# Variation of Parameters

• February 25th 2010, 05:17 PM
palmaas
Variation of Parameters
Hello everyone. I have tried this problem a couple times and seem to get it wrong.

Find a particular solution to
Any help would be appreciated
• February 26th 2010, 07:13 AM
Mauritzvdworm
its a bit tedeous...

firstly we have to consider the homogenous equation

$y^{\prime\prime}-4y^{\prime}+4=0$

which has solution

$y_{c}(t)=c_1 e^{2t}+c_2 t e^{2t}$

using the variation of parameters theorem we see that

$y_{p}(t)=-e^{2t}\int\frac{te^{2t}f(t)}{W(t)}dt+te^{2t}\int\f rac{e^{2t}f(t)}{W(t)}dt$

with $W(t)=e^{4x}$ calculted using the Wronskian determinant

so we can then see that

$y_{p}(t)=-e^{2x}\int\frac{te^{2t}f(t)}{e^{4t}}dt+te^{2t}\int \frac{e^{2t}f(t)}{e^{4t}}dt$

and after substituting $f(t)=\frac{e^{2t}}{t^{2}+1}$ and turning the crank we end up with

$y_{p}(t)=-e^{2t}\int\frac{t}{t^{2}+1}dt+te^{2t}\int\frac{1}{ t^{2}+1}dt$

which is simple to integrate, we then find the final answer of

$y_{p}(t)=\frac{1}{2}e^{2t}\ln \left(t^{2}+1\right)+te^{2t}\tan^{-1}\left(x\right)+C$

The final anwer will then just be the superposition of the particular and complimentary solutions
• February 26th 2010, 10:21 AM
palmaas
Thanks for the help!
I assume that the arctan(x) is supposed to be arctan(t) eh? I still get the wrong answer with that, but it does give me something to build off of. Thanks again!
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