Linear homogeneous differential equations of the second order with constant coefficients. Linear homogeneous differential equations Particular solution of 2nd order lode

Theorem. If and is linear independent solutions equation (2.3), then their linear combination , where and are arbitrary constants, will be the general solution of this equation.

Proof. The fact that there is a solution to equation (2.3) follows from the theorem on the properties of solutions to 2nd order Lodo. We just need to show that the solution will be general, i.e. it is necessary to show that for any initial conditions, one can choose arbitrary constants in such a way as to satisfy these conditions. Let us write the initial conditions in the form:

The constants and from this system of linear algebraic equations are determined uniquely, since the determinant of this system is the value of the Wronski determinant for linearly independent solutions to Lodu at: ,

and such a determinant, as we saw in the previous paragraph, is nonzero. The theorem has been proven.

Construction general solution LOD of II order with constant coefficients in case

13. simple roots of the characteristic equation (case D>0) (with documentation).

14. multiple roots of the characteristic equation (case D=0) (with document).

15. complex conjugate roots of the characteristic equation (case D<0) (c док-вом).

Given a 2nd order lode with constant coefficients (5.1), where , . According to the previous paragraph, the general solution to a 2nd order lode is easily determined if two linearly independent partial solutions of this equation are known. A simple method for finding partial solutions to an equation with constant coefficients was proposed by L. Euler. This method, which is called Euler's method, consists in the fact that partial solutions are sought in the form.

Substituting this function into equation (5.1), after reducing by , we obtain an algebraic equation, which is called characteristic: (5.2)

The function will be a solution to equation (5.1) only for those values of k that are the roots of the characteristic equation (5.2). Depending on the value of the discriminant, three cases are possible.

1. . Then the roots of the characteristic equation are different: . The solutions will be linearly independent, because and the general solution (5.1) can be written as .

2. . In this case and . As a second linearly independent solution, we can take the function . Let us check that this function satisfies equation (5.1). Really, , . Substituting these expressions into equation (5.1), we obtain

Or, because And .

Particular solutions are linearly independent, because . Therefore, the general solution (5.1) has the form:

3. . In this case, the roots of the characteristic equation are complex conjugate: , where , . It can be verified that linearly independent solutions of equation (5.1) will be the functions and . Let us make sure that equation (5.1) is satisfied, for example, by the function y 1 . Really, , . Substituting these expressions into equation (5.1), we obtain

Both brackets on the left side of this equality are identically equal to zero. Really, ,

Thus, the function satisfies equation (5.1). Similarly, it is not difficult to verify that there is a solution to equation (5.1). Since , then the general solution will look like: .

16. Theorem on the structure of the general solution of second-order LNDDE (with proof).

Theorem 1. The general solution to the 2nd order lndu f(x) (6.1) is represented as the sum of the general solution of the corresponding homogeneous equation (6.2) and any particular solution to the lndu (6.1).

Proof. Let us first prove what the solution to equation (6.1) will be. To do this, let’s substitute f(x) into equation (6.1). This equality is an identity, because and f(x). Consequently, there is a solution to equation (6.1).

Let us now prove that this solution is general, i.e. you can choose the arbitrary constants included in it in such a way that any initial conditions of the form: , (6.3) will be satisfied. According to the theorem on the structure of the general solution of a linear homogeneous differential equation (Lod), the general solution of equation (6.2) can be represented in the form , where and are linearly independent solutions of this equation. Thus: and, therefore, the initial conditions (6.3) can be written as: or (6.4)

Arbitrary constants and are determined from this system of linear algebraic equations uniquely for any right-hand side, because the determinant of this system = is the value of the Wronski determinant for linearly independent solutions of equation (6.2) for , and such a determinant, as we saw above, is nonzero. By determining the constants and from the system of equations (6.4) and substituting them into the expression , we obtain a particular solution to equation (6.1) that satisfies the given initial conditions. The theorem has been proven.

17. Construction of a particular solution of a second-order LNDDE in the case of the right-hand side of the form

Let the coefficients in equation (6.1) be constant, i.e. the equation has the form: f(x) (7.1) where .

