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

Printable View

- Feb 25th 2010, 04:17 PMpalmaasVariation 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 - Feb 26th 2010, 06:13 AMMauritzvdworm
its a bit tedeous...

firstly we have to consider the homogenous equation

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

which has solution

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

using the variation of parameters theorem we see that

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

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

so we can then see that

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

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

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

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

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

The final anwer will then just be the superposition of the particular and complimentary solutions - Feb 26th 2010, 09:21 AMpalmaas
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!