the form
, depending on an arbitrary constant WITH, each of which is a solution to a given differential equation for any admissible value of an arbitrary constant WITH. Thus, the differential equation has an infinite number of solutions.

Private decision differential equation is a solution obtained from the general solution formula for a specific value of an arbitrary constant WITH, including
.

Cauchy problem and its geometric interpretation

Equation (2) has an infinite number of solutions. In order to select one solution from this set, which is called a private one, you need to set some additional conditions.

The problem of finding a particular solution to equation (2) under given conditions is called Cauchy problem . This problem is one of the most important in the theory of differential equations.

The Cauchy problem is formulated as follows: among all solutions of equation (2) find such a solution
, in which the function
takes the given numeric value , if the independent variable x takes the given numeric value , i.e.

,
, (5)

Where D– domain of definition of the function
.

Meaning called the initial value of the function , A – initial value of the independent variable . Condition (5) is called initial condition or Cauchy condition .

WITH geometric point From a perspective, the Cauchy problem for differential equation (2) can be formulated as follows: from the set of integral curves of equation (2), select the one that passes through a given point
.

Differential equations with separable variables

One of the simplest types of differential equations is a first-order differential equation that does not contain the desired function:

. (6)

Considering that
, we write the equation in the form
or
. Integrating both sides of the last equation, we get:
or

. (7)

Thus, (7) is a general solution to equation (6).

Example 1 . Find the general solution to the differential equation
.

Solution . Let's write the equation in the form
or
. Let's integrate both sides of the resulting equation:
,
. We'll finally write it down
.

Example 2 . Find the solution to the equation
given that
.

Solution . Let's find a general solution to the equation:
,
,
,
. By condition
,
. Let's substitute into the general solution:
or
. We substitute the found value of an arbitrary constant into the formula for the general solution:
. This is a particular solution of the differential equation that satisfies the given condition.

The equation

(8)

Called a first order differential equation that does not contain an independent variable . Let's write it in the form
or
. Let's integrate both sides of the last equation:
or
- general solution of equation (8).

Example . Find the general solution to the equation
.

Solution . Let's write this equation in the form:
or
. Then
,
,
,
. Thus,
is the general solution of this equation.

Equation of the form

(9)

integrates using separation of variables. To do this, we write the equation in the form
, and then using the operations of multiplication and division we bring it to such a form that one part includes only the function of X and differential dx, and in the second part – the function of at and differential dy. To do this, both sides of the equation need to be multiplied by dx and divide by
. As a result, we obtain the equation

, (10)

in which the variables X And at separated. Let's integrate both sides of equation (10):
. The resulting relation is the general integral of equation (9).

Example 3 . Integrate Equation
.

Solution . Let's transform the equation and separate the variables:
,
. Let's integrate:
,
or is the general integral of this equation.
.

Let the equation be given in the form

This equation is called first order differential equation with separable variables in a symmetrical form.

To separate the variables, you need to divide both sides of the equation by
:

. (12)

The resulting equation is called separated differential equation . Let's integrate equation (12):

.(13)

Relation (13) is the general integral of differential equation (11).

Example 4 . Integrate a differential equation.

Solution . Let's write the equation in the form

and divide both parts by
,
. The resulting equation:
is a separated variable equation. Let's integrate it:

,
,

,
. The last equality is the general integral of this differential equation.

Example 5 . Find a particular solution to the differential equation
, satisfying the condition
.

Solution . Considering that
, we write the equation in the form
or
. Let's separate the variables:
. Let's integrate this equation:
,
,
. The resulting relation is the general integral of this equation. By condition
. Let's substitute it into the general integral and find WITH:
,WITH=1. Then the expression
is a partial solution of a given differential equation, written as a partial integral.

Linear differential equations of the first order

The equation

(14)

called linear differential equation of the first order . Unknown function
and its derivative enter into this equation linearly, and the functions
And
continuous.

If
, then the equation

(15)

called linear homogeneous . If
, then equation (14) is called linear inhomogeneous .

To find a solution to equation (14) one usually uses substitution method (Bernoulli) , the essence of which is as follows.

We will look for a solution to equation (14) in the form of a product of two functions

, (16)

Where
And
- some continuous functions. Let's substitute
and derivative
into equation (14):

Function v we will select in such a way that the condition is satisfied
.
Then

. Thus, to find a solution to equation (14), it is necessary to solve the system of differential equations
,
,
,
,
The first equation of the system is a linear homogeneous equation and can be solved by the method of separation of variables:
. As a function WITH=1:
you can take one of the partial solutions of the homogeneous equation, i.e. at
or
. Let's substitute into the second equation of the system:
.Then
.

. Thus, the general solution to a first-order linear differential equation has the form . Solve the equation
.

