# Thread: Laplace Transform [LEVEL 2 MATHS]

1. ## Laplace Transform [LEVEL 2 MATHS]

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!

2. You have an error in your Laplace transform

$x''-4x'+4x=4, \quad x(0)=0 x'(0)=-2$

$s^2X-s\cdot 0-(-2)-4(sX-0)+4X=\frac{4}{s}$

$(s^2-4s+4)X=-2+\frac{4}{s}=\frac{-2s+4}{s}$

$X=\frac{-2s+4}{s(s-2)^2}=-\frac{-2(s-2)}{s(s-2)^2}=\frac{-2}{s(s-2)}$

Now from here you can use partial fractions or notice that

$\frac{-2}{s(s-2)}=\frac{-s+(s-2)}{s(s-2)}=\frac{-s}{s(s-2)}+\frac{s-2}{s(s-2)}=\frac{-1}{s-2}+\frac{1}{s}$

both of these are known Laplace transforms

$-e^{2t}+1$

3. Thank you for showing me my mistake for the Laplace Transform! I've got the answer now.

4. Could you help me with another equation:
I don't think I can solve it using partial fractions?

EDIT: I don't know if you've noticed this post but I managed to solve it

5. You can avoid partial fractions by using

$\mathcal{L}\left\{t^nf(t) \right\}=(-1)^n\frac{d^n}{ds^n}F(s)$

$V(s)=\frac{8}{s^2+16}+\frac{8s}{(s^2+16)^2}=2\frac {4}{s^2+4^2}-\frac{d}{ds}\left( \frac{4}{s^2+4^2}\right)$

Now finding the inverse gives

$2\sin(4t)+t\sin(4t)$