Ordinary differential equation called an equation that relates an independent variable, an unknown function of this variable and its derivatives (or differentials) of various orders.
The order of the differential equation is the order of the highest derivative contained in it.
In addition to ordinary ones, partial differential equations are also studied. These are equations relating independent variables, an unknown function of these variables and its partial derivatives with respect to the same variables. But we will only consider ordinary differential equations and therefore we will omit the word "ordinary" for brevity.
Examples of differential equations:
(1) 
;
(3) 
;
(4) 
;
Equation (1) is of the fourth order, equation (2) is of the third order, equations (3) and (4) are of the second order, equation (5) is of the first order.
Differential equation n order does not have to explicitly contain a function, all its derivatives from first to n th order and an independent variable. It may not explicitly contain derivatives of some orders, a function, an independent variable.
For example, in equation (1) there are clearly no derivatives of the third and second orders, as well as functions; in equation (2) - second-order derivative and function; in equation (4) - independent variable; in equation (5) - functions. Only equation (3) explicitly contains all derivatives, the function, and the independent variable.
By solving the differential equation any function is called y = f(x), substituting which into the equation, it turns into an identity.
The process of finding a solution to a differential equation is called its integration.
Example 1 Find a solution to the differential equation.
Solution. We write this equation in the form . The solution is to find the function by its derivative. The original function, as is known from the integral calculus, is the antiderivative for, i.e.
That's what it is solution of the given differential equation . changing in it C, we will get different solutions. We found out that there are an infinite number of solutions to a first-order differential equation.
General solution of the differential equation n th order is its solution expressed explicitly with respect to the unknown function and containing n independent arbitrary constants, i.e.
The solution of the differential equation in example 1 is general.
Partial solution of the differential equation its solution is called, in which specific numerical values are assigned to arbitrary constants.
Example 2 Find the general solution of the differential equation and a particular solution for 
.
Solution. We integrate both parts of the equation such a number of times that the order of the differential equation is equal.
,
.
As a result, we got the general solution -
given third-order differential equation.
Now let's find a particular solution under the specified conditions. To do this, we substitute their values instead of arbitrary coefficients and obtain
.
If, in addition to the differential equation, the initial condition is given in the form , then such a problem is called Cauchy problem . The values and are substituted into the general solution of the equation and the value of an arbitrary constant is found C, and then a particular solution of the equation for the found value C. This is the solution to the Cauchy problem.
Example 3 Solve the Cauchy problem for the differential equation from Example 1 under the condition .
Solution. We substitute into the general solution the values from the initial condition y = 3, x= 1. We get
We write down the solution of the Cauchy problem for the given differential equation of the first order:
Solving differential equations, even the simplest ones, requires good skills in integrating and taking derivatives, including complex functions. This can be seen in the following example.
Example 4 Find the general solution of the differential equation.
Solution. The equation is written in such a form that both sides can be integrated immediately.
.
We apply the method of integration by changing the variable (substitution). Let , then .
Required to take dx and now - attention - we do it according to the rules of differentiation of a complex function, since x and there is a complex function ("apple" - extracting the square root or, which is the same - raising to the power "one second", and "minced meat" - the expression itself under the root):
We find the integral:
Returning to the variable x, we get:
.
This is the general solution of this differential equation of the first degree.
Not only skills from the previous sections of higher mathematics will be required in solving differential equations, but also skills from elementary, that is, school mathematics. As already mentioned, in a differential equation of any order there may not be an independent variable, that is, a variable x. The knowledge about proportions that has not been forgotten (however, anyone has it like) from the school bench will help to solve this problem. This is the next example.
Recall the problem that we faced when finding definite integrals:
or dy = f(x)dx. Her solution:
and it reduces to the calculation of an indefinite integral. In practice, a more difficult task is more common: to find a function y, if it is known that it satisfies a relation of the form
This relation 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 order of the highest is called the 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 . To solve it means to find all its solutions. If for the desired function y succeeded in obtaining a formula that gives all solutions, then we say that we have found its general solution , or general integral .
Common decision
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 private integral. To obtain particular solutions (integrals) from general ones, it is necessary to attach specific numerical values to the constants.
The graph of a particular solution is called an integral curve. The general solution, which contains all particular 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 to the equation n th order, satisfying n initial conditions:
which determine n constants с 1 , с 2 ,..., c n.
1st order differential equations
For an unresolved with respect to the derivative, the differential equation of the 1st order has the form
or for permitted relatively
Example 3.46. Find a general solution to the equation
Solution. Integrating, we get
where C is an arbitrary constant. If we give C specific numerical values, then we get 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 after the expiration x years. When 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 successively the values 0, 1/n, 2/n, 3/n,..., then
Denote 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 interest accrual:
Thus, it can be seen that with a continuous change x the law of change in the money supply is expressed by a differential equation of the 1st order. Where Y x is an unknown function, x- independent variable, r- constant. We solve this equation, for this we rewrite it as follows:
where 
, or 
, where P stands for e C .
