# y''+y=e^{-2t}sin t , y(0)=y′(0)=0 solution of given initial value problem with Laplace transform

Question
Laplace transform
$$y''+y=e^{-2t}\sin t , y(0)=y′(0)=0$$
solution of given initial value problem with Laplace transform

2020-11-03
Given:
$$y''+y=e^{-2t}\sin t , y(0)=y′(0)=0$$
To find: solution of given initial value problem with Laplace transform
Step 2 Solution
$$y''+y=e^{-2t}\sin t , y(0)=y′(0)=0$$
Taking Laplace transform of the differential equation , we get:
$$[s^2Y(s)-sy(0)-y′(0)]+Y(s)= L[e^{-2t}\sin t]$$ where $$Y(s) = L(y(t))$$
using initial values we get
$$[s^2Y(s)-0]+Y(s) =L[e^{-2t}\sin t]$$
$$\text{Applying } L[e^{at}f(t)] = \bar f(s-a)$$
$$L(\sin t) = \frac{1}{s^2+1}$$
$$L(e^{-2t}\sin t) = \frac{1}{(s+2)^2 +1}$$
Step 3
$$\Rightarrow Y(s)(s^2+1) = \frac{1}{(s+2)^2+1}$$
$$\Rightarrow Y(s) = \frac{1}{(s^2+1)((s+2)^2 +1)}$$
after partial fraction we get
$$Y(s) = \frac{1}{8}\left[-\frac{s}{s^2+1}+\frac{1}{s^2+1} +\frac{s}{(s+2)^2 +1} +\frac{3}{(s+2)^2 +1}\right]$$
using inverse Laplace transform to get y(t) :
$$y(t) = L^{-1}[Y(s)]$$
$$=\frac{1}{8}\left[-\cos t + \sin t +e^{-2t}\cos t +3e^{-2t}\sin t\right]$$
required solution of differential equation .

### Relevant Questions

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y(t) - ?
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$$\text{where the forcing } f(t) \text{ is given by }$$
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