Given y1(x) and y2(x) be two linearly independent solutions of

y'' + p(x)y' + q(x)y = 0

If p(x) and q(x) are continuous functions , show that the Wronskian , W , is given by W[y1,y2] = Aexp(-int[p(x)dx)])

where A is a constant

Question

Given y1(x) and y2(x) be two linearly independent solutions of

y'' + p(x)y' + q(x)y = 0

If p(x) and q(x) are continuous functions , show that the Wronskian , W , is given by W[y1,y2] = Aexp(-int[p(x)dx)])

where A is a constant

turingtest · Answer

You want to prove Abel's theorem?

anonymous · Answer

Just  proving the Wronskian . By starting from the equation above . I have no idea what to do ...

turingtest · Answer

We have to prove that the Wronskian is$$W(y_1,y_2)(x)=\det\left[\begin{matrix}y_1(x) & y_2(x) \ y_1'(x) & y_2'(x)\end{matrix}\right]$$first?
I can't do that, but I can take it from there to prove Abel's identity. Would that help?

anonymous · Answer

Trying to google what an Abel's identity is....

The question actually specifically stated that "show that...... is given by ...."

So I'm really blur now . Sorry

anonymous · Answer

i have got a website on Abel's identity . Let me read it up . =)

anonymous · Answer

Facepalm . I believe I got it . Thanks a lot!!!!

turingtest · Answer

$$y''+p(x)y'+q(x)y=0\to y''=-(py'+qy)$$$$W(y_1,y_2)(x)=\det\left[\begin{matrix}y_1(x) & y_2(x) \ y_1'(x) & y_2'(x)\end{matrix}\right]=y_1y_2'-y_1'y_2$$$$W'=y_1'y_2'+y_1y_2''-y_1''y_2-y_1'y_2'=y_1y_2''-y_1''y_2$$subbing in what we had for y'' above we will get a first order linear DE which we can rewrite in terms of the Wronskian. A few steps more will yield Abel's identity.
...but you already got it. Good luck!