Linear homogeneous differential equations of the second order with constant coefficients. Second Order Linear Differential Equations

Fundamentals of solving linear inhomogeneous differential equations of the second order (LNDE-2) with constant coefficients(PC)

A second-order CLDE with constant coefficients $p$ and $q$ has the form $y""+p\cdot y"+q\cdot y=f\left(x\right)$, where $f\left(x \right)$ is a continuous function.

The following two statements are true with respect to the 2nd LNDE with PC.

Assume that some function $U$ is an arbitrary particular solution of an inhomogeneous differential equation. Let us also assume that some function $Y$ is a general solution (OR) of the corresponding linear homogeneous differential equation (LODE) $y""+p\cdot y"+q\cdot y=0$. Then the OR of LNDE-2 is equal to the sum of the indicated private and general solutions, i.e. $y=U+Y$.

If a right part The 2nd order LDE is the sum of functions, i.e. $f\left(x\right)=f_(1) \left(x\right)+f_(2) \left(x\right)+...+ f_(r) \left(x\right)$, then first you can find the PDs $U_(1) ,U_(2) ,...,U_(r) $ that correspond to each of the functions $f_(1) \ left(x\right),f_(2) \left(x\right),...,f_(r) \left(x\right)$, and after that write the LNDE-2 PD as $U= U_(1) +U_(2) +...+U_(r) $.

Solution of 2nd order LNDE with PC

Obviously, the form of one or another PD $U$ of a given LNDE-2 depends on the specific form of its right-hand side $f\left(x\right)$. The simplest cases of searching for the PD of LNDE-2 are formulated as the following four rules.

Rule number 1.

The right side of LNDE-2 has the form $f\left(x\right)=P_(n) \left(x\right)$, where $P_(n) \left(x\right)=a_(0) \cdot x^(n) +a_(1) \cdot x^(n-1) +...+a_(n-1) \cdot x+a_(n) $, that is, it is called a polynomial of degree $n$. Then its PR $U$ is sought in the form $U=Q_(n) \left(x\right)\cdot x^(r) $, where $Q_(n) \left(x\right)$ is another polynomial of the the same degree as $P_(n) \left(x\right)$, and $r$ is the number of roots of the characteristic equation of the corresponding LODE-2, zero. The coefficients of the polynomial $Q_(n) \left(x\right)$ are found by the method of indefinite coefficients (NC).

Rule number 2.

The right side of LNDE-2 has the form $f\left(x\right)=e^(\alpha \cdot x) \cdot P_(n) \left(x\right)$, where $P_(n) \left( x\right)$ is a polynomial of degree $n$. Then its PD $U$ is sought in the form $U=Q_(n) \left(x\right)\cdot x^(r) \cdot e^(\alpha \cdot x) $, where $Q_(n) \ left(x\right)$ is another polynomial of the same degree as $P_(n) \left(x\right)$, and $r$ is the number of roots of the characteristic equation of the corresponding LODE-2 equal to $\alpha $. The coefficients of the polynomial $Q_(n) \left(x\right)$ are found by the NK method.

Rule number 3.

The right part of LNDE-2 has the form $f\left(x\right)=a\cdot \cos \left(\beta \cdot x\right)+b\cdot \sin \left(\beta \cdot x\right) $, where $a$, $b$ and $\beta $ are known numbers. Then its PD $U$ is searched for in the form $U=\left(A\cdot \cos \left(\beta \cdot x\right)+B\cdot \sin \left(\beta \cdot x\right)\right )\cdot x^(r) $, where $A$ and $B$ are unknown coefficients, and $r$ is the number of roots of the characteristic equation of the corresponding LODE-2 equal to $i\cdot \beta $. The coefficients $A$ and $B$ are found by the NDT method.

Rule number 4.

