Question

find the laplace transform of y''+y=sin(t)

y(0)=y'(0)=0

Answer 1

use Laplace and plug those conditions in

Answer 2

and I have yet to learn Laplace...getting there

Answer 3

\[\qquad y''+y=\sin (t)\]
\[\quad\mathcal L\big\{y''+y\big\}=\mathcal L\big\{\sin (t) \big\}\]
\[\mathcal L\big\{y''\}+\mathcal L\big\{y\big\}=\mathcal L\big\{\sin (t) \big\}\]

Answer 4

\[\mathcal L\big\{y''\}=s^2Y(s)-sy(0)-y'(0)\]
\[\mathcal L\big\{y\big\}=Y(s)\]
\[\mathcal L\big\{\sin (nt) \big\}=\frac{n}{s^2+n^2}\]

Answer 5

\[\mathcal L\big\{y''\}+\mathcal L\big\{y\big\}=\mathcal L\big\{\sin (t) \big\}\\
[s^2Y(s)−sy(0)−y'(0)]+[Y(s)]=\frac{n}{s^2+n^2}\\
(s^2+1)Y(s)=\frac{n}{s^2+n^2}\\
Y(s)=\frac{n}{(s^2+n^2)}\times\frac1{(s^2+1)}\\
y(t)=\mathcal L^{-1}\Big\{\frac{n}{(s^2+n^2)}\times\frac1{(s^2+1)}\Big\}\]