Let us consider a method for finding a particular solution to equation (7.1) in the case when the right-hand side f(x) has special type. This method is called the method of indefinite coefficients and consists of selecting a particular solution depending on the type of the right-hand side f(x). Consider the right-hand sides of the following form:

1. f(x) , where is a polynomial of degree , and some coefficients, except , may be equal to zero. Let us indicate the form in which a particular solution must be taken in this case.

a) If the number is not the root of the characteristic equation for equation (5.1), then we write the partial solution in the form: , where are the undetermined coefficients, which must be determined by the method of indefinite coefficients.

b) If is the root of the multiplicity of the corresponding characteristic equation, then we look for a particular solution in the form: , where are the undetermined coefficients.

18.f(x) , where and are polynomials of degree and, respectively, and one of these polynomials may be equal to zero. Let us indicate the type of particular solution in this general case.

A) If the number is not the root of the characteristic equation for equation (5.1), then the form of the particular solution will be: , (7.2) where are the undetermined coefficients, and .

B) If the number is the root of the characteristic equation for equation (5.1) of multiplicity , then a particular solution to lndu will have the form: , (7.3) i.e. a particular solution of the form (7.2) must be multiplied by . In expression (7.3) - polynomials with undetermined coefficients, and their degree .

19. Variation method for solving second-order LDDEs (Lagrange method).

Directly finding a particular solution to an equation, except in the case of an equation with constant coefficients and with special free terms, is very difficult. Therefore, to find a general solution to the equation, the method of variation of arbitrary constants is usually used, which always makes it possible to find the general solution to the equation in quadratures if the fundamental system of solutions to the corresponding homogeneous equation is known. This method is as follows.

According to the above, the general solution to a linear homogeneous equation is:

where are linearly independent Lodu solutions on a certain interval X, and are arbitrary constants. We will look for a particular solution to lnd in the form (8.1), assuming that they are not constant, but some, as yet unknown, functions of : . (8.2) Let us differentiate equality (8.2): . (8.3)

Let us select the functions so that the equality holds: . Then instead of (8.3) we will have:

Let us differentiate this expression again with respect to . As a result we get: . (8.5) Let us substitute (8.2), (8.4), (8.5) into the 2nd order lnd f(x):

Or f(x). (8.6)

Since - solutions to Lod, the last equality (8.6) takes the form: f(x).

Thus, function (8.2) will be a solution to lndu if the functions and satisfy the system of equations:

(8.7)

Since the determinant of this system is the Wronski determinant for two solutions corresponding to the lod linearly independent on X, it does not vanish at any point in the interval X. Therefore, solving system (8.7), we find and : and . Integrating, you get , , where is the prod. fast.

Returning to equality (8.2), we obtain a general solution to the inhomogeneous equation: .

Rows

1. Number series. Basic concepts, properties of convergent series. Necessary sign of convergence (with proof).

Basic definitions. Let us be given an infinite number sequence . Number series is called a record made up of members of this sequence. Or .Numbers called members of the series;, is called the common term of the series. As a result of calculating the values of this function at n =1, n =2,n =3, ... the terms of the series should be obtained.

Let the series (18.1.1) be given. Let us compile from its members finite sums called partial sums of a series:

Definition. If there is a finite limit S sequences of partial sums of the series (18.1.1) for , then the series is said to converge; number S called the sum of the series and written or .

If does not exist (including infinite), the series is called divergent.

Properties of convergent series. A necessary sign of convergence of a series. Common term of a convergent series tends to zero as : Proof. If , then and , but , therefore .

We must begin solving any problem to study the convergence of a series by checking the fulfillment of the condition: if this condition is not met, then the series obviously diverges. This condition is necessary, but not sufficient for the convergence of the series: the general term of the harmonic series is (18.1.2), but this series diverges.

Definition. The rest of the row after n the th term is called the series .

The 2nd order linear differential equation (LDE) has the following form:

where , , and are given functions that are continuous on the interval on which the solution is sought. Assuming that a 0 (x) ≠ 0, we divide (2.1) by and, after introducing new notations for the coefficients, we write the equation in the form:

Let us accept without proof that (2.2) has a unique solution on some interval that satisfies any initial conditions , , if on the interval under consideration the functions , and are continuous. If , then equation (2.2) is called homogeneous, and equation (2.2) is called inhomogeneous otherwise.

Let us consider the properties of solutions to the 2nd order lode.