The right side of LNDE-2 has the form $f\left(x\right)=e^(\alpha \cdot x) \cdot \left$, where $P_(n) \left(x\right)$ is a polynomial of degree $ n$, and $P_(m) \left(x\right)$ is a polynomial of degree $m$. Then its PD $U$ is searched for in the form $U=e^(\alpha \cdot x) \cdot \left\cdot x^(r) $, where $Q_(s) \left(x\right)$ and $ R_(s) \left(x\right)$ are polynomials of degree $s$, the number $s$ is the maximum of two numbers $n$ and $m$, and $r$ is the number of roots of the characteristic equation of the corresponding LODE-2, equal to $\alpha +i\cdot \beta $. The coefficients of the polynomials $Q_(s) \left(x\right)$ and $R_(s) \left(x\right)$ are found by the NK method.

The NK method consists in applying the following rule. In order to find the unknown coefficients of the polynomial, which are part of the particular solution of the inhomogeneous differential equation LNDE-2, it is necessary:

substitute the PD $U$ written in general view, to the left side of LNDU-2;
on the left side of LNDE-2, perform simplifications and group terms with the same powers $x$;
in the resulting identity, equate the coefficients of the terms with the same powers $x$ of the left and right sides;
solve the resulting system of linear equations for unknown coefficients.

Example 1

Task: find the OR LNDE-2 $y""-3\cdot y"-18\cdot y=\left(36\cdot x+12\right)\cdot e^(3\cdot x) $. Also find the PR , satisfying the initial conditions $y=6$ for $x=0$ and $y"=1$ for $x=0$.

Write the corresponding LODA-2: $y""-3\cdot y"-18\cdot y=0$.

Characteristic equation: $k^(2) -3\cdot k-18=0$. The roots of the characteristic equation: $k_(1) =-3$, $k_(2) =6$. These roots are real and distinct. Thus, the OR of the corresponding LODE-2 has the form: $Y=C_(1) \cdot e^(-3\cdot x) +C_(2) \cdot e^(6\cdot x) $.

The right part of this LNDE-2 has the form $\left(36\cdot x+12\right)\cdot e^(3\cdot x) $. It is necessary to consider the coefficient of the exponent of the exponent $\alpha =3$. This coefficient does not coincide with any of the roots of the characteristic equation. Therefore, the PR of this LNDE-2 has the form $U=\left(A\cdot x+B\right)\cdot e^(3\cdot x) $.

We will look for the coefficients $A$, $B$ using the NK method.

We find the first derivative of the CR:

$U"=\left(A\cdot x+B\right)^((") ) \cdot e^(3\cdot x) +\left(A\cdot x+B\right)\cdot \left( e^(3\cdot x) \right)^((") ) =$

$=A\cdot e^(3\cdot x) +\left(A\cdot x+B\right)\cdot 3\cdot e^(3\cdot x) =\left(A+3\cdot A\ cdot x+3\cdot B\right)\cdot e^(3\cdot x) .$

We find the second derivative of the CR:

$U""=\left(A+3\cdot A\cdot x+3\cdot B\right)^((") ) \cdot e^(3\cdot x) +\left(A+3\cdot A\cdot x+3\cdot B\right)\cdot \left(e^(3\cdot x) \right)^((") ) =$

$=3\cdot A\cdot e^(3\cdot x) +\left(A+3\cdot A\cdot x+3\cdot B\right)\cdot 3\cdot e^(3\cdot x) =\left(6\cdot A+9\cdot A\cdot x+9\cdot B\right)\cdot e^(3\cdot x) .$

We substitute the functions $U""$, $U"$ and $U$ instead of $y""$, $y"$ and $y$ into the given LNDE-2 $y""-3\cdot y"-18\cdot y=\left(36\cdot x+12\right)\cdot e^(3\cdot x).$ At the same time, since the exponent $e^(3\cdot x) $ is included as a factor in all components, then its can be omitted.

$6\cdot A+9\cdot A\cdot x+9\cdot B-3\cdot \left(A+3\cdot A\cdot x+3\cdot B\right)-18\cdot \left(A\ cdot x+B\right)=36\cdot x+12.$

We perform actions on the left side of the resulting equality:

$-18\cdot A\cdot x+3\cdot A-18\cdot B=36\cdot x+12.$

We use the NC method. We get a system of linear equations with two unknowns:

$-18\cdot A=36;$

$3\cdot A-18\cdot B=12.$

The solution to this system is: $A=-2$, $B=-1$.

