Using Wronskian to solve nonhomegeneous ODE 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
 A: It is generally easier to solve linear ODE with constant coefficients this way:


*

*your homogeneous equation has root $r$ with multiplicity $m$ .

*the full equation has a RHS of the form $P(x)e^{rx}$ with $P$ polynomial.

 Then you need to search for a particular solution in the form
$Q(x)e^{rx}$ with $Q$ polynomial and $$\deg(Q)=\deg(P)+m$$
Although since the homogeneous solution will already have vanishing
terms $(C_0+C_1x+\cdots+C_{m-1}x^{m-1})e^{rx}$, you can ignore them in
the polynomial Q.


So here your homogeneous equation is $y''+2y'+2y=0$ of characteristic equation $$r^2+2r+2=0$$
It has roots $r=-1\pm i$ of multiplicity $m=1$.
The sinus can be rewritten as a combination of $e^{ix}$ and $e^{-ix}$, therefore you RHS is $$\underbrace{\frac 1{2i}}_{P_1}e^{-x+ix}-\underbrace{\frac 1{2i}}_{P_2}e^{-x-ix}$$
Both terms of your RHS collide with the roots of the characteristic equation so since $P_1,\ P_2$ are constant polynomials, you need to search for particular solutions of the form $(ax+b)e^{-x+ix}$ and $(cx+d)e^{-x-ix}$
(i.e. $\deg(Q)=\deg(P)+m=0+1=1$)
As said coefficients $b$ and $d$ can be ignored because we know already reporting in the ODE, the associated part will just vanish.
Therefore we can search for $ax\,e^{-x+ix}+cx\,e^{-x-ix}$ or directly in recomposed trig form $$\big(A\cos(x)+B\sin(x)\big)xe^{-x}$$
Solving gives $A=-\frac 12$ and $B=0$.
A: $$y''+2y'+2y=0$$
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.