Definition. A linear combination of functions is the expression , where are arbitrary numbers.

Theorem. If and – solution

then their linear combination will also be a solution to this equation.

Proof.

Let us put the expression in (2.3) and show that the result is the identity:

Let's rearrange the terms:

Since the functions are solutions of equation (2.3), then each of the brackets in the last equation is identically equal to zero, which is what needed to be proved.

Corollary 1. From the proven theorem it follows that if is a solution to equation (2.3), then there is also a solution to this equation.

Corollary 2. Assuming , we see that the sum of two solutions to Lod is also a solution to this equation.

Comment. The property of solutions proven in the theorem remains valid for problems of any order.

§3. Vronsky's determinant.

Definition. A system of functions is said to be linearly independent on a certain interval if none of these functions can be represented as a linear combination of all the others.

In the case of two functions this means that , i.e. . The last condition can be rewritten as or . The determinant in the numerator of this expression is is called the Wronski determinant for the functions and . Thus, the Wronski determinant for two linearly independent functions cannot be identically equal to zero.

Let is the Wronski determinant for linearly independent solutions and equation (2.3). Let us make sure by substitution that the function satisfies the equation. (3.1)

Really, . Since the functions and satisfy equation (2.3), then, i.e. – solution of equation (3.1). Let's find this solution: ; . Where , . , , .

On the right side of this formula you need to take the plus sign, since only in this case is identity obtained. Thus,

(3.2)

This formula is called the Liouville formula. It was shown above that the Wronski determinant for linearly independent functions cannot be identically equal to zero. Consequently, there is a point at which the determinant for linearly independent solutions of equation (2.3) is different from zero. Then it follows from Liouville’s formula that the function will be nonzero for all values in the interval under consideration, since for any value both factors on the right side of formula (3.2) are nonzero.

§4. Structure of the general solution to the 2nd order lode.

Theorem. If and are linearly independent solutions of equation (2.3), then their linear combination , where and are arbitrary constants, will be the general solution of this equation.

Proof.

What is a solution to equation (2.3), follows from the theorem on the properties of solutions to 2nd order Lodo. We just need to show that the solution will general, i.e. it is necessary to show that for any initial conditions, one can choose arbitrary constants in such a way as to satisfy these conditions. Let us write the initial conditions in the form:

and such a determinant, as we saw in the previous paragraph, is nonzero. The theorem is proven.

Example. Prove that the function , where and are arbitrary constants, is a general solution to Lod.

Solution.

It is easy to verify by substitution that the functions and satisfy this equation. These functions are linearly independent, since . Therefore, according to the theorem on the structure of the general solution, the 2nd order lode is a general solution to this equation.

Homogeneous linear differential equations second order with constant coefficients have the form

where p and q are real numbers. Let's look at examples of how homogeneous second-order differential equations with constant coefficients are solved.

The solution of a second order linear homogeneous differential equation depends on the roots of the characteristic equation. The characteristic equation is the equation k²+pk+q=0.

1) If the roots of the characteristic equation are different real numbers:

then the general solution of a linear homogeneous second-order differential equation with constant coefficients has the form

2) If the roots of the characteristic equation are equal real numbers

(for example, with a discriminant equal to zero), then the general solution of a homogeneous second-order differential equation is

3) If the roots of the characteristic equation are complex numbers

(for example, with a discriminant equal to a negative number), then the general solution of a homogeneous second-order differential equation is written in the form

Examples of solving linear homogeneous second order differential equations with constant coefficients

Find general solutions of homogeneous second order differential equations:

We make up the characteristic equation: k²-7k+12=0. Its discriminant is D=b²-4ac=1>0, so the roots are different real numbers.

Hence, the general solution of this homogeneous 2nd order DE is

Let's compose and solve the characteristic equation:

The roots are real and distinct. Hence we have a general solution to this homogeneous differential equation:

In this case, the characteristic equation

The roots are different and valid. Therefore, the general solution to a homogeneous differential equation of the 2nd order is here

Characteristic equation

Since the roots are real and equal, for this differential equation we write the general solution as

The characteristic equation is here

Since the discriminant is negative number, the roots of the characteristic equation are complex numbers.

The general solution of this homogeneous second-order differential equation has the form

Characteristic equation

From here we find the general solution to this differential. equations:

Examples for self-test.

