General particular solution of differential equations. First order differential equations

Let us recall the task that confronted us when finding definite integrals:

or dy = f(x)dx. Her solution:

and it comes down to calculating the indefinite integral. In practice, more often occurs difficult task: find function y, if it is known that it satisfies a relation of the form

This relationship relates the independent variable x, unknown function y and its derivatives up to the order n inclusive, are called .

A differential equation includes a function under the sign of derivatives (or differentials) of one order or another. The highest order is called order (9.1) .

Differential equations:

- first order,

Second order

- fifth order, etc.

The function that satisfies a given differential equation is called its solution , or integral . Solving it means finding all its solutions. If for the required function y managed to obtain a formula that gives all the solutions, then we say that we have found it general solution, or general integral .

General solution contains n arbitrary constants and looks like

If a relation is obtained that relates x, y And n arbitrary constants, in a form not permitted with respect to y -

then such a relation is called the general integral of equation (9.1).

Cauchy problem

Each specific solution, i.e., each specific function that satisfies a given differential equation and does not depend on arbitrary constants, is called a particular solution , or a partial integral. To obtain particular solutions (integrals) from general ones, it is necessary to give specific constants numeric values.

The graph of a particular solution is called an integral curve. The general solution, which contains all the partial solutions, is a family of integral curves. For a first-order equation this family depends on one arbitrary constant, for the equation n-th order - from n arbitrary constants.

The Cauchy problem is to find a particular solution for the equation n-th order, satisfying n initial conditions:

by which n constants c 1, c 2,..., c n are determined.

1st order differential equations

For a 1st order differential equation that is unresolved with respect to the derivative, it has the form

or for permitted relatively

Example 3.46. Find the general solution to the equation

Solution. Integrating, we get

where C is an arbitrary constant. If we assign specific numerical values ​​to C, we obtain particular solutions, for example,

Example 3.47. Consider an increasing amount of money deposited in the bank subject to the accrual of 100 r compound interest per year. Let Yo be the initial amount of money, and Yx - at the end x years. If interest is calculated once a year, we get

where x = 0, 1, 2, 3,.... When interest is calculated twice a year, we get

where x = 0, 1/2, 1, 3/2,.... When calculating interest n once a year and if x takes sequential values ​​0, 1/n, 2/n, 3/n,..., then

Designate 1/n = h, then the previous equality will look like:

With unlimited magnification n(at ) in the limit we come to the process of increasing the amount of money with continuous accrual of interest:

Thus it is clear that with continuous change x the law of change in the money supply is expressed by a 1st order differential equation. Where Y x is an unknown function, x- independent variable, r- constant. Let's solve this equation, to do this we rewrite it as follows:

where , or , where P denotes e C .

From the initial conditions Y(0) = Yo, we find P: Yo = Pe o, from where, Yo = P. Therefore, the solution has the form:

Let's consider the second economic problem. Macroeconomic models are also described by linear differential equations of the 1st order, describing changes in income or output Y as functions of time.

Example 3.48. Let national income Y increase at a rate proportional to its value:

and let the deficit in government spending be directly proportional to income Y with the proportionality coefficient q. A spending deficit leads to an increase in national debt D:

Initial conditions Y = Yo and D = Do at t = 0. From the first equation Y= Yoe kt. Substituting Y we get dD/dt = qYoe kt . The general solution has the form
D = (q/ k) Yoe kt +С, where С = const, which is determined from the initial conditions. Substituting the initial conditions, we get Do = (q/ k)Yo + C. So, finally,

D = Do +(q/ k)Yo (e kt -1),

this shows that the national debt is increasing at the same relative rate k, the same as national income.

Let us consider the simplest differential equations n th order, these are equations of the form

Its general solution can be obtained using n times integrations.

Example 3.49. Consider the example y """ = cos x.

