Recent content by jeannea

Sorry for the confusion, I should have written it differently. I used L{tf(t)}=-F'(s). I had the other one written down as part of the inverse L to find f(t), which was not as relevant here

I got L{tx''}=-\frac{1}{t}\frac{d}{ds}tL{x''}=-\frac{d}{ds}(s^{2}X-sx(0)-x'(0))=-(2sX+s^{2}X'). I treated x(0) and x'(0) as constants to take the derivative with respect to s

I got X(s)=A\frac{1}{(s+1)^{4}}, so x(t)=Bt^{3}e^{-t}, which turned out to be correct in the back of my book. When I plugged in the derivatives of x(t) it worked out. I wasn't given any other initial conditions.

Yes. Thanks

Oh, I see. I knew I was doing the separation wrong. Wouldn't the s cancel so I wouldn't need to do partial fractions? Also, for the next step I'm not sure what to integrate with respect to on each side. The notation is throwing me off. Should I do: ∫XdX=∫-\frac{s+1}{4}ds ? Thank you

The differential equation would be y^{(4)}+2y^{(3)}-2y'-y Remember that r=r^{1} and 1=r^{0}

Homework Statement Transform the given DE to find a nontrivial solution such that x(0)=0. tx''+(t-2)x'+x=0Homework Equations The Attempt at a Solution Using L{f(t)}=-\frac{1}{t}F'(s), I got 4sX(s)+s(s+1)X'(s)=0. I see that it is separable, but I do not know how to go about separating it...

Not quite. If you multiply (r-1)(r+1)^{3}, you should get (r-1)(r^{3}+3r^{2}+3r+1)=r^{4}+2r^{3}-2r-1

The general solution implies that the characteristic equation has one distinct real root and one repeated real root, it can be factored as (r-1)(r+1)^3