Calculate Wronksian of Second Order Differential Equation Use variation of parameters to find a particular solution to:
$\frac{d^{2}y}{dx^{x}} + 2 \frac{dy}{dx} + y = \frac{1}{x^{4}e^{4}}.$
There are no solutions given so finding a wronskian that way is nil.
But since it is still in the order $p(x)y'' + q(x)y' + r(x)y = g(x)$ I think there is still a way to calculate a Wronskian. I have not worked with second order differential equations before and some hints/tips/help would be appreciated. 
 A: you can see that the homogeneous equation $$y'' + 2y' + y = 0 $$ has the fundamental solutions $$\left\{  e^{-x}, xe^{-x} \right\}.$$ the wronskian $w$ is $$w = e^{-x} (xe^{-x})'-xe^{-x}(e^{-x})'=e^{-2x}\left( -x+1+x\right) = e^{-2x} $$  so that the wronskian satisfies $$w' = -2w $$
wronskian of $$ay'' + by' + cy = 0 $$ satisfies the abel's equation $$ aw' + b w = 0$$
A: Since the discriminant of the differential equation $y'' + 2y' + y = 0$ is $2^{2} - 4 = 0,$ it follows that $$u_{1} := e^{-x},\ u_{2} := xe^{-x}$$ are the basis solutions. If $x \mapsto w$ is the Wronskain of $u_{1}$ and $u_{2}$, then 
$$w = u_{1}u_{2}' - u_{2}u_{1}' = e^{-2x}.$$
Let $R(x) := 1/x^{4}e^{4},$ let $t_{1} := -D^{-1}u_{2}R(x)/w,$ and let $t_{2} := D^{-1}u_{1}R(x)/w,$ where $D^{-1}$ means the primitive "operator". Then
$$t_{1} = -D^{-1}x^{-3}e^{x-4},\ t_{2} = D^{-1}x^{-4}e^{x-4}.$$ Then the particular solution $y_{1}$ is simply
$$y_{1} = t_{1}u_{1} + t_{2}u_{2}.$$
As to the underlying theorems, please simply check any book on ordinary differential equations.