The CR $U=\left(A\cdot x+B\right)\cdot e^(3\cdot x) $ for our problem looks like this: $U=\left(-2\cdot x-1\right) \cdot e^(3\cdot x) $.

The OR $y=Y+U$ for our problem looks like this: $y=C_(1) \cdot e^(-3\cdot x) +C_(2) \cdot e^(6\cdot x) +\ left(-2\cdot x-1\right)\cdot e^(3\cdot x) $.

In order to search for a PD that satisfies the given initial conditions, we find the derivative $y"$ OR:

$y"=-3\cdot C_(1) \cdot e^(-3\cdot x) +6\cdot C_(2) \cdot e^(6\cdot x) -2\cdot e^(3\ cdot x) +\left(-2\cdot x-1\right)\cdot 3\cdot e^(3\cdot x) .$

We substitute in $y$ and $y"$ the initial conditions $y=6$ for $x=0$ and $y"=1$ for $x=0$:

$6=C_(1) +C_(2) -1; $

$1=-3\cdot C_(1) +6\cdot C_(2) -2-3=-3\cdot C_(1) +6\cdot C_(2) -5.$

We got a system of equations:

$C_(1) +C_(2) =7;$

$-3\cdot C_(1) +6\cdot C_(2) =6.$

We solve it. We find $C_(1) $ using Cramer's formula, and $C_(2) $ is determined from the first equation:

$C_(1) =\frac(\left|\begin(array)(cc) (7) & (1) \\ (6) & (6) \end(array)\right|)(\left|\ begin(array)(cc) (1) & (1) \\ (-3) & (6) \end(array)\right|) =\frac(7\cdot 6-6\cdot 1)(1\ cdot 6-\left(-3\right)\cdot 1) =\frac(36)(9) =4; C_(2) =7-C_(1) =7-4=3.$

Thus, the PD of this differential equation is: $y=4\cdot e^(-3\cdot x) +3\cdot e^(6\cdot x) +\left(-2\cdot x-1\right )\cdot e^(3\cdot x) $.

In this section, we will consider a special case of second-order linear equations, when the coefficients of the equation are constant, i.e., they are numbers. Such equations are called equations with constant coefficients. This type of equation finds particularly wide application.

1. Linear homogeneous differential equations

second order with constant coefficients

Consider the equation

where the coefficients are constant. Assuming that dividing all terms of the equation by and denoting

we write this equation in the form

As is known, to find the general solution of a linear homogeneous equation of the second order, it suffices to know its fundamental system private decisions. Let us show how the fundamental system of particular solutions is found for a homogeneous linear differential equation with constant coefficients. We will look for a particular solution of this equation in the form

Differentiating this function two times and substituting the expressions for into Eq. (59), we obtain

Since , then, reducing by we get the equation

From this equation, those values of k are determined for which the function will be a solution to equation (59).

The algebraic equation (61) for determining the coefficient k is called the characteristic equation of the given differential equation (59).

The characteristic equation is an equation of the second degree and therefore has two roots. These roots can be either real different, or real and equal, or complex conjugate.

Let us consider the form of the fundamental system of partial solutions in each of these cases.

1. The roots of the characteristic equation are real and different: . In this case, according to formula (60), we find two particular solutions:

These two particular solutions form a fundamental system of solutions on the whole number line, since the Wronsky determinant never vanishes:

Therefore, the general solution of the equation according to formula (48) has the form

2. The roots of the characteristic equation are equal: . In this case both roots will be real. By formula (60) we obtain only one particular solution

Let us show that the second particular solution, which together with the first one forms a fundamental system, has the form

First of all, we check that the function is a solution of Eq. (59). Really,

But , since is the root of the characteristic equation (61). In addition, according to the Vieta theorem, therefore . Therefore, , i.e., the function is indeed a solution of Eq. (59).

Let us now show that the found particular solutions form a fundamental system of solutions. Really,

Thus, in this case, the general solution of the homogeneous linear equation has the form

3. The roots of the characteristic equation are complex. As you know, the complex roots of a quadratic equation with real coefficients are conjugate complex numbers, i.e. they have the form: . In this case, particular solutions of equation (59), according to formula (60), will have the form:

Using the Euler formulas (see Ch. XI, § 5 p. 3), the expressions for can be written in the form:

These solutions are complex. To get real solutions, consider the new functions

They are linear combinations of solutions and, therefore, are themselves solutions of equation (59) (see § 3, item 2, theorem 1).

