<p>How do I solve ty''+(1-2t)y'-2y=0
I don't know how to get the t's out of the equation. I am supposed to use laplace transforms, but I can't when I have t's. Thanks.</p>

<p>wait do you want t to be isolated on one side or completely eliminated from the entire thing?</p>

<p>isolaled on one side. I just need to be able to solve for y. Thanks</p>

<p>ty''+(1-2t)y'-2y=0
ty''-2ty' = 2y-y'
t(y''-2y') = 2y-y'
t = (2y-y') / (y''-2y')</p>

<p>can I solve for y using laplace transforms? I didn't think you could have a fraction that included two different types of derivatives. Thanks again.</p>

<p>does anyone know how to find the equation y=...?</p>

<p>So, it's late, meaning this may not be right, but here's a start:
If y(t) <-> Y(s), then
t y(t) <-> -dY(s)/ds
using the "Multiplication By Time" property (<em>such</em> a clever name). What kinds of initial conditions do you have? you will need them since
y'(t) <-> sY(s) - y(0-)
y''(t) <-> s^2Y(s) - sy(0-) - y'(0-)</p>

<p>y(0)=1 and y'(0)=2
I dont remember the multiplication by time property??
Thank you very much for the help</p>

<p>does anyone know how i would use the time property on y'. any extra help would be appreciated as well. thanks</p>

<p>haha! little rusty here but Duke guy was correct;
you must change first ty'' and ty' terms. Theses are steps</p>

<p>1) L(ty') = -d/ds(L(y'))= -d/ds(sY(s)-y(0)) = -sY'(s)-Y(s)
2) L(ty'') = -d/ds(L(y''))= -d/ds(s^2Y(s)-sy(0)-y'(0)) = -s^2Y'(s)-2sY(s) +y(0)
3) Substituting 1) and 2) into the original eq. with L(y') = sY(s)-y(0) and L(y) =Y(s)
4) Solve for 1st order linear diff. eq. for Y(s)
5) inv. laplace Y(s) to get y(t), then apply initial value conditions
6) the answer should be exp(2t)</p>

<p>Using xmaple, got exp(2t) as well.</p>

<p>thanks for the help</p>