# Wronskian

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• October 8th 2008, 05:31 PM
Oblivionwarrior
Wronskian
I am stuck on this question, I would appreciate any help.

If y1 and y2 are linearly independent solutions of

$ty^{''} +2y^{'} +te^{4t}y =0$

And if W(y1,y2)(1) = 2, find W(y1,y2)(3)

Thanks for any help.
• October 9th 2008, 05:26 PM
ThePerfectHacker
Quote:

Originally Posted by Oblivionwarrior
$ty^{''} +2y^{'} +te^{4t}y =0$

And if W(y1,y2)(1) = 2, find W(y1,y2)(3)

I assume $t>0$ and in that case write:
$y'' + \frac{2}{t}y' + e^{4t}y=0$

Then $W(y_1,y_2) = a \cdot e^{-\int \frac{2}{t} dt} = a\cdot e^{ - 2\ln t} = \frac{a}{t^2}$.

Since $W(y_1,y_2)(1) = 2 \implies \frac{a}{1^2} = 2 \implies a=2$.

Now you can find $W(y_1,y_2)(3)$
• October 9th 2008, 05:48 PM
Oblivionwarrior
Abel's Theorem, of course!