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
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