SOLVED Question:
My workings so far :
As you can see, I have a problem finding the laplace transform of the middle equation of the last image. Would you please give me tips/advise as how to obtain the answer? Thank you!
SOLVED Question:
My workings so far :
As you can see, I have a problem finding the laplace transform of the middle equation of the last image. Would you please give me tips/advise as how to obtain the answer? Thank you!
You have an error in your Laplace transform
$\displaystyle x''-4x'+4x=4, \quad x(0)=0 x'(0)=-2$
$\displaystyle s^2X-s\cdot 0-(-2)-4(sX-0)+4X=\frac{4}{s}$
$\displaystyle (s^2-4s+4)X=-2+\frac{4}{s}=\frac{-2s+4}{s}$
$\displaystyle X=\frac{-2s+4}{s(s-2)^2}=-\frac{-2(s-2)}{s(s-2)^2}=\frac{-2}{s(s-2)}$
Now from here you can use partial fractions or notice that
$\displaystyle \frac{-2}{s(s-2)}=\frac{-s+(s-2)}{s(s-2)}=\frac{-s}{s(s-2)}+\frac{s-2}{s(s-2)}=\frac{-1}{s-2}+\frac{1}{s}$
both of these are known Laplace transforms
$\displaystyle -e^{2t}+1$
You can avoid partial fractions by using
$\displaystyle \mathcal{L}\left\{t^nf(t) \right\}=(-1)^n\frac{d^n}{ds^n}F(s)$
$\displaystyle V(s)=\frac{8}{s^2+16}+\frac{8s}{(s^2+16)^2}=2\frac {4}{s^2+4^2}-\frac{d}{ds}\left( \frac{4}{s^2+4^2}\right)$
Now finding the inverse gives
$\displaystyle 2\sin(4t)+t\sin(4t)$