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
.
From a geometric point of view, 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.
Equation
(8)
Called first order differential equation not containing 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
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 
you can take one of the partial solutions of the homogeneous equation, i.e. at WITH=1:
. Let's substitute into the second equation of the system: 
or 
.Then 
. Thus, the general solution to a first-order linear differential equation has the form 
.
Example 6
. 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) 
; G) 
.
2. Solve first order linear differential equations:
A) 
; b) 
; V) 
;
G) 
; d) 
.
Often just a mention differential equations makes students feel uncomfortable. Why is this happening? Most often, because when studying the basics of the material, a gap in knowledge arises, due to which further study of difurs becomes simply torture. It’s not clear what to do, how to decide, where to start?
However, we will try to show you that difurs are not as difficult as it seems.
Basic concepts of the theory of differential equations
From school we know the simplest equations in which we need to find the unknown x. Essentially differential equations only slightly different from them - instead of a variable X you need to find a function in them y(x) , which will turn the equation into an identity.
D differential equations are of great practical importance. This is not abstract mathematics that has no relation to the world around us. Many real natural processes are described using differential equations. For example, the vibrations of a string, the movement of a harmonic oscillator, using differential equations in problems of mechanics, find the speed and acceleration of a body. Also DU are widely used in biology, chemistry, economics and many other sciences.
Differential equation (DU) is an equation containing derivatives of the function y(x), the function itself, independent variables and other parameters in various combinations.
There are many types of differential equations: ordinary differential equations, linear and nonlinear, homogeneous and inhomogeneous, first and higher order differential equations, partial differential equations, and so on.
The solution to a differential equation is a function that turns it into an identity. There are general and particular solutions of the remote control.
A general solution to a differential equation is a general set of solutions that turn the equation into an identity. A partial solution of a differential equation is a solution that satisfies additional conditions specified initially.
The order of a differential equation is determined by the highest order of its derivatives.
Ordinary differential equations
Ordinary differential equations are equations containing one independent variable.
Let's consider the simplest ordinary differential equation of the first order. It looks like:
Such an equation can be solved by simply integrating its right-hand side.
Examples of such equations:
Separable equations
In general, this type of equation looks like this:
Here's an example:
When solving such an equation, you need to separate the variables, bringing it to the form:
After this, it remains to integrate both parts and obtain a solution.
Linear differential equations of the first order
Such equations look like:
Here p(x) and q(x) are some functions of the independent variable, and y=y(x) is the desired function. Here is an example of such an equation:
When solving such an equation, most often they use the method of varying an arbitrary constant or represent the desired function as a product of two other functions y(x)=u(x)v(x).
To solve such equations, certain preparation is required and it will be quite difficult to take them “at a glance”.
An example of solving a differential equation with separable variables
So we looked at the simplest types of remote control. Now let's look at the solution to one of them. Let this be an equation with separable variables.
First, let's rewrite the derivative in a more familiar form:
Then we divide the variables, that is, in one part of the equation we collect all the “I’s”, and in the other - the “X’s”:
Now it remains to integrate both parts:
We integrate and obtain a general solution to this equation:
Of course, solving differential equations is a kind of art. You need to be able to understand what type of equation it is, and also learn to see what transformations need to be made with it in order to lead to one form or another, not to mention just the ability to differentiate and integrate. And to succeed in solving DE, you need practice (as in everything). And if you currently don’t have time to understand how differential equations are solved or the Cauchy problem has stuck like a bone in your throat, or you don’t know, contact our authors. In a short time, we will provide you with a ready-made and detailed solution, the details of which you can understand at any time convenient for you. In the meantime, we suggest watching a video on the topic “How to solve differential equations”:
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 obtain: , .
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 a 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) 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 solution 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
Let us consider a linear homogeneous equation of the second order, i.e. equation
and establish some properties of its solutions.
Property 1
If is a solution to a linear homogeneous equation, then C, Where C- an arbitrary constant, is a solution to the same equation.
Proof.
Substituting into the left side of the equation under consideration C, we get: ,
but, because is a solution to the original equation.
Hence,
and the validity of this property has been proven.
Property 2
The sum of two solutions to a linear homogeneous equation is a solution to the same equation.
Proof.