It is easy to show that the Wronsky determinant for these solutions is different from zero and, therefore, the solutions form a fundamental system of solutions.

Thus, the general solution of a homogeneous linear differential equation in the case of complex roots of the characteristic equation has the form

In conclusion, we give a table of formulas for the general solution of equation (59) depending on the form of the roots of the characteristic equation.

In some problems of physics, a direct connection between the quantities describing the process cannot be established. But there is a possibility to obtain an equality containing the derivatives of the functions under study. This is how differential equations and the need to solve them to find the unknown function.

This article is intended for those who are faced with the problem of solving a differential equation in which the unknown function is a function of one variable. The theory is built in such a way that with a zero understanding of differential equations, you can do your job.

Each type of differential equations is associated with a solution method with detailed explanations and solutions to specific examples and problems. You just have to determine the type of differential equation for your problem, find a similar analyzed example and carry out similar actions.

For successful solution differential equations on your part, you will also need the ability to find sets of antiderivatives ( indefinite integrals) of various functions. If necessary, we recommend that you refer to the section.

First, we consider the types of ordinary differential equations of the first order that can be solved with respect to the derivative, then we move on to second-order ODEs, then we dwell on higher-order equations and finish with systems of differential equations.

Recall that if y is a function of the argument x .

First order differential equations.

The simplest differential equations of the first order of the form .

Let us write down several examples of such DE .

Differential Equations can be resolved with respect to the derivative by dividing both sides of the equality by f(x) . In this case, we arrive at the equation , which will be equivalent to the original one for f(x) ≠ 0 . Examples of such ODEs are .

If there are values of the argument x for which the functions f(x) and g(x) simultaneously vanish, then additional solutions appear. Additional solutions to the equation given x are any functions defined for those argument values. Examples of such differential equations are .

Second order differential equations.

Second Order Linear Homogeneous Differential Equations with Constant Coefficients.

LODE with constant coefficients is a very common type of differential equations. Their solution is not particularly difficult. First, the roots of the characteristic equation are found . For different p and q, three cases are possible: the roots of the characteristic equation can be real and different, real and coinciding or complex conjugate. Depending on the values of the roots of the characteristic equation, the general solution of the differential equation is written as , or , or respectively.

For example, consider a second-order linear homogeneous differential equation with constant coefficients. The roots of his characteristic equation are k 1 = -3 and k 2 = 0. The roots are real and different, therefore, the general solution to the LDE with constant coefficients is

Linear Nonhomogeneous Second Order Differential Equations with Constant Coefficients.

Common decision A second-order LIDE with constant coefficients y is sought as the sum of the general solution of the corresponding LODE and a particular solution of the original inhomogeneous equation, i.e, . The previous paragraph is devoted to finding a general solution to a homogeneous differential equation with constant coefficients. And a particular solution is determined either by the method of indefinite coefficients for a certain form of the function f (x) , standing on the right side of the original equation, or by the method of variation of arbitrary constants.

As examples of second-order LIDEs with constant coefficients, we present

Understand the theory and familiarize yourself with detailed decisions examples we offer you on the page of linear inhomogeneous differential equations of the second order with constant coefficients.

Linear Homogeneous Differential Equations (LODEs) and second-order linear inhomogeneous differential equations (LNDEs).

A special case of differential equations of this type are LODE and LODE with constant coefficients.

The general solution of the LODE on a certain interval is represented by a linear combination of two linearly independent particular solutions y 1 and y 2 of this equation, that is, .

The main difficulty lies precisely in finding linearly independent partial solutions of this type of differential equation. Usually, particular solutions are chosen from the following systems of linearly independent functions:

However, particular solutions are not always presented in this form.

An example of a LODU is .

The general solution of the LIDE is sought in the form , where is the general solution of the corresponding LODE, and is a particular solution of the original differential equation. We just talked about finding, but it can be determined using the method of variation of arbitrary constants.

