I have the given ODE:
$$y''+2y'+2y=e^{-x}\sin x$$
This has the homogeneous solution $y_h=C_1\cos(i-1)x+C_2\sin(-i-1)x$.
The particular solution, in the form $y_p=uy_1+vy_2$, we seek the Ansatz: $y_p=uy_1+vy_2=e^{-x}(\sin x+\cos x)$. So $y_1=e^{-x}\sin x$ and $y_2=e^{-x}\cos x$
Then we aim to solve for $u$ and $v$ by use of the variation of parameters formula:
$$u'y_1+v'y_2=0$$
$$u'y_1'+v'y_2'=f(x)$$
where $f(x)=e^{-x}\sin x$.
So here I should use the Wronskian to facilitate the process. The Wronskian is naturally dependent on $y_1$ and $y_2$ and are $y_1=e^{-x}\sin x$ and $y_2=e^{-x}\cos x$.
So the Wronskian would be
\begin{equation}
\text{Det}\begin{vmatrix}
e^{-x}\sin x & e^{-x}\cos x\\
e^{-x}\cos x-e^{-x}\sin x & -e^{-x}\cos x-e^{-x}\sin x
\end{vmatrix}
\end{equation}
My calculation gives:
$Det=-e^{-2x}\cos2x$
So how is this useful to solve the ODE, when I could just use the formula for variation of parameters?
Thanks
Best Answer
$$y''+2y'+2y=0$$ The solution of the homogeneous should be: $$(r+1)^2-i^2=(r+1-i)(r+1+i)=0$$ $$y_h=e^{-x}(c_1\cos x+c_2\sin x)$$ Note that it's easier to first rewrite the DE as: $$y''+2y'+2y=e^{-x}\sin x$$ $$(ye^x)''+ye^x=\sin x$$ $$v''+v=\sin x$$ Then apply the method of variation of parameters.