Linear homogeneous differential equations of the 2nd order. Homogeneous Second Order Differential Equations with Constant Coefficients

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 second-order homogeneous differential equations with constant coefficients are solved.

The solution of a linear homogeneous homogeneous differential equation of the second order 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 differential equation of the second order 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 second-order homogeneous 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 second-order homogeneous differential equation is written as

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

Find general solutions of second-order homogeneous differential equations:

We make 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 DE of the 2nd order is

We compose and solve the characteristic equation:

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

In this case, the characteristic equation

The roots are distinct and real. Therefore, the general solution of the homogeneous differential equation of the 2nd order is here

Characteristic equation

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

The characteristic equation is here

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

The general solution of this second-order homogeneous differential equation is

Characteristic equation

From here we find the general solution of this dif. equations:

Examples for self-test.

2nd order differential equations

§one. Methods for lowering 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 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 look like: 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 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 separable differential equation: 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 required 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="> - equation order downgraded.

§2. Linear differential equation of the 2nd order.

The linear differential equation (LDE) of the 2nd order 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 notation 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.

Consequence 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 proved in the theorem remains valid for the case 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="> is identical. Thus,

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. The structure of the general solution to the 2nd order lod.

Theorem. If https://pandia.ru/text/78/516/images/image074_11.gif" width="42" height="25 src="> are linearly independent solutions of 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 2nd order lodu solutions..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 uniquely determined, 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 lodu 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 suggested by L. Euler..gif" width="25" height="26 src=">, we get an algebraic equation, which is called the 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=">. 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=">.

Private 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 brackets 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 of equation (5.1) ..gif" width="129" height="25 src="> will look like this:

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

represented 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. In this way:

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 different from 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 be solution of 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="> of equation (7.1) in the case when right part 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 form of the right side of f(x). Consider the right parts of the following form:

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

a) If the number is 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 shorten both parts by https://pandia.ru/text/78/516/images/image009_41.gif" height="25 src="> in the left and right parts 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 to the given equation 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=">. Finally 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 form of a particular solution in this case.

a) If the number is 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 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.

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

Bringing like terms, equating 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 of the given equation is: 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 can be equal to zero. Let us indicate the form of a particular solution in this general case.

a) If the number is 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 is https://pandia.ru/text/78/516/images/image210_5.gif" width="80" height="25 src=">, then a particular solution 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 the lod 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).

The direct finding of a particular solution to a line, except for the case of an equation with constant coefficients, and moreover with special constant terms, presents great difficulties. Therefore, in order to find the general solution to the lindu, the method of variation of arbitrary constants is usually used, which always makes it possible to find the general solution to the lindu in quadratures, if the fundamental system of solutions of the corresponding homogeneous equation is known. This method is as follows.

According to the above, the general solution of the 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 constant, but some, yet unknown, functions of f(x). . must be taken from the interval. In fact, in this case, the Wronsky determinant is nonzero at all points of the interval, i.e., in the entire space, it is the complex root of the characteristic equation..gif" width="20" height="25 src="> linearly independent particular solutions of the form :

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

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

Determination of second-order LODE with constant coefficients

Characteristic equation:

Case 1. Discriminant greater than zero

Case 2. The discriminant is zero

Case 3. Discriminant less than zero

Algorithm for Finding a General Solution to a Second-Order LODE with Constant Coefficients

§ 10. Linear inhomogeneous second-order differential equations with constant coefficients

Determination of second order LIDE with constant coefficients

Constant Variation Method

Method for solving LIDE with a special right hand side

Theorem on the structure of the general solution of LIDE

1. Function r (x) is a polynomial of degree T

2. Function r (x) is the product of a number by exponential function

3. Function r (x) - sum trigonometric functions

Algorithm for Finding a General Solution to LIDE with a Special Right-Hand Side

Appendix

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

The second order differential equation is called linear homogeneous differential equation (LODE) with constant coefficients if it looks like:

where p and q

To find a general solution to the LODE, it suffices to find two of its different particular solutions and . Then the general solution to the LODE will have the form

where WITH 1 and WITH

Leonhard Euler proposed to look for particular solutions of LODE in the form

where k- some number.

Differentiating this function two times and substituting expressions for at, at" and at" into the equation, we get:

The resulting equation is called characteristic equation LODU. To compile it, it is enough to replace in the original equation at", at" and at respectively on k 2 , k and 1:

Having solved the characteristic equation, i.e. finding roots k 1 and k 2 , we will also find particular solutions to the original LODE.

The characteristic equation is a quadratic equation, its roots are found through the discriminant

In this case, the following three cases are possible.

Case 1. Discriminant greater than zero , hence the roots k 1 and k 2 valid and different:

k 1¹ k 2

where WITH 1 and WITH 2 are arbitrary independent constants.

Case 2. The discriminant is zero , hence the roots k 1 and k 2 real and equal:

k 1 = k 2 = k

In this case, the general solution to the LODE has the form

where WITH 1 and WITH 2 are arbitrary independent constants.

Case 3. Discriminant less than zero . In this case, the equation has no real roots:

There are no roots.

In this case, the general solution to the LODE has the form

where WITH 1 and WITH 2 are arbitrary independent constants,

Thus, finding a general solution to a second-order LODE with constant coefficients is reduced to finding the roots of the characteristic equation and using formulas for the general solution of the equation (without resorting to calculating integrals).

Algorithm for Finding a General Solution to a Second-Order LODE with Constant Coefficients:

1. Bring the equation to the form , where p and q are some real numbers.

2. Compose a characteristic equation.

3. Find the discriminant of the characteristic equation.

4. Using the formulas (see Table 1), depending on the sign of the discriminant, write down the general solution.

Table 1

Table of possible general solutions

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

where , , and – predefined functions, 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 notation for the coefficients, we write the equation in the form:

Let us assume without proof that (2.2) has a unique solution on some interval that satisfies any initial conditions , , if the functions , and are continuous on the considered interval. 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 lodu.

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

Theorem. If and is a lodu solution

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

Proof.

We put the expression in (2.3) and show that the result is an identity:

Let's rearrange the terms:

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

Consequence 1. From the proved theorem it follows for , that if is a solution of equation (2.3), then it is also a solution of this equation.

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

Comment. The property of solutions proved in the theorem remains valid for the case of any order.

§3. Vronsky's determinant.

Definition. A system of functions is called linearly independent on some 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 in the form or . The determinant in the numerator of this expression is called the Wronsky determinant for the functions and . Thus, the Wronsky determinant for two linearly independent functions cannot be identically equal to zero.

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

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

On the right side of this formula, you must take the plus sign, since only in this case is an identity obtained. In this way,

(3.2)

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

§4. The structure of the general solution to the 2nd order lod.

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

Proof.

What is a solution of Eq. (2.3), follows from the theorem on the properties of solutions to a second-order lodu. 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 and so as to satisfy these conditions. We write the initial conditions in the form:

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

,

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

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

Solution.

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

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 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 condition on what properties particular solutions should have y1 (x) and y2 (x) .

And this condition is called condition linear independence private decisions.

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.

Solution. 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 certainty, 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 of a linear homogeneous differential equation of the second order with constant coefficients has the form

.

Example 2. Solve a linear homogeneous differential equation

.

Example 3. Solve a linear homogeneous differential equation

.

Solution. 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

That is, . In this case, the solution of a linear homogeneous differential equation of the second order with constant coefficients has the form

.

Example 4. Solve a linear homogeneous differential equation

.

Solution. 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

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