An example of an LNDE is .

Higher order differential equations.

Differential equations admitting order reduction.

Order of differential equation , which does not contain the desired function and its derivatives up to k-1 order, can be reduced to n-k by replacing .

In this case , and the original differential equation reduces to . After finding its solution p(x), it remains to return to the replacement and determine the unknown function y .

For example, the differential equation after the replacement becomes a separable equation , and its order is reduced from the third to the first.

Linear homogeneous differential equation of the second order with constant coefficients has a general solution
, where and linearly independent particular solutions of this equation.

General form of solutions of a second-order homogeneous differential equation with constant coefficients
, depends on the roots of the characteristic equation
.

The roots of the characteristic equations	Kind of general solution
Roots and valid and various
Roots == valid and identical
Complex roots ,

Example

Find the general solution of linear homogeneous differential equations of the second order with constant coefficients:

Decision:
.

Having solved it, we will find the roots
,
valid and different. Therefore, the general solution is:
.

Decision: Let's make the characteristic equation:
.

Having solved it, we will find the roots

valid and identical. Therefore, the general solution is:
.

Decision: Let's make the characteristic equation:
.

Having solved it, we will find the roots
complex. Therefore, the general solution is:

Linear inhomogeneous second-order differential equation with constant coefficients has the form

Where
. (1)

The general solution of a linear inhomogeneous second-order differential equation has the form
, where
is a particular solution of this equation, is a general solution of the corresponding homogeneous equation, i.e. equations.

Type of private decision
inhomogeneous equation (1) depending on the right side
:

Right part	Private decision
– degree polynomial	, where is the number of roots of the characteristic equation equal to zero.
	, where = is the root of the characteristic equation.
	Where - number, equal to the number roots of the characteristic equation coinciding with .
	where is the number of roots of the characteristic equation coinciding with .

Consider different types of right-hand sides of a linear non-homogeneous differential equation:

1.
, where is a polynomial of degree . Then a particular solution
can be searched in the form
, where

, a is the number of roots of the characteristic equation equal to zero.

Example

Find a general solution
.

Decision:

B) Since the right side of the equation is a polynomial of the first degree and none of the roots of the characteristic equation
not equal to zero (
), then we look for a particular solution in the form where and are unknown coefficients. Differentiating twice
and substituting
,
and
into the original equation, we find.

Equating the coefficients at the same powers on both sides of the equation
,
, we find
,
. So, a particular solution given equation has the form
, and its general solution.

2. Let the right side look like
, where is a polynomial of degree . Then a particular solution
can be searched in the form
, where
is a polynomial of the same degree as
, a - a number indicating how many times is the root of the characteristic equation.

Example

Find a general solution
.

Decision:

A) Find the general solution of the corresponding homogeneous equation
. To do this, we write the characteristic equation
. Let's find the roots of the last equation
. Therefore, the general solution of the homogeneous equation has the form
.

characteristic equation

, where is an unknown coefficient. Differentiating twice
and substituting
,
and
into the original equation, we find. Where
, i.e
or
.

So, a particular solution of this equation has the form
, and its general solution
.

3. Let the right side look like , where
and - given numbers. Then a particular solution
can be searched in the form where and are unknown coefficients, and is a number equal to the number of roots of the characteristic equation coinciding with
. If in a function expression
include at least one of the functions
or
, then in
should always be entered both functions.

Example

Find a general solution .

Decision:

B) Since the right side of the equation is a function
, then the control number of this equation, it does not coincide with the roots
characteristic equation
. Then we look for a particular solution in the form

Where and are unknown coefficients. Differentiating twice, we get. Substituting
,
and
into the original equation, we find

Bringing like terms together, we get

We equate the coefficients at
and
on the right and left sides of the equation, respectively. We get the system
. Solving it, we find
,
.

So, a particular solution of the original differential equation has the form .

The general solution of the original differential equation has the form .

Linear differential equation of the second order is called an equation of the form

y"" + p(x)y" + q(x)y = f(x) ,

where y is the function to be found, and p(x) , q(x) and f(x) are continuous functions on some interval ( a, b) .