In this article we will examine the principles of solving linear homogeneous second-order differential equations with constant coefficients, where p and q are arbitrary real numbers. First, let's focus on the theory, then apply the results obtained in solving examples and problems.

If you come across unfamiliar terms, then refer to the section on definitions and concepts of the theory of differential equations.

Let us formulate a theorem that indicates in what form to find the general solution of the LOD.

Theorem.

The general solution of a linear homogeneous differential equation with coefficients continuous on the integration interval X is determined by a linear combination , Where are linearly independent partial solutions of the LDE on X, and are arbitrary constants.

Thus, the general solution of a linear homogeneous differential equation of the second order with constant coefficients has the form y 0 =C 1 ⋅y 1 +C 2 ⋅y 2, where y 1 and y 2 are partial linearly independent solutions, and C 1 and C 2 are arbitrary constants. It remains to learn how to find partial solutions y 1 and y 2.

Euler suggested looking for particular solutions in the form .

If we take a partial solution of a second-order LODE with constant coefficients, then when substituting this solution into the equation we should obtain the identity:

So we got the so-called characteristic equation linear homogeneous differential equation of the second order with constant coefficients. Solutions k 1 and k 2 of this characteristic equation determine partial solutions of our second-order LODE with constant coefficients.

Depending on the coefficients p and q, the roots of the characteristic equation can be:

In the first case linearly independent partial solutions of the original differential equation are and , the general solution of a second-order LODE with constant coefficients is .

The functions and are indeed linearly independent, since the Wronski determinant is nonzero for any real x for .

In the second case one particular solution is the function . As the second particular solution we take . Let us show what really is a particular solution to a second-order LODE with constant coefficients and prove linear independence y 1 and y 2.

Since k 1 = k 0 and k 2 = k 0 are the same roots of the characteristic equation, it has the form . Therefore, is the original linear homogeneous differential equation. Let’s substitute it into it and make sure that the equation becomes an identity:

Thus, it is a partial solution of the original equation.

Let us show the linear independence of the functions and . To do this, let's calculate the Wronski determinant and make sure that it is different from zero.

Conclusion: linearly independent partial solutions of second-order LODEs with constant coefficients are and , and the general solution exists for .

In the third case we have a pair of complex partial solutions of the LDE and . The general solution will be written as . These particular solutions can be replaced by two real functions and , corresponding to the real and imaginary parts. This can be clearly seen if we transform the general solution , using the formulas from theory of function of a complex variable type:

where C 3 and C 4 are arbitrary constants.

So, let's summarize the theory.

Algorithm for finding a general solution to a second order linear homogeneous differential equation with constant coefficients.

Let's look at examples for each case.

Example.

Find the general solution to a second order linear homogeneous differential equation with constant coefficients .

Differential equations of 2nd order

§1. Methods for reducing the order of an equation.

The 2nd order differential equation has the form:

https://pandia.ru/text/78/516/images/image002_107.gif" width="19" height="25 src=">.gif" width="119" height="25 src="> ( or Differential" href="/text/category/differentcial/" rel="bookmark">2nd order differential equation). Cauchy problem for a 2nd order differential equation (1..gif" width="85" height= "25 src=">.gif" width="85" height="25 src=">.gif" height="25 src=">.

Let the 2nd order differential equation have the form: https://pandia.ru/text/78/516/images/image009_41.gif" height="25 src=">..gif" width="39" height=" 25 src=">.gif" width="265" height="28 src=">.

Thus, the 2nd order equation https://pandia.ru/text/78/516/images/image015_28.gif" width="34" height="25 src=">.gif" width="118" height ="25 src=">.gif" width="117" height="25 src=">.gif" width="34" height="25 src=">. Solving it, we obtain the general integral of the original differential equation, depending on two arbitrary constants: DIV_ADBLOCK219">

Example 1. Solve the differential equation https://pandia.ru/text/78/516/images/image021_18.gif" width="70" height="25 src=">.gif" height="25 src=">.gif" width="39" height="25 src=">.gif" width="157" height="25 src=">.gif" width="112" height="25 src=">.