Let and be solutions of the equation under consideration, then
And .
Now substituting + into the equation under consideration we will have:
, i.e. + is the solution to the original equation.
From the proven properties it follows that, knowing two particular solutions of a linear homogeneous equation of the second order, we can obtain the solution 
, depending on two arbitrary constants, i.e. from the number of constants that the second order equation must contain a general solution. But will this decision be general, i.e. Is it possible to satisfy arbitrarily given initial conditions by choosing arbitrary constants?
When answering this question, we will use the concept of linear independence of functions, which can be defined as follows.
The two functions are called linearly independent on a certain interval, if their ratio on this interval is not constant, i.e. If
.
Otherwise the functions are called linearly dependent.
In other words, two functions are said to be linearly dependent on a certain interval if on the entire interval.
Examples
1. Functions y 1
= e x and y 2
= e - x are linearly independent for all values of x, because 
.
2. Functions y 1
= e x and y 2
= 5 e x are linearly dependent, because 
.
Theorem 1.
If the functions and are linearly dependent on a certain interval, then the determinant is called Vronsky's determinant given functions is identically equal to zero on this interval.
Proof.
If
,
where , then and .
Hence,
.
The theorem has been proven.
Comment.
The Wronski determinant, which appears in the theorem considered, is usually denoted by the letter W or symbols 
.
If the functions are solutions of a linear homogeneous equation of the second order, then the following converse and, moreover, stronger theorem is valid for them.
Theorem 2.
If the Wronski determinant, compiled for solutions and a linear homogeneous equation of the second order, vanishes at least at one point, then these solutions are linearly dependent.
Proof.
Let the Wronski determinant vanish at the point, i.e. =0,
and let and .
Consider a linear homogeneous system 
relatively unknown and .
The determinant of this system coincides with the value of the Wronski determinant at
x=, i.e. coincides with , and therefore equals zero. Therefore, the system has a non-zero solution and ( and are not equal to zero). Using these values and , consider the function . This function is a solution to the same equation as the and functions. In addition, this function satisfies zero initial conditions: , because And .
On the other hand, it is obvious that the solution to the equation satisfying the zero initial conditions is the function y=0.
Due to the uniqueness of the solution, we have: 
. Whence it follows that 
,
those. functions and are linearly dependent. The theorem has been proven.
Consequences.
1. If the Wronski determinant appearing in the theorems is equal to zero for some value x=, then it is equal to zero for any value xfrom the considered interval.
2. If the solutions are linearly independent, then the Wronski determinant does not vanish at any point in the interval under consideration.
3. If the Wronski determinant is nonzero at least at one point, then the solutions are linearly independent.
Theorem 3.
If and are two linearly independent solutions of a homogeneous second-order equation, then the function , where and are arbitrary constants, is a general solution to this equation.
Proof.
As is known, the function is a solution to the equation under consideration for any values of and . Let us now prove that whatever the initial conditions
And ,
it is possible to select the values of arbitrary constants and so that the corresponding particular solution satisfies the given initial conditions.
Substituting the initial conditions into the equalities, we obtain a system of equations 
.
From this system it is possible to determine and , since determinant of this system 
there is a Wronski determinant for x= and, therefore, is not equal to zero (due to the linear independence of the solutions and ).
;
.
A particular solution with the obtained values and satisfies the given initial conditions. Thus, the theorem is proven.
Examples
Example 1.
The general solution to the equation is the solution .
Really,
.
Therefore, the functions sinx and cosx are linearly independent. This can be verified by considering the relationship of these functions:
.
Example 2.
Solution y = C 1
e x +C 2
e -x equation is general, because 
.
Example 3.
Equation 
, whose coefficients and
continuous on any interval not containing the point x = 0, admits partial solutions 
(easy to check by substitution). Therefore, its general solution has the form: 
.
Comment
We have established that the general solution of a linear homogeneous second-order equation can be obtained by knowing any two linearly independent partial solutions of this equation. However, there are no general methods for finding such partial solutions in final form for equations with variable coefficients. For equations with constant coefficients, such a method exists and will be discussed later.
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, a more complex task is more often encountered: finding the 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.
A 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 solutions, then we say that we have found its 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, the constants must be given specific numerical 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) are given 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 to (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:
