You make that choice for because any derivative of will have that form.

So that part was correct.

so and

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- March 13th 2011, 10:09 AM #1

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- Jan 2011
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## y'' + 4y' + 5y = xe^x

Hi I have some problem with the following d.e

y'' + 4y' + 5y = xe^x.

I will solve it as much as I can

Yh

fives yh = c1 cosx + c2 sinx)*e^-2x

Yp

hard to know what to choose, but I looked at an example

Yp = (A+Bx)e^x**(I don't really know why we take it)**

Yp' = (A + B + Bx)e^x

Yp'' = (A + 2b + Bx)e^x

Right lets put it into the d.e

((A + 2b + Bx)e^x) + 4((A + B + Bx)e^x) + 5((A+Bx)e^x) = x^x

I dont whats happenening but i dont get the real, constants, I get B = -1/4 etc

would be nice if someone could correct my solution, and help me

Thanks

- March 13th 2011, 10:35 AM #2

- March 13th 2011, 10:45 AM #3