Recent content by frggr

I know, that's one of the problems. The inequalities didn't switch when I tried to do it algebraicaly :. I'm doing something wrong. The first inequality wrong as well. it should be \frac{1}{cos \theta} \leq r \leq 2cos \theta

Alright, so I'm having a problem converting to polar co-ordinates Never been confused by polar before :(I'm trying to convert the domain of an integral to polar co-ordinates The domain would be something like D:{(x,y)| 1<x<2 , 0<y<(2x-x2) 1/2} First I recognize that y=(2x-x2) 1/2 is the...

I could be wrong (Still just a student) but even w/o initial conditions the series method should work. Just you won't know a0 and a1

Ahh, thanks. Not sure why I thought (-2)^2 would be -4. My bad. Do have one last question though I don;t really care if it's answered. Would e^(3t) cos(2i) + i e(3t)sin(2i) be a valid solution, or are Laplace transforms only defined for Real Numbers?

I'm more lost now than when I started. -> I must be doing Laplace transforms completely wrong. Or am using the wrong definition to calculate them because : I tried computing exp(3t)sin(2t) by definition and got 2/((s-3)^2-4), Any explanations would be appreciated I've supplied my...

thank you both for the reply, and yea I was just trying to get practice with Laplace transforms. Heh, guess I shouldn't have missed that. Question now is why did my solution come out wrong? -- Simplifying shouldn't really change the outcome just how difficult it is to get there right?

I'm not sure where I am making the mistake was hoping you guys could help. ------------------------------------------------------------------------------------------------------- Q: x' = 3x - 2y; x0=1 y' = 3y - 2x; y0=1 My Sol: 1) Take Laplace Transform of both sides. X:=L{x}, Y := L{y} Xs -...