From the initial conditions Y(0) = Yo , we find P: Yo = Pe o , whence, Yo = P. Therefore, the solution looks like:
Consider the second economic problem. Macroeconomic models are also described by linear differential equations of the 1st order, describing the change in income or output Y as a function of time.
Example 3.48. Let the national income Y increase at a rate proportional to its value:
and let, the deficit in government spending is directly proportional to income Y with a proportionality coefficient q. The deficit in spending leads to an increase in the 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 obtain 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, which is the national income.
Consider the simplest differential equations n order, these are equations of the form
Its general solution can be obtained using n times of integration.
Example 3.49. Consider the example y """ = cos x.
Solution. Integrating, we find
The general solution has the form
Linear differential equations
In economics, they are of great use, consider the solution of such equations. If (9.1) has the form:
then it is called linear, where po(x), p1(x),..., pn(x), f(x) are given functions. If f(x) = 0, then (9.2) is called homogeneous, otherwise it is called non-homogeneous. 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 p o (x), p 1 (x),..., p n (x) are constants, then (9.2)
(9.4) is called a linear differential equation with constant coefficients of order n .
For (9.4) it has the form:
We can set without loss of generality 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 . Substitute the obtained 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 
are particular solutions (9.7), while the general
Consider a linear homogeneous differential equation of the second order 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 is:
y = C 1 cos 3x + C 2 sin 3x.
Second-order linear differential equations are used to study a web-like economic model with stocks of goods, where the rate of change of price P depends on the size of the stock (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 a differential equation:
For a particular solution, you can take a constant
which has the meaning of the equilibrium price. Deviation 
satisfies the homogeneous equation
(9.10)
The characteristic equation will be the following:
In case, the term is positive. Denote 
. The roots of the characteristic equation k 1,2 = ± i w, so the general solution (9.10) has the form:
where C and arbitrary constants, they are determined from the initial conditions. We have obtained the law of price change in time:
Either already solved with respect to the derivative, or they can be solved with respect to the derivative 
.
Common decision differential equations type on the interval X, which is given, can be found by taking the integral of both sides of this equality.
Get 
.
If we look at the properties of the indefinite integral, we find the desired general solution:
y = F(x) + C,
where F(x)- one of the antiderivatives of the function f(x) in between X, a FROM is an arbitrary constant.
Please note that in most tasks the interval X do not indicate. This means that a solution must be found for everyone. x, for which and the desired function y, and the original equation make sense.
If you need to calculate a particular solution of a differential equation that satisfies the initial condition y(x0) = y0, then after calculating the general integral y = F(x) + C, it is still necessary to determine the value of the constant C=C0 using the initial condition. That is, a constant C=C0 determined from the equation F(x 0) + C = y 0, and the desired particular solution of the differential equation will take the form:
y = F(x) + C0.
Consider an example:
Find the general solution of the differential equation , check the correctness of the result. Let's find a particular solution of this equation that would satisfy the initial condition .
Solution:
After we have integrated the given differential equation, we get:
.
We take this integral by the method of integration by parts:
That., 
is a general solution of the differential equation.
Let's check to make sure the result is correct. To do this, we substitute the solution that we found into the given equation:
.
That is, at 
the original equation turns into an identity:
therefore, the general solution of the differential equation was determined correctly.
The solution we have found is the general solution of the differential equation for each valid argument values x.
It remains to calculate a particular solution of the ODE that would satisfy the initial condition . In other words, it is necessary to calculate the value of the constant FROM, at which the equality will be true:
.
.
Then, substituting C = 2 into the general solution of the ODE, we obtain a particular solution of the differential equation that satisfies the initial condition:
.
Ordinary differential equation 
can be solved with respect to the derivative by dividing the 2 parts of the equation by f(x). This transformation will be equivalent if f(x) does not go to zero for any x from the interval of integration of the differential equation X.
Situations are likely when, for some values of the argument x ∈ X functions f(x) and g(x) turn to zero at the same time. For similar values x the general solution of the differential equation is any function y, which is defined in them, because .
If for some values of the argument x ∈ X the condition is satisfied, which means that in this case the ODE has no solutions.
For all others x from interval X the general solution of the differential equation is determined from the transformed equation.
Let's look at examples:
Example 1
Let us find the general solution of the ODE: 
.
Solution.
From the properties of the basic elementary functions, it is clear that the natural logarithm function is defined for non-negative values of the argument, therefore, the domain of the expression log(x+3) there is an interval x > -3 . Hence, the given differential equation makes sense for x > -3 . With these values of the argument, the expression x + 3 does not vanish, so one can solve the ODE with respect to the derivative by dividing the 2 parts by x + 3.