If the right side of the equation is zero ( f(x) = 0 ), then the equation is called linear homogeneous equation . Such equations will be mainly devoted to the practical part of this lesson. If the right side of the equation is not equal to zero ( f(x) ≠ 0 ), then the equation is called .

In tasks, we are required to solve the equation with respect to y"" :

y"" = −p(x)y" − q(x)y + f(x) .

Second-order linear differential equations have a unique solution Cauchy problems .

Linear homogeneous differential equation of the second order and its solution

Consider a linear homogeneous differential equation of the second order:

y"" + p(x)y" + q(x)y = 0 .

If a y1 (x) and y2 (x) are particular solutions of this equation, then the following statements are true:

1) y1 (x) + y 2 (x) - is also a solution to this equation;

2) Cy1 (x) , where C- an arbitrary constant (constant), is also a solution to this equation.

It follows from these two statements that the function

C1 y 1 (x) + C 2 y 2 (x)

is also a solution to this equation.

A fair question arises: is this solution general solution of a linear homogeneous differential equation of the second order , that is, such a solution in which, for various values C1 and C2 is it possible to get all possible solutions of the equation?

The answer to this question is: it can, but under certain conditions. This is condition on what properties particular solutions should have y1 (x) and y2 (x) .

And this condition is called the condition of linear independence of particular solutions.

Theorem. Function C1 y 1 (x) + C 2 y 2 (x) is a general solution of a second-order linear homogeneous differential equation if the functions y1 (x) and y2 (x) are linearly independent.

Definition. Functions y1 (x) and y2 (x) are called linearly independent if their ratio is a non-zero constant:

y1 (x)/y 2 (x) = k ; k = const ; k ≠ 0 .

However, establishing by definition whether these functions are linearly independent is often very difficult. There is a way to establish linear independence using the Wronsky determinant W(x) :

If the Wronsky determinant is not equal to zero, then the solutions are linearly independent . If the Wronsky determinant is equal to zero, then the solutions are linearly dependent.

Example 1 Find the general solution of a linear homogeneous differential equation.

Decision. We integrate twice and, as it is easy to see, in order for the difference of the second derivative of the function and the function itself to be equal to zero, the solutions must be associated with an exponent whose derivative is equal to itself. That is, private solutions are and .

Since the Vronsky determinant

is not equal to zero, then these solutions are linearly independent. Therefore, the general solution of this equation can be written as

Linear homogeneous differential equations of the second order with constant coefficients: theory and practice

Linear homogeneous differential equation of the second order with constant coefficients is called an equation of the form

y"" + py" + qy = 0 ,

where p and q are constant values.

The fact that this is a second-order equation is indicated by the presence of the second derivative of the desired function, and its homogeneity is indicated by zero on the right side. The quantities already mentioned above are called constant coefficients.

To solve a second-order linear homogeneous differential equation with constant coefficients , you must first solve the so-called characteristic equation of the form

k² + pq + q = 0 ,

which, as can be seen, is an ordinary quadratic equation.

Depending on the solution of the characteristic equation, three different options are possible solution of a linear homogeneous differential equation of the second order with constant coefficients , which we will now analyze. For complete definiteness, we will assume that all particular solutions have been tested by the Vronsky determinant and in all cases it is not equal to zero. Doubters, however, can check it for themselves.

Roots of the characteristic equation - real and different

In other words, . In this case, the solution to a second-order linear homogeneous differential equation with constant coefficients has the form

Example 2. Solve a linear homogeneous differential equation

Example 3. Solve a linear homogeneous differential equation

Decision. The characteristic equation has the form , its roots and are real and different. The corresponding particular solutions of the equation: and . The general solution of this differential equation has the form

Roots of the characteristic equation - real and equal

I.e, . In this case, the solution to a second-order linear homogeneous differential equation with constant coefficients has the form

Example 4. Solve a linear homogeneous differential equation

Decision. Characteristic equation has equal roots. The corresponding particular solutions of the equation: and . The general solution of this differential equation has the form

Example 5. Solve a linear homogeneous differential equation

Decision. The characteristic equation has equal roots. The corresponding particular solutions of the equation: and . The general solution of this differential equation has the form