VARIATION OF PARAMETERS EXAMPLE

VARIATION OF PARAMETERS EXAMPLE

Problem:

Find the general solution of

y00 ? 6y0 + 9y = x?3 e3x

Solution:

The general solution for an inhomogeneous equation is split into two parts: y p (x), the particular solution, and yh (x),

the homogeneous solution (solution of the associated homogeneous equation). After finding both yh (x) and y p (x) we

recover the general solution,

y(x) = yh (x) + y p (x).

First let��s find the homogeneous solution, yh (x). We do this first b/c: (1) If we��re going to use the method of undetermined coefficients to find the particular solution, we want to make sure that our trial expression or ��guess�� is not a

homogeneous solution, and (2) if we use variation of parameters to find the particular solution (and for this example

we will), we require the two linearly independent homogeneous solutions. The homogeneous solution yh (x) satisfies

y00h ? 6yh + 9yh = 0.

To find the solution, use the substitution yh (x) = erx . Then y0h = rerx , y00h = r2 erx and our ode becomes r2 erx ? 6rerx +

9erx = 0. Since erx 6= 0, we can divide by erx and recover the characteristic equation,

r2 ? 6r + 9 = 0.

Solving for r (either by factoring or by using the quadratic equation) we find that r = 3 is a repeated root. Therefore

the homogeneous solution is

yh (x) = C1 e3x +C2 xe3x

and y1 (x) = e3x , y2 (x) = xe3x are the two linearly independent solutions to the homogeneous equation.

Next let��s find the particular solution y p . We use variation of parameters. Let

y p (x) = u1 (x)y1 (x) + u2 (x)y2 (x)

where y1 (x) and y2 (x) are the two linearly independent solutions to the homogeneous equation, found above. We could

insert our computed y1 and y2 at this point if desired. Your choice - we��ll do it both ways here. Then u1 (x) and u2 (x)

must satisfy



 0  



 3x

 0  



0

y1 y2

u1

0

e

xe3x

u1

=

OR

=

.

y01 y02

u02

g(x)

u02

x?3 e3x

3e3x e3x + 3xe3x

where g(x) = x?3 e3x is our inhomogeneity. Solving for u01 , u02 (performing a matrix inversion) we find

 0





 0 

 3x





 0 

1

1

0

y2 ?y2

0

u1

e + 3xe3x ?xe3x

u1

OR

= 6x

=

g(x)

u02

u02

x?3 e3x

?3e3x

e3x

W [y1 , y2 ] ?y01 y1

e

where the determinant of the inverted matrix is y1 y02 ?y2 y01 = W [y1 , y2 ] , the Wronskian. Note that if correctly computed

y1 (x) = e3x and y2 (x), they��re linearly independent, and the Wronskian is nonzero. Thus we find

u01 =

?x?2 e6x

?y2 g

OR u01 =

= ?x?2

W [y1 , y2 ]

e6x

u02 =

y1 g

x?3 e6x

OR u02 =

= x?3 .

W [y1 , y2 ]

e6x

If we hadn��t inserted values yet, we could do so now. y1 = e3x , y2 = xe3x , g = x?3 e3x , and W [y1 , y2 ] = y1 y02 ? y2 y01 =

(e3x )(e3x + 3xe3x ) ? (xe3x )(3e3x ) = e6x . Then,

u01 = ?x?2 ; solving, u1 = x?1

1

u02 = x?3 ; solving, u2 = ?x?2 /2.

(We drop constants b/c they would only multiply, and become part of, the homogeneous solution). Thus the particular

solution y p (x) = u1 (x)y1 (x) + u2 (x)y2 (x) is

y p (x) = (x?1 )(e3x ) + (?x?2 /2)(xe3x )

= x?1 e3x ? x?1 e3x /2

y p (x) = x?1 e3x /2.

We can check our particular solution by plugging it back into the ODE.

That��s it, we��re done! We��ve found the homogeneous solution yh (x) and - using variation of parameters - the particular

solution y p (x). The general solution is there sum y(x) = yh (x) + y p (x), or

y(x) = C1 e3x +C2 xe3x +

2

e3x

.

2x

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