This is a differential equation with separable variables: https://pandia.ru/text/78/516/images/image026_19.gif" width="99" height="41 src=">, i.e..gif" width= "96" height="25 src=">.gif" width="53" height="25 src=">.gif" width="48" height="38 src=">..gif" width=" 99" height="38 src=">..gif" width="95" height="25 src=">.

2..gif" width="117" height="25 src=">, i.e..gif" width="102" height="25 src=">..gif" width="117" height= "25 src=">.gif" width="106" height="25 src=">.gif" width="34" height="25 src=">.gif" width="117" height="25 src=">.gif" width="111" height="27 src=">

Solution.

This 2nd order equation clearly does not include the desired function https://pandia.ru/text/78/516/images/image043_16.gif" width="98" height="25 src=">.gif" width= "33" height="25 src=">.gif" width="105" height="36 src=">, which is a linear equation..gif" width="109" height="36 src=">.. gif" width="144" height="36 src=">.gif" height="25 src="> from some functions..gif" width="25" height="25 src=">.gif " width="127" height="25 src=">.gif" width="60" height="25 src="> – the order of the equation is lowered.

§2. Linear differential equation of 2nd order.

The 2nd order linear differential equation (LDE) has the following form:

https://pandia.ru/text/78/516/images/image059_12.gif" width="42" height="25 src=">.gif" width="42" height="25 src=">. gif" width="42" height="25 src="> and, after introducing new notations for the coefficients, we write the equation in the form:

https://pandia.ru/text/78/516/images/image064_12.gif" width="76" height="25 src=">.gif" width="35" height="25 src=">. gif" width="30" height="25 src="> continuous..gif" width="165" height="25 src=">.gif" width="95" height="25 src="> – arbitrary numbers.

Theorem. If https://pandia.ru/text/78/516/images/image074_11.gif" width="42" height="25 src="> - the solution is

https://pandia.ru/text/78/516/images/image076_10.gif" width="182" height="25 src="> will also be a solution to this equation.

Proof.

Let's put the expression https://pandia.ru/text/78/516/images/image077_11.gif" width="420" height="25 src=">.

Let's rearrange the terms:

https://pandia.ru/text/78/516/images/image073_10.gif" width="42" height="25 src=">.gif" width="54" height="25 src=">. gif" width="94" height="25 src="> is also a solution to this equation.

Corollary 2. Assuming https://pandia.ru/text/78/516/images/image083_11.gif" width="58" height="25 src="> is also a solution to this equation.

Comment. The property of solutions proven in the theorem remains valid for problems of any order.

§3. Vronsky's determinant.

Definition. System of functions https://pandia.ru/text/78/516/images/image084_10.gif" width="61" height="25 src=">.gif" width="110" height="47 src=" >..gif" width="106" height="42 src=">..gif" width="42" height="25 src=">.gif" width="181" height="47 src=" >.gif" width="42" height="25 src="> equations (2.3)..gif" width="182" height="25 src="> (3.1)

Indeed, ..gif" width="18" height="25 src="> satisfy the equation (2..gif" width="42" height="25 src="> is a solution to equation (3.1)..gif" width="87" height="28 src=">..gif" width="182" height="34 src=">..gif" width="162" height="42 src=">.gif" width="51" height="25 src="> the identity is obtained.

https://pandia.ru/text/78/516/images/image107_7.gif" width="18" height="25 src=">, in which the determinant for linearly independent solutions of the equation (2..gif" width= "42" height="25 src=">.gif" height="25 src="> both factors on the right side of formula (3.2) are non-zero.

§4. Structure of the general solution to the 2nd order lode.

Theorem. If https://pandia.ru/text/78/516/images/image074_11.gif" width="42" height="25 src="> are linearly independent solutions to the equation (2..gif" width="19" height="25 src=">.gif" width="129" height="25 src=">is a solution to equation (2.3), follows from the theorem on the properties of solutions to 2nd order lode..gif" width="85 " height="25 src=">.gif" width="19" height="25 src=">.gif" width="220" height="47">