Solution. Integrating, we find

The general solution has the form

Linear differential equations

They are widely used in economics; let’s consider solving such equations. If (9.1) has the form:

then it is called linear, where рo(x), р1(x),..., рn(x), f(x) - specified functions. If f(x) = 0, then (9.2) is called homogeneous, otherwise it is called inhomogeneous. The general solution of equation (9.2) is equal to the sum of any of its particular solutions y(x) and the general solution of the homogeneous equation corresponding to it:

If the coefficients р o (x), р 1 (x),..., р n (x) are constant, then (9.2)

(9.4) is called a linear differential equation with constant coefficients of order n .

For (9.4) has the form:

Without loss of generality, we can set p o = 1 and write (9.5) in the form

We will look for a solution (9.6) in the form y = e kx, where k is a constant. We have: ; y " = ke kx , y "" = k 2 e kx , ..., y (n) = kne kx . Substituting the resulting expressions into (9.6), we will have:

(9.7) is an algebraic equation, its unknown is k, it is called characteristic. The characteristic equation has degree n And n roots, among which there can be both multiple and complex. Let k 1 , k 2 ,..., k n be real and distinct, then - particular solutions (9.7), and general

Consider a linear homogeneous second-order differential equation with constant coefficients:

Its characteristic equation has the form

(9.9)

its discriminant D = p 2 - 4q, depending on the sign of D, three cases are possible.

1. If D>0, then the roots k 1 and k 2 (9.9) are real and different, and the general solution has the form:

Solution. Characteristic equation: k 2 + 9 = 0, whence k = ± 3i, a = 0, b = 3, the general solution has the form:

y = C 1 cos 3x + C 2 sin 3x.

Linear differential equations of the 2nd order are used when studying a web-type economic model with inventories of goods, where the rate of change in price P depends on the size of the inventory (see paragraph 10). If supply and demand are linear functions of price, that is

a is a constant that determines the reaction rate, then the process of price change is described by the differential equation:

For a particular solution we can take a constant

meaningful equilibrium price. Deviation satisfies the homogeneous equation

(9.10)

The characteristic equation will be as follows:

In case the term is positive. Let's denote . The roots of the characteristic equation k 1,2 = ± i w, therefore the general solution (9.10) has the form:

where C and are arbitrary constants, they are determined from the initial conditions. We obtained the law of price change over time:

I. Ordinary differential equations

1.1. Basic concepts and definitions

A differential equation is an equation that relates an independent variable x, the required function y and its derivatives or differentials.

Symbolically, the differential equation is written as follows:

F(x,y,y")=0, F(x,y,y")=0, F(x,y,y",y",.., y (n))=0

A differential equation is called ordinary if the required function depends on one independent variable.

Solving a differential equation is called a function that turns this equation into an identity.

The order of the differential equation is the order of the highest derivative included in this equation

Examples.

1. Consider a first order differential equation

The solution to this equation is the function y = 5 ln x. Indeed, substituting y" into the equation, we get the identity.

And this means that the function y = 5 ln x– is a solution to this differential equation.

2. Consider the second order differential equation y" - 5y" +6y = 0. The function is the solution to this equation.

Really, .

Substituting these expressions into the equation, we obtain: , – identity.

And this means that the function is the solution to this differential equation.

Integrating differential equations is the process of finding solutions to differential equations.

General solution of the differential equation called a function of the form , which includes as many independent arbitrary constants as the order of the equation.

Partial solution of the differential equation is a solution obtained from a general solution for various numerical values ​​of arbitrary constants. The values ​​of arbitrary constants are found at certain initial values ​​of the argument and function.

The graph of a particular solution to a differential equation is called integral curve.

Examples

1. Find a particular solution to a first order differential equation

xdx + ydy = 0, If y= 4 at x = 3.

Solution. Integrating both sides of the equation, we get

Comment. An arbitrary constant C obtained as a result of integration can be represented in any form convenient for further transformations. In this case, taking into account the canonical equation of a circle, it is convenient to represent an arbitrary constant C in the form .

- general solution of the differential equation.

Particular solution of the equation satisfying the initial conditions y = 4 at x = 3 is found from the general by substituting the initial conditions into the general solution: 3 2 + 4 2 = C 2 ; C=5.

Substituting C=5 into the general solution, we get x 2 +y 2 = 5 2 .

This is a particular solution to a differential equation obtained from a general solution under given initial conditions.

