# If the differential equation $t^2 y'' - 2y' + (3 + t)y = 0$ has $y_1$ and $y_2$ as a fundamental set of solutions…

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?

• What are $y_1,y_2$? – user170231 Feb 2 at 21:30
• y1, and y2 aren't given. – Not2Scary Feb 2 at 21:35
• what is $t$ here? independent variable or some scalar? – Kate Feb 2 at 21:43
• t is just an input. Since I'm not given any initial conditions, I'm not quite sure how to solve this problem. – Not2Scary Feb 2 at 22:05

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.