We get 
.
Next, we integrate the resulting differential equation, solved with respect to the derivative: 
. To take this integral, we use the method of subsuming under the sign of the differential.
6.1. BASIC CONCEPTS AND DEFINITIONS
When solving various problems of mathematics and physics, biology and medicine, quite often it is not possible to immediately establish a functional dependence in the form of a formula linking the variables that describe the process under study. Usually, one has to use equations containing, in addition to the independent variable and the unknown function, also its derivatives.
Definition. An equation relating an independent variable, an unknown function, and its derivatives of various orders is called differential.
The unknown function is usually denoted y(x) or simply y, and its derivatives are y", y" etc.
Other notations are also possible, for example: if y= x(t), then x"(t), x""(t) are its derivatives, and t is an independent variable.
Definition. If the function depends on one variable, then the differential equation is called ordinary. General form ordinary differential equation:
or
Functions F and f may not contain some arguments, but in order for the equations to be differential, the presence of a derivative is essential.
Definition.The order of the differential equation is the order of the highest derivative included in it.
For example, x 2 y"- y= 0, y" + sin x= 0 are first-order equations, and y"+ 2 y"+ 5 y= x is a second order equation.
When solving differential equations, the integration operation is used, which is associated with the appearance of an arbitrary constant. If the integration action is applied n times, then, obviously, the solution will contain n arbitrary constants.
6.2. FIRST ORDER DIFFERENTIAL EQUATIONS
General form first order differential equation is defined by the expression
The equation may not explicitly contain x and y, but necessarily contains y".
If the equation can be written as
then we get a first-order differential equation solved with respect to the derivative.
Definition. The general solution of the first order differential equation (6.3) (or (6.4)) is the set of solutions 
, where FROM is an arbitrary constant.
The graph for solving a differential equation is called integral curve.
Giving an arbitrary constant FROM different values, it is possible to obtain particular solutions. On surface xOy the general solution is a family of integral curves corresponding to each particular solution.
If you set a point A(x0, y0), through which the integral curve must pass, then, as a rule, from the set of functions 
one can be singled out - a particular solution.
Definition.Private decision of a differential equation is its solution that does not contain arbitrary constants.
If a 
is a general solution, then from the condition
you can find a permanent FROM. The condition is called initial condition.
The problem of finding a particular solution of a differential equation (6.3) or (6.4) that satisfies the initial condition 
at 
called the Cauchy problem. Does this problem always have a solution? The answer is contained in the following theorem.
Cauchy's theorem(theorem of existence and uniqueness of the solution). Let in the differential equation y"= f(x, y) function f(x, y) and her
partial derivative 
defined and continuous in some
areas D, containing a dot 
Then in the area D exists
the only solution to the equation that satisfies the initial condition 
at 
Cauchy's theorem states that under certain conditions there is a unique integral curve y= f(x), passing through a point 
Points where the conditions of the theorem are not satisfied
Cats are called special. Breaks at these points f(x, y) or.
Either several integral curves pass through a singular point, or none.
Definition. If the solution (6.3), (6.4) is found in the form f(x, y, c)= 0 not permitted with respect to y, then it is called common integral differential equation.
Cauchy's theorem only guarantees that a solution exists. Since there is no single method for finding a solution, we will consider only some types of first-order differential equations that are integrable in squares.
Definition. The differential equation is called integrable in quadratures, if the search for its solution is reduced to the integration of functions.
6.2.1. First order differential equations with separable variables
Definition. A first order differential equation is called an equation with separable variables,
The right side of equation (6.5) is the product of two functions, each of which depends on only one variable.
For example, the equation 
is an equation with separating
passing variables 
and the equation 
cannot be represented in the form (6.5).
Given that 
, we rewrite (6.5) as 
From this equation we obtain a differential equation with separated variables, in which the differentials contain functions that depend only on the corresponding variable:
Integrating term by term, we have
where C= C 2 - C 1 is an arbitrary constant. Expression (6.6) is the general integral of equation (6.5).
Dividing both parts of equation (6.5) by , we can lose those solutions for which, 
Indeed, if 
at 
then 
is obviously a solution of equation (6.5).
Example 1 Find a solution to the equation satisfying
condition: y= 6 at x= 2 (y(2) = 6).
Solution. Let's replace at" for then 
. Multiply both sides by
dx, since in further integration it is impossible to leave dx in the denominator:
and then dividing both parts by 
we get the equation,
which can be integrated. We integrate: 
Then 
; potentiating, we get y = C . (x + 1) - ob-
solution.
Based on the initial data, we determine an arbitrary constant by substituting them into the general solution
Finally we get y= 2(x + 1) is a particular solution. Consider a few more examples of solving equations with separable variables.
