# SecondOrder Nonhomogenous Linear Differential Equations

1. Jan 28, 2004

### wubie

Hello,

First here is the question that I am supposed to solve:

Solve the following nonhomogeneous differential equations:

b) y'' + 2y' + 2y = e^-x

c) 2y'' + y' = cos 2x.

I am supposed to be using the method of variation of parameters to solve these equations. What my problem is I end up getting to a point where I have two equations in which I should be able to solve for the derivative of parameter one (u'sub1) and the derivative of parameter two (u'sub2). Unfortunately I am getting stuck. And I am not sure why.

For b) I have the following two equations:

u'sub1 ysub1 + u'sub2 ysub2 = 0 = u'sub1 e^-x + u'sub2 xe^-x

and the particular equation

u'sub2 - u'sub2 x + usub2 x - u'sub1 + usub1 = 1

From these I am supposed to find u'sub1 and u'sub2 and eventually come to find usub1 and usub2.

Now when I solve for

u'sub1 e^-x + u'sub2 xe^-x = 0

I get u'sub1 = (-u'sub2 xe^-x)/e^-x = -u'sub2 x

I then sub. into the other equation for u'sub1

u'sub2 - u'sub2 x + usub2 x - u'sub1 + usub1 = 1

becomes

u'sub2 - u'sub2 x + usub2 x + u'sub2 x + usub1 = 1

which becomes

u'sub2 + usub2 x + usub1 = 1

But I am stumped here. How do I solve for u'sub2 when I still have usub2 and usub1? I know I am missing something incredibly obvious. I just can't seem to know what.

For question b) I am having similar problems - still trying to solve for u'sub1 and u'sub2.

Update: I have figured out question c). I am still working on part b however.

2nd Update: I figured out question b as well. Thanks to all who took the time to look at my post.

Cheers.

Last edited by a moderator: Jan 28, 2004
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