I am stuck on this question, I would appreciate any help.

If y1 and y2 are linearly independent solutions of

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

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

Thanks for any help.

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- Oct 8th 2008, 05:31 PMOblivionwarriorWronskian
I am stuck on this question, I would appreciate any help.

If y1 and y2 are linearly independent solutions of

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

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

Thanks for any help. - Oct 9th 2008, 05:26 PMThePerfectHacker
I assume $\displaystyle t>0$ and in that case write:

$\displaystyle y'' + \frac{2}{t}y' + e^{4t}y=0$

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

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

Now you can find $\displaystyle W(y_1,y_2)(3)$ - Oct 9th 2008, 05:48 PMOblivionwarrior
Abel's Theorem, of course!