Example 2 Find a solution to the equation 
Solution. Given that 
, we get 
.
Integrating both sides of the equation, we have
where 
Example 3 Find a solution to the equation Solution. We divide both parts of the equation by those factors that depend on a variable that does not coincide with the variable under the differential sign, i.e., by 
and integrate. Then we get
and finally
Example 4 Find a solution to the equation
Solution. Knowing what we'll get. Section-
lim variables. Then 
Integrating, we get
Comment. In examples 1 and 2, the desired function y expressed explicitly (general solution). In examples 3 and 4 - implicitly (general integral). In the future, the form of the decision will not be specified.
Example 5 Find a solution to the equation Solution.
Example 6 Find a solution to the equation 
satisfying
condition y(e)= 1.
Solution. We write the equation in the form 
Multiplying both sides of the equation by dx and on, we get
Integrating both sides of the equation (the integral on the right side is taken by parts), we obtain 
But by condition y= 1 at x= e. Then
Substitute the found values FROM into a general solution:
The resulting expression is called a particular solution of the differential equation.
6.2.2. Homogeneous differential equations of the first order
Definition. The first order differential equation is called homogeneous if it can be represented as
We present an algorithm for solving a homogeneous equation.
1. Instead y introduce a new function Then 
and hence 
2. In terms of function u equation (6.7) takes the form
i.e., the replacement reduces the homogeneous equation to an equation with separable variables.
3. Solving equation (6.8), we first find u, and then y= ux.
Example 1 solve the equation 
Solution. We write the equation in the form
We make a substitution: 
Then 
Let's replace 
Multiply by dx: 
Divide by x and on 
then
Integrating both parts of the equation with respect to the corresponding variables, we have
or, returning to the old variables, we finally get
Example 2solve the equation 
Solution.Let 
then
Divide both sides of the equation by x2: 
Let's open the brackets and rearrange the terms:
Moving on to the old variables, we arrive at the final result:
Example 3Find a solution to the equation 
on condition
Solution.Performing a standard replacement 
we get
or 
or
So the particular solution has the form 
Example 4 Find a solution to the equation
Solution.
Example 5Find a solution to the equation 
Solution.
Independent work
Find a solution to differential equations with separable variables (1-9).
Find a solution to homogeneous differential equations (9-18).
6.2.3. Some applications of first order differential equations
The problem of radioactive decay
The decay rate of Ra (radium) at each moment of time is proportional to its available mass. Find the law of radioactive decay of Ra if it is known that at the initial moment there was Ra and the half-life of Ra is 1590 years.
Solution. Let at the moment the mass Ra be x= x(t) g, and 
Then the decay rate of Ra is
According to the task 
where k
Separating the variables in the last equation and integrating, we get
where 
For determining C we use the initial condition: 
.
Then 
and, therefore, 
Proportionality factor k determined from the additional condition:
We have
From here 
and the desired formula
The problem of the rate of reproduction of bacteria
The rate of reproduction of bacteria is proportional to their number. At the initial moment there were 100 bacteria. Within 3 hours their number doubled. Find the dependence of the number of bacteria on time. How many times will the number of bacteria increase within 9 hours?
Solution. Let x- the number of bacteria at the moment t. Then, according to the condition, 
where k- coefficient of proportionality.
From here 
It is known from the condition that 
. Means,
From the additional condition 
. Then
Required function: 
So, at t= 9 x= 800, i.e. within 9 hours the number of bacteria increased by 8 times.
The task of increasing the amount of the enzyme
In the culture of brewer's yeast, the growth rate of the active enzyme is proportional to its initial amount. x. Initial amount of enzyme a doubled within an hour. Find dependency
x(t).
Solution. By condition, the differential equation of the process has the form 
from here
But 
. Means, C= a and then 
It is also known that 
Consequently,
6.3. SECOND ORDER DIFFERENTIAL EQUATIONS
6.3.1. Basic concepts
Definition.Second order differential equation is called the relation connecting the independent variable, the desired function and its first and second derivatives.
In special cases, x may be absent in the equation, at or y". However, the second-order equation must necessarily contain y". In the general case, the second-order differential equation is written as:
or, if possible, in the form allowed for the second derivative:
As in the case of a first-order equation, a second-order equation can have a general and a particular solution. The general solution looks like:
Finding a private solution
under initial conditions - given
number) is called the Cauchy problem. Geometrically, this means that it is required to find the integral curve at= y(x), passing through a given point 
and having a tangent at this point, which is about
forks with positive axis direction Ox given angle. e. 
(Fig. 6.1). The Cauchy problem has a unique solution if the right side of equation (6.10), 
unpre-
is discontinuous and has continuous partial derivatives with respect to u, u" in some neighborhood of the starting point
To find constant 
included in a particular solution, it is necessary to allow the system
Rice. 6.1. integral curve
