Question

Find the solution of the given initial value problem (Laplace Transform):

Answer 1

\[\large y''+4y= \sin(t)-u_{2\pi}(t) \sin(t-2\pi)\]
\[y(0)=0,~~~~~~y'(0)=0\]

I was able to make the proper substitutions for y'' and 4y such that:\[\large F(s)*(s^2+4)= \sin(t)-u_{2\pi}(t) \sin(t-2\pi)\]\[\large F(s)=\frac{\sin(t)-u_{2\pi}(t) \sin(t-2\pi)}{s^2+4}\]Where do I go from here? =/

Answer 2

**I used:
\[\mathcal{L}[y'']=s^2F(s)-s f(0)-f'(0)~~~\text{and}~~~\mathcal{L}[y]=F(s)\]

Answer 3

Oh wait I think I see my mistake -- Kinda forget to take the LT of the RHS....

Answer 4

If I did this correctly... then,
\[F(s)*(s^2+4)=\frac{1}{s^2+1}-\frac{e^{-2\pi s}}{s^2+1}=\frac{1-e^{-2\pi s}}{s^2+1}\]
\[F(s)=\frac{1-e^{-2\pi s}}{(s^2+1)(s^2+4)}\]

Answer 5

@tkhunny

Answer 6

Use partial fractions decomposition

Answer 7

for \[\frac{ 1 }{ (s^2+1)(s^2+4) }\]

Answer 8

assuming you did everything correctly :P

Answer 9

Aw :( What happens to the exponential?

Answer 10

You have \[Y(s)=\frac{1-e^{-2\pi s}}{(s^2+1)(s^2+4)}\]
which is right

The solution will be
\[y(t)=\mathcal L^{-1}\left\{\frac{1-e^{-2\pi s}}{(s^2+1)(s^2+4)}\right\}\]

Answer 11

Yes, you take the inverse, but first do partial fraction decomposition, leave the numerator alone, this will make it much simpler to solve

Answer 12

\[\frac1{(s^2+1)(s^2+4)}=\frac A{s^2+1}+\frac B{s^2+4}\]
\[1=A(s^2+4)+B(s^2+1)\]

to find the constants \(A\), and \(B\), take the cases that \(s=i\), and \(s=2i\)

Answer 13

ahhh, I see. Thanks!!

Answer 14

what do you get for \(A\) and \(B\)?

Answer 15

@UnkleRhaukus A = 1/3, B = -1/3

Answer 16

good,
so you have
\[\frac1{(s^2+1)(s^2+4)}=\frac13\left(\frac 1{s^2+1}-\frac1{s^2+4}\right)\]

Our solution becomes
\[y(t)=\mathcal L^{-1}\left\{\frac13\left(\frac{1-e^{-2\pi s}}{(s^2+1)}-\frac{1-e^{-2\pi s}}{(s^2+4)}\right)\right\}\\
\qquad =\frac13\left(\mathcal L^{-1}\left\{\frac{1-e^{-2\pi s}}{s^2+1}\right\}-\mathcal L^{-1}\left\{\frac{1-e^{-2\pi s}}{s^2+4}\right\}\right)\\
\qquad=\]

Answer 17

@UnkleRhaukus Hey! I think you made a slight mistake during the PFD. Shouldn't the two numerators be As + B and Cs + D? Or does it not matter?

Answer 18

It should be As+B and Cs+D

Answer 19

We needn't consider the clauses \((s^2+1)\) and \((s^2+4)\) as quadratics.

If we simplify the expression with \(p = s^2\)
the composition shows that:
\[\frac1{(p+1)(p+4)}=\frac13\left(\frac 1{p+1}-\frac1{p+4}\right)\]which agrees with the above.

Answer 20

Nicee