Solution . We will look for a solution to the equation in the form
. Then
. Let's substitute into the equation:

or
. Function v choose in such a way that the equality holds
. Then
. Let's solve the first of these equations using the method of separation of variables:
,
,
,
,. Function v Let's substitute into the second equation:
,
,
,
. The general solution to this equation is
.

Questions for self-control of knowledge

What is a differential equation?

What is the order of a differential equation?

Which differential equation is called a first order differential equation?

How is a first order differential equation written in differential form?

What is the solution to a differential equation?

What is an integral curve?

What is the general solution of a first order differential equation?

What is called a partial solution of a differential equation?

How is the Cauchy problem formulated for a first order differential equation?

What is the geometric interpretation of the Cauchy problem?

How to write a differential equation with separable variables in symmetric form?

Which equation is called a first order linear differential equation?

What method can be used to solve a first-order linear differential equation and what is the essence of this method?

Tasks for independent work

Solve differential equations with separable variables:

A)
;
;

b)
V)
.

;

A)
;
G)
;

2. Solve first order linear differential equations:
;
.

V)

G)

;

d)

1. The first order differential equation has the form

If this equation can be solved with respect to then it can be written as

In this case we say that the differential equation is resolved with respect to the derivative. For such an equation the following theorem is valid, which is called the theorem on the existence and uniqueness of a solution to a differential equation. Theorem. If in Eq.

function and its partial derivative with respect to y are continuous in some domain D on the plane containing some point, then there is a unique solution to this equation

The condition that when the function y must equal a given number is called the initial condition. It is often written in the form

Definition 1. The general solution of a first order differential equation is the function

which depends on one arbitrary constant C and satisfies the following conditions:

a) it satisfies the differential equation for any specific value of the constant C;

b) whatever the initial condition, it is possible to find a value such that the function satisfies the given initial condition. In this case, it is assumed that the values ​​belong to the region of variation of the variables x and y in which the conditions of the theorem of existence and uniqueness of the solution are satisfied.

2. In the process of finding a general solution to a differential equation, we often come to a relation of the form

not permitted regarding y. Resolving this relation for y, we obtain a general solution. However, it is not always possible to express y from relation (2) in elementary functions; in such cases the general solution is left implicit. An equality of the form that implicitly specifies a general solution is called the general integral of a differential equation.

Definition 2. A particular solution is any function that is obtained from the general solution if in the latter an arbitrary constant C is given a certain value. The relation is called in this case a partial integral of the equation.

Example 1. For a first order equation

the general solution will be a family of functions; this can be verified by simple substitution into the equation.

Let us find a particular solution that satisfies the following initial condition: when Substituting these values ​​into the formula, we obtain or Therefore, the desired particular solution will be the function

From a geometric point of view, the general integral is a family of curves on the coordinate plane, depending on one arbitrary constant C (or, as they say, on one parameter C).

These curves are called integral curves of a given differential equation. The partial integral corresponds to one curve of this family, passing through some given point of the plane.

Thus, in the last example, the general integral is geometrically represented by a family of hyperbolas, and the particular integral defined by the indicated initial condition is represented by one of these hyperbolas passing through the point in Fig. 251 shows the curves of the family corresponding to some values ​​of the parameter: etc.

To make the reasoning more clear, we will henceforth call the solution of the equation not only the function that satisfies the equation, but also the corresponding integral curve. In this regard, we will talk, for example, about a solution passing through the point .

Comment. The equation has no solution passing through a point lying on the axis of Fig. 251), since the right-hand side of the equation for is not defined and, therefore, is not continuous.

Solving or, as they often say, integrating a differential equation means:

a) find its general solution or general integral (if the initial conditions are not given) or

b) find that particular solution of the equation that satisfies the given initial conditions (if any).

3. Let us give a geometric interpretation of the first order differential equation.

Let a differential equation be given that is resolved with respect to the derivative:

and let there be a general solution to this equation. This general solution defines a family of integral curves on the plane

Equation (G) for each point M with coordinates x and y determines the value of the derivative, i.e., the angular coefficient of the tangent to the integral curve passing through this point. Thus, the differential equation (D) gives a set of directions or, as they say, determines the field of directions on the plane

Therefore, from a geometric point of view, the problem of integrating a differential equation is to find curves whose tangents have the same direction as the field at the corresponding points.

For differential equation (1), the locus of points at which the relation is satisfied is called the isocline of this differential equation.

For different values ​​of k we obtain different isoclines. The equation of the isocline corresponding to the value of k will obviously be By constructing a family of isoclines, one can approximately construct a family of integral curves. They say that, knowing isoclines, one can qualitatively determine the location of integral curves on the plane.