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Find the general solution to the following differential equations.

  1. y 5 y + 4 y = 3 e x
  2. y + y 6 y = 52 cos 2 t
  1. y ( x ) = c 1 e 4 x + c 2 e x x e x
  2. y ( t ) = c 1 e −3 t + c 2 e 2 t 5 cos 2 t + sin 2 t
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Variation of parameters

Sometimes, r ( x ) is not a combination of polynomials, exponentials, or sines and cosines. When this is the case, the method of undetermined coefficients does not work, and we have to use another approach to find a particular solution to the differential equation. We use an approach called the method of variation of parameters    .

To simplify our calculations a little, we are going to divide the differential equation through by a , so we have a leading coefficient of 1. Then the differential equation has the form

y + p y + q y = r ( x ) ,

where p and q are constants.

If the general solution to the complementary equation is given by c 1 y 1 ( x ) + c 2 y 2 ( x ) , we are going to look for a particular solution of the form y p ( x ) = u ( x ) y 1 ( x ) + v ( x ) y 2 ( x ). In this case, we use the two linearly independent solutions to the complementary equation to form our particular solution. However, we are assuming the coefficients are functions of x , rather than constants. We want to find functions u ( x ) and v ( x ) such that y p ( x ) satisfies the differential equation. We have

y p = u y 1 + v y 2 y p = u y 1 + u y 1 + v y 2 + v y 2 y p = ( u y 1 + v y 2 ) + u y 1 + u y 1 + v y 2 + v y 2 .

Substituting into the differential equation, we obtain

y p + p y p + q y p = [ ( u y 1 + v y 2 ) + u y 1 + u y 1 + v y 2 + v y 2 ] + p [ u y 1 + u y 1 + v y 2 + v y 2 ] + q [ u y 1 + v y 2 ] = u [ y 1 + p y 1 + q y 1 ] + v [ y 2 + p y 2 + q y 2 ] + ( u y 1 + v y 2 ) + p ( u y 1 + v y 2 ) + ( u y 1 + v y 2 ) .

Note that y 1 and y 2 are solutions to the complementary equation, so the first two terms are zero. Thus, we have

( u y 1 + v y 2 ) + p ( u y 1 + v y 2 ) + ( u y 1 + v y 2 ) = r ( x ).

If we simplify this equation by imposing the additional condition u y 1 + v y 2 = 0 , the first two terms are zero, and this reduces to u y 1 + v y 2 = r ( x ). So, with this additional condition, we have a system of two equations in two unknowns:

u y 1 + v y 2 = 0 u y 1 + v y 2 = r ( x ).

Solving this system gives us u and v , which we can integrate to find u and v .

Then, y p ( x ) = u ( x ) y 1 ( x ) + v ( x ) y 2 ( x ) is a particular solution to the differential equation. Solving this system of equations is sometimes challenging, so let’s take this opportunity to review Cramer’s rule, which allows us to solve the system of equations using determinants.

Rule: cramer’s rule

The system of equations

a 1 z 1 + b 1 z 2 = r 1 a 2 z 1 + b 2 z 2 = r 2

has a unique solution if and only if the determinant of the coefficients is not zero. In this case, the solution is given by

z 1 = | r 1 b 1 r 2 b 2 | | a 1 b 1 a 2 b 2 | and z 2 = | a 1 r 1 a 2 r 2 | | a 1 b 1 a 2 b 2 | .

Using cramer’s rule

Use Cramer’s rule to solve the following system of equations.

x 2 z 1 + 2 x z 2 = 0 z 1 3 x 2 z 2 = 2 x

We have

a 1 ( x ) = x 2 a 2 ( x ) = 1 b 1 ( x ) = 2 x b 2 ( x ) = −3 x 2 r 1 ( x ) = 0 r 2 ( x ) = 2 x .

Then,

| a 1 b 1 a 2 b 2 | = | x 2 2 x 1 −3 x 2 | = −3 x 4 2 x

and

| r 1 b 1 r 2 b 2 | = | 0 2 x 2 x −3 x 2 | = 0 4 x 2 = −4 x 2 .

Thus,

z 1 = | r 1 b 1 r 2 b 2 | | a 1 b 1 a 2 b 2 | = −4 x 2 −3 x 4 2 x = 4 x 3 x 3 + 2 .

In addition,

| a 1 r 1 a 2 r 2 | = | x 2 0 1 2 x | = 2 x 3 0 = 2 x 3 .

Thus,

z 2 = | a 1 r 1 a 2 r 2 | | a 1 b 1 a 2 b 2 | = 2 x 3 −3 x 4 2 x = −2 x 2 3 x 3 + 2 .
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Use Cramer’s rule to solve the following system of equations.

2 x z 1 3 z 2 = 0 x 2 z 1 + 4 x z 2 = x + 1

z 1 = 3 x + 3 11 x 2 , z 2 = 2 x + 2 11 x

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Problem-solving strategy: method of variation of parameters

  1. Solve the complementary equation and write down the general solution
    c 1 y 1 ( x ) + c 2 y 2 ( x ).
  2. Use Cramer’s rule or another suitable technique to find functions u ( x ) and v ( x ) satisfying
    u y 1 + v y 2 = 0 u y 1 + v y 2 = r ( x ).
  3. Integrate u and v to find u ( x ) and v ( x ). Then, y p ( x ) = u ( x ) y 1 ( x ) + v ( x ) y 2 ( x ) is a particular solution to the equation.
  4. Add the general solution to the complementary equation and the particular solution found in step 3 to obtain the general solution to the nonhomogeneous equation.
Practice Key Terms 4

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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