Recent content by Sol-chan

Okay, arbitrary... meaning I can just set them equal to whatever I want, right? But I still don't see how that helps me...

Doesn't that just give 0 = 0? Because if y=C1x then y'=C1 and y''=0, then you get 0-x*C1 + x*C1 = 0 0 = 0 And I don't really understand the recurrence relation, either. I'm not sure if I'm doing this right, but taking the derivative of both sides I get y''' - xy'' +2y' = 0 And I'm not...

Ah, right, I solved that wrong. So, with Cn+2 = (n-1)Cn/(n+1)(n+2) I get that C4 = -C0/24 C6 = -C0/240 C8 = -C0/2688 What am I plugging y=C1x into?

Solve by the method of series y''-xy'+y=0 From what I understand, I need to substitute in y(x)=SUM(Cn*xn) from n=0 to infinity I get that: y'(x)=SUM(n*Cn*xn-1) from n=1 to infinity and y''(x)=SUM(n*(n-1)*Cn*xn-2) from n=2 to infinity When I substitute in and collect the terms, I...

Okay, so then I would have (s-2)/(s-2)2 + 2/(s-2)2 which is 1/(s-2) + 2/(s-2)2 so then I'd get e2t + 2e2tt

Homework Statement Find the inverse laplace transform of: 1/(s-2)3 + 25/(s+1)(s-2)2 + s/(s-2)2 The attempt at a solution I get (1/2)e2tt2 + (25/7)e-t-(25/7)e2t+(75/7)te2t for the first two, but I'm not even sure where to start for s/(s-2)2. I was thinking it might use...