2. Find the general solution to the differential equation

The solution to this equation is any function of the form , where C is an arbitrary constant. Indeed, substituting into the equations, we get: , .

Consequently, this differential equation has an infinite number of solutions, since for different values ​​of the constant C, equality determines different solutions to the equation.

For example, by direct substitution you can verify that the functions are solutions to the equation.

A problem in which you need to find a particular solution to the equation y" = f(x,y) satisfying the initial condition y(x 0) = y 0, is called the Cauchy problem.

Solving the equation y" = f(x,y), satisfying the initial condition, y(x 0) = y 0, is called a solution to the Cauchy problem.

The solution to the Cauchy problem has a simple geometric meaning. Indeed, according to these definitions, to solve the Cauchy problem y" = f(x,y) given that y(x 0) = y 0, means to find the integral curve of the equation y" = f(x,y) which passes through given point M 0 (x 0,y 0).

II. First order differential equations

2.1. Basic Concepts

A first order differential equation is an equation of the form F(x,y,y") = 0.

A first order differential equation includes the first derivative and does not include higher order derivatives.

Equation y" = f(x,y) is called a first-order equation solved with respect to the derivative.

The general solution of a first-order differential equation is a function of the form , which contains one arbitrary constant.

Example. Consider a first order differential equation.

The solution to this equation is the function.

Indeed, replacing this equation with its value, we get

that is 3x=3x

Therefore, the function is a general solution to the equation for any constant C.

Find a particular solution to this equation that satisfies the initial condition y(1)=1 Substituting initial conditions x = 1, y =1 into the general solution of the equation, we get from where C=0.

Thus, we obtain a particular solution from the general one by substituting into this equation the resulting value C=0– private solution.

2.2. Differential equations with separable variables

A differential equation with separable variables is an equation of the form: y"=f(x)g(y) or through differentials, where f(x) And g(y)– specified functions.

For those y, for which , the equation y"=f(x)g(y) is equivalent to the equation, in which the variable y is present only on the left side, and the variable x is only on the right side. They say, "in Eq. y"=f(x)g(y Let's separate the variables."

Equation of the form called a separated variable equation.

Integrating both sides of the equation By x, we get G(y) = F(x) + C is the general solution of the equation, where G(y) And F(x)– some antiderivatives, respectively, of functions and f(x), C arbitrary constant.

Algorithm for solving a first order differential equation with separable variables

Example 1

Solve the equation y" = xy

Solution. Derivative of a function y" replace it with

let's separate the variables

Let's integrate both sides of the equality:

Example 2

2yy" = 1- 3x 2, If y 0 = 3 at x 0 = 1

This is a separated variable equation. Let's imagine it in differentials. To do this, we rewrite this equation in the form From here

Integrating both sides of the last equality, we find

Substituting the initial values x 0 = 1, y 0 = 3 we'll find WITH 9=1-1+C, i.e. C = 9.

Therefore, the required partial integral will be or

Example 3

Write an equation for a curve passing through a point M(2;-3) and having a tangent with an angular coefficient

Solution. According to the condition

This is an equation with separable variables. Dividing the variables, we get:

Integrating both sides of the equation, we get:

Using the initial conditions, x = 2 And y = - 3 we'll find C:

Therefore, the required equation has the form

2.3. Linear differential equations of the first order

A linear differential equation of the first order is an equation of the form y" = f(x)y + g(x)

Where f(x) And g(x)- some specified functions.

If g(x)=0 then the linear differential equation is called homogeneous and has the form: y" = f(x)y

If then the equation y" = f(x)y + g(x) is called heterogeneous.

General solution of a linear homogeneous differential equation y" = f(x)y is given by the formula: where WITH– arbitrary constant.

In particular, if C =0, then the solution is y = 0 If a linear homogeneous equation has the form y" = ky Where k is some constant, then its general solution has the form: .

General solution of a linear inhomogeneous differential equation y" = f(x)y + g(x) is given by the formula ,

those. is equal to the sum of the general solution of the corresponding linear homogeneous equation and the particular solution of this equation.

