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If the differential equation $t^2 y'' - 2y' + (3 + t)y = 0$ has $y_1$ and $y_2$ as a fundamental set of solutions and if $W(y_1, y_2)(2) = 3$, find $W(y_1, y_2)(4)$.

Is it possible for me to solve this problem as such:

If $W(y_1, y_2)(2) = 3$,

then $W(y_1, y_2)(4) = W(y_1, y_2)(2^2) = 3^2$.

Therefore, $W(y_1, y_2)(4) = 9$

I'm not sure if this is an acceptable way to solve this question or not, and if it's not, could someone please explain why it would be wrong, and how I could go about solving it correctly?

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  • $\begingroup$ What are $y_1,y_2$? $\endgroup$ – user170231 Feb 2 at 21:30
  • $\begingroup$ y1, and y2 aren't given. $\endgroup$ – Not2Scary Feb 2 at 21:35
  • $\begingroup$ what is $t$ here? independent variable or some scalar? $\endgroup$ – Kate Feb 2 at 21:43
  • $\begingroup$ t is just an input. Since I'm not given any initial conditions, I'm not quite sure how to solve this problem. $\endgroup$ – Not2Scary Feb 2 at 22:05
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You know that for $W=\det\pmatrix{y_1&y_2\\y_1'&y_2'}$ you get $$ W'=\det\pmatrix{y_1&y_2\\y_1''&y_2''}=\frac2{t^2}W $$ so that $W(t)=Ce^{-2/t}=3e^{1-2/t}$. So no, your solution for $W(4)$ is wrong.

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  • $\begingroup$ Great, thank you for the response! $\endgroup$ – Not2Scary Feb 2 at 22:41

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