The constants https://pandia.ru/text/78/516/images/image003_79.gif" width="19" height="25 src="> from this system of linear algebraic equations are determined uniquely, since the determinant of this system is https: //pandia.ru/text/78/516/images/image006_56.gif" width="51" height="25 src=">:

https://pandia.ru/text/78/516/images/image116_7.gif" width="138" height="25 src=">.gif" width="19" height="25 src=">. gif" width="69" height="25 src=">.gif" width="235" height="48 src=">..gif" width="143" height="25 src="> (5 ..gif" width="77" height="25 src=">. According to the previous paragraph, the general solution to the 2nd order Lod is easily determined if two linearly independent partial solutions of this equation are known. A simple method for finding partial solutions to an equation with constant coefficients proposed by L. Euler..gif" width="25" height="26 src=">, we obtain an algebraic equation, which is called characteristic:

https://pandia.ru/text/78/516/images/image124_5.gif" width="59" height="26 src="> will be a solution to equation (5.1) only for those values of k that are the roots of the characteristic equation (5.2)..gif" width="49" height="25 src=">..gif" width="76" height="28 src=">.gif" width="205" height="47 src ="> and the general solution (5..gif" width="45" height="25 src=">..gif" width="74" height="26 src=">..gif" width="83 " height="26 src=">. Let's check that this function satisfies equation (5.1)..gif" width="190" height="26 src=">. Substituting these expressions into equation (5.1), we get

https://pandia.ru/text/78/516/images/image141_6.gif" width="328" height="26 src=">, because..gif" width="137" height="26 src=">.

Particular solutions https://pandia.ru/text/78/516/images/image145_6.gif" width="86" height="28 src="> are linearly independent, because..gif" width="166" height="26 src=">.gif" width="45" height="25 src=">..gif" width="65" height="33 src=">.gif" width="134" height ="25 src=">.gif" width="267" height="25 src=">.gif" width="474" height="25 src=">.

Both parentheses on the left side of this equality are identically equal to zero..gif" width="174" height="25 src=">..gif" width="132" height="25 src="> is the solution to equation (5.1) ..gif" width="129" height="25 src="> will look like:

https://pandia.ru/text/78/516/images/image162_6.gif" width="179" height="25 src="> f(x) (6.1)

is presented as the sum of the general solution https://pandia.ru/text/78/516/images/image164_6.gif" width="195" height="25 src="> (6.2)

and any particular solution https://pandia.ru/text/78/516/images/image166_6.gif" width="87" height="25 src="> will be a solution to equation (6.1)..gif" width=" 272" height="25 src="> f(x). This equality is an identity, because..gif" width="128" height="25 src="> f(x). Therefore.gif" width="85" height="25 src=">.gif" width="138" height="25 src=">.gif" width="18" height="25 src="> are linearly independent solutions to this equation. Thus:

https://pandia.ru/text/78/516/images/image173_5.gif" width="289" height="48 src=">

https://pandia.ru/text/78/516/images/image002_107.gif" width="19" height="25 src=">.gif" width="11" height="25 src=">. gif" width="51" height="25 src=">, and such a determinant, as we saw above, is non-zero..gif" width="19" height="25 src="> from the system of equations (6 ..gif" width="76" height="25 src=">.gif" width="76" height="25 src=">.gif" width="140" height="25 src="> will solving the equation

https://pandia.ru/text/78/516/images/image179_5.gif" width="91" height="25 src="> into equation (6.5), we get

https://pandia.ru/text/78/516/images/image181_5.gif" width="140" height="25 src=">.gif" width="128" height="25 src="> f (x) (7.1)

where https://pandia.ru/text/78/516/images/image185_5.gif" width="34" height="25 src="> equation (7.1) in the case when the right-hand side f(x) has a special form. This method is called the method of indefinite coefficients and consists in selecting a particular solution depending on the type of the right-hand side f(x). Consider the right-hand sides of the following form:

1..gif" width="282" height="25 src=">.gif" width="53" height="25 src=">, can be zero. Let us indicate the form in which a particular solution must be taken in this case.

a) If the number https://pandia.ru/text/78/516/images/image191_5.gif" width="393" height="25 src=">.gif" width="157" height="25 src =>>.

Solution.

For the equation https://pandia.ru/text/78/516/images/image195_4.gif" width="86" height="25 src=">..gif" width="62" height="25 src= ">..gif" width="101" height="25 src=">.gif" width="153" height="25 src=">.gif" width="383" height="25 src=" >.