For a linear inhomogeneous equation of the form y" = kx + b,

Where k And b- some numbers and a particular solution will be a constant function. Therefore, the general solution has the form .

Example. Solve the equation y" + 2y +3 = 0

Solution. Let's represent the equation in the form y" = -2y - 3 Where k = -2, b= -3 The general solution is given by the formula.

Therefore, where C is an arbitrary constant.

2.4. Solving linear differential equations of the first order by the Bernoulli method

Finding a General Solution to a First Order Linear Differential Equation y" = f(x)y + g(x) reduces to solving two differential equations with separated variables using substitution y=uv, Where u And v- unknown functions from x. This solution method is called Bernoulli's method.

Algorithm for solving a first order linear differential equation

y" = f(x)y + g(x)

1. Enter substitution y=uv.

2. Differentiate this equality y" = u"v + uv"

3. Substitute y And y" into this equation: u"v + uv" =f(x)uv + g(x) or u"v + uv" + f(x)uv = g(x).

4. Group the terms of the equation so that u take it out of brackets:

5. From the bracket, equating it to zero, find the function

This is a separable equation:

Let's divide the variables and get:

Where . .

6. Substitute the resulting value v into the equation (from step 4):

and find the function This is an equation with separable variables:

7. Write the general solution in the form: , i.e. .

Example 1

Find a particular solution to the equation y" = -2y +3 = 0 If y=1 at x = 0

Solution. Let's solve it using substitution y=uv,.y" = u"v + uv"

Substituting y And y" into this equation, we get

By grouping the second and third terms on the left side of the equation, we take out the common factor u out of brackets

We equate the expression in brackets to zero and, having solved the resulting equation, we find the function v = v(x)

We get an equation with separated variables. Let's integrate both sides of this equation: Find the function v:

Let's substitute the resulting value v into the equation we get:

This is a separated variable equation. Let's integrate both sides of the equation: Let's find the function u = u(x,c) Let's find a general solution: Let us find a particular solution to the equation that satisfies the initial conditions y = 1 at x = 0:

III. Higher order differential equations

3.1. Basic concepts and definitions

A second-order differential equation is an equation containing derivatives of no higher than second order. In the general case, a second-order differential equation is written as: F(x,y,y",y") = 0

The general solution of a second-order differential equation is a function of the form , which includes two arbitrary constants C 1 And C 2.

A particular solution to a second-order differential equation is a solution obtained from a general one for certain values ​​of arbitrary constants C 1 And C 2.

3.2. Linear homogeneous differential equations of the second order with constant coefficients.

Linear homogeneous differential equation of the second order with constant coefficients called an equation of the form y" + py" +qy = 0, Where p And q- constant values.

Algorithm for solving homogeneous second-order differential equations with constant coefficients

1. Write the differential equation in the form: y" + py" +qy = 0.

2. Create its characteristic equation, denoting y" through r 2, y" through r, y in 1: r 2 + pr +q = 0

Given online calculator allows you to solve differential equations online. It is enough to enter your equation in the appropriate field, denoting the derivative of the function through an apostrophe and click on the “solve equation” button. And the system, implemented on the basis of the popular WolframAlpha website, will give detailed solving a differential equation absolutely free. You can also define the Cauchy problem so that from the entire set possible solutions select the quotient corresponding to the given initial conditions. The Cauchy problem is entered in a separate field.

Differential equation

By default, the function in the equation y is a function of a variable x. However, you can specify your own designation for the variable; if you write, for example, y(t) in the equation, the calculator will automatically recognize that y there is a function from a variable t. With the help of a calculator you can solve differential equations of any complexity and type: homogeneous and inhomogeneous, linear or nonlinear, first order or second and higher orders, equations with separable or nonseparable variables, etc. Solution diff. equation is given in analytical form, has detailed description. Differential equations are very common in physics and mathematics. Without calculating them, it is impossible to solve many problems (especially in mathematical physics).

One of the stages of solving differential equations is integrating functions. There are standard methods for solving differential equations. It is necessary to reduce the equations to a form with separable variables y and x and separately integrate the separated functions. To do this, sometimes a certain replacement must be made.