We reduce both parts to https://pandia.ru/text/78/516/images/image009_41.gif" height="25 src="> on the left and right sides of the equality

https://pandia.ru/text/78/516/images/image206_5.gif" width="111" height="40 src=">

From the resulting system of equations we find: https://pandia.ru/text/78/516/images/image208_5.gif" width="189" height="25 src=">, and the general solution given equation There is:

https://pandia.ru/text/78/516/images/image190_5.gif" width="11" height="25 src=">.gif" width="423" height="25 src=">,

where https://pandia.ru/text/78/516/images/image212_5.gif" width="158" height="25 src=">.

Solution.

The corresponding characteristic equation has the form:

https://pandia.ru/text/78/516/images/image214_6.gif" width="53" height="25 src=">.gif" width="85" height="25 src=">. gif" width="45" height="25 src=">.gif" width="219" height="25 src=">..gif" width="184" height="35 src=">. Final we have the following expression for the general solution:

https://pandia.ru/text/78/516/images/image223_4.gif" width="170" height="25 src=">.gif" width="13" height="25 src="> excellent from zero. Let us indicate the type of particular solution in this case.

a) If the number https://pandia.ru/text/78/516/images/image227_5.gif" width="204" height="25 src=">,

where https://pandia.ru/text/78/516/images/image226_5.gif" width="16" height="25 src="> is the root of the characteristic equation for the equation (5..gif" width="229 " height="25 src=">,

where https://pandia.ru/text/78/516/images/image229_5.gif" width="147" height="25 src=">.

Solution.

Roots of the characteristic equation for the equation https://pandia.ru/text/78/516/images/image231_4.gif" width="58" height="25 src=">.gif" width="203" height="25 src=">.

The right side of the equation given in example 3 has a special form: f(x) https://pandia.ru/text/78/516/images/image235_3.gif" width="50" height="25 src=">.gif " width="55" height="25 src=">.gif" width="229" height="25 src=">.

To determine https://pandia.ru/text/78/516/images/image240_2.gif" width="11" height="25 src=">.gif" width="43" height="25 src=" > and substitute it into the given equation:

Citing similar terms, equating the coefficients at https://pandia.ru/text/78/516/images/image245_2.gif" width="46" height="25 src=">.gif" width="100" height= "25 src=">.

The final general solution to the given equation has the form: https://pandia.ru/text/78/516/images/image249_2.gif" width="281" height="25 src=">.gif" width="47" height ="25 src=">.gif" width="10" height="25 src="> respectively, and one of these polynomials may be equal to zero. Let us indicate the type of particular solution in this general case.

a) If the number https://pandia.ru/text/78/516/images/image255_2.gif" width="605" height="51">, (7.2)

where https://pandia.ru/text/78/516/images/image257_2.gif" width="121" height="25 src=">.

b) If the number https://pandia.ru/text/78/516/images/image210_5.gif" width="80" height="25 src=">, then the particular solution to lndu will look like:

https://pandia.ru/text/78/516/images/image259_2.gif" width="17" height="25 src=">. In the expression (7..gif" width="121" height=" 25 src=">.

Example 4. Indicate the type of particular solution for the equation

https://pandia.ru/text/78/516/images/image262_2.gif" width="129" height="25 src=">..gif" width="95" height="25 src="> . The general solution to Lodu has the form:

https://pandia.ru/text/78/516/images/image266_2.gif" width="183" height="25 src=">..gif" width="42" height="25 src="> ..gif" width="36" height="25 src=">.gif" width="351" height="25 src=">.

Further coefficients https://pandia.ru/text/78/516/images/image273_2.gif" width="34" height="25 src=">.gif" width="42" height="28 src=" > there is a particular solution for the equation with the right side f1(x), and Variation" href="/text/category/variatciya/" rel="bookmark">variations of arbitrary constants (Lagrange method).

According to the above, the general solution to a linear homogeneous equation is:

https://pandia.ru/text/78/516/images/image278_2.gif" width="46" height="25 src=">.gif" width="51" height="25 src="> – not constants, but some, as yet unknown, functions of f(x). . must be taken from the interval. In fact, in this case, the Wronski determinant is nonzero at all points of the interval, i.e., throughout the entire space - the complex root of the characteristic equation..gif" width="20" height="25 src="> linearly independent partial solutions of the form :

In the general solution formula, this root corresponds to an expression of the form.