# Solve the following initial value problems using Laplace Transforms: a) frac{d^2y}{dt^2}+4frac{dy}{dt}+3y=1, y(0=0) , y'(0)=0 b) frac{d^2y}{dt^2}+4frac{dy}{dt}=1, y(0=0) , y'(0)=0 c) 2frac{d^2y}{dt^2}+3frac{dy}{dt}-2y=te^{-2t}, y(0=0) , y'(0)=-2

Question
Laplace transform
Solve the following initial value problems using Laplace Transforms:
$$a) \frac{d^2y}{dt^2}+4\frac{dy}{dt}+3y=1,$$
$$y(0=0) , y'(0)=0$$
$$b) \frac{d^2y}{dt^2}+4\frac{dy}{dt}=1,$$
$$y(0=0) , y'(0)=0$$
$$c) 2\frac{d^2y}{dt^2}+3\frac{dy}{dt}-2y=te^{-2t},$$
$$y(0=0) , y'(0)=-2$$

2021-01-11
Step 1 We need to solve the given differential equations using Laplace transformation.
$$a) \frac{d^2y}{dt^2}+4\frac{dy}{dt}+3y=1,$$
$$y(0=0) , y'(0)=0$$
Now , taking Laplace transformation both of the sides.
$$L\left\{\frac{d^2y}{dt^2}+4\frac{dy}{dt}+3y\right\}=L\left\{1\right\}$$
$$\Rightarrow s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))+3Y(s)=\frac{1}{s}$$
$$\Rightarrow (s^2+4s+3)Y(s)=\frac{1}{s} (\text{ Using } y(0)=0 , y'(0)=0)$$
\Rightarrow Y(s)=\frac{1}{s^2+4s+3}\)
$$\Rightarrow Y(s)=\frac{1}{s(s+1)(s+3)}$$
$$\Rightarrow Y(s)= \frac{1}{3s}-\frac{1}{2(s+1)}+\frac{1}{6(s+3)}$$ (By the method of partial fraction)
Step 2
Now, taking inverse Laplace transformation to get the final answer.
$$\Rightarrow y(t)=L^{-1}\left\{\frac{1}{3s}-\frac{1}{2(s+1)}+\frac{1}{6(s+3)}\right\}$$
$$\Rightarrow y(t)=\frac{1}{3}L^{-1}\left\{\frac{1}{s}\right\}-\frac{1}{2}L^{-1}\left\{\frac{1}{s+1}\right\}+\frac{1}{6}L^{-1}\left\{\frac{1}{s+3}\right\}$$
$$\Rightarrow y(t)=\frac{1}{3}\cdot 1 -\frac{1}{2} \cdot e^{-t}+\frac{1}{6} \cdot e^{-3t}$$
$$\Rightarrow y(t)=\frac{1}{3}-\frac{e^{-t}}{2}+\frac{e^{-3t}}{6}$$
Answer $$y(t)=\frac{1}{3}-\frac{e^{-t}}{2}+\frac{e^{-3t}}{6}$$
Step 3
$$b) \frac{d^2y}{dt^2}+4\frac{dy}{dt}=1,$$
$$y(0=0) , y'(0)=0$$
Now , taking Laplace transformation both of the sides. $$L\left\{\frac{d^2y}{dt^2}+4\frac{dy}{dt}\right\}=L\left\{1\right\}$$
$$\Rightarrow s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))=\frac{1}{s}$$
$$\Rightarrow (s^2+4s)Y(s)=\frac{1}{s} (\text{Using } y(0)=0 , y'(0)=0)$$
$$\Rightarrow Y(s)=\frac{1}{s(s^2+4s)}$$
$$\Rightarrow Y(s)=\frac{1}{s(s+1)(s+3)}$$
$$\Rightarrow Y(s)= \frac{1}{4s^2}+\frac{1}{16(s+4)}-\frac{1}{16s}$$ (By the method of partial fraction)
Step 4
Now, taking inverse Laplace transformation to get the final answer.
$$\Rightarrow y(t)=L^{-1}\left\{\frac{1}{4s^2}+\frac{1}{16(s+4)}-\frac{1}{16s}\right\}$$
$$\Rightarrow y(t)=\frac{1}{4}L^{-1}\left\{\frac{1}{s^2}\right\}+\frac{1}{16}L^{-1}\left\{\frac{1}{s+4}\right\}-\frac{1}{16}L^{-1}\left\{\frac{1}{s}\right\}$$
$$\Rightarrow y(t)=\frac{1}{4}\cdot t +\frac{1}{16} \cdot e^{-4t}-\frac{1}{16} \cdot 1$$
$$\Rightarrow y(t)=\frac{t}{4}+\frac{e^{-4t}}{16} - \frac{1}{16}$$
Answer $$y(t)=\frac{t}{4}+\frac{e^{-4t}}{16} - \frac{1}{16}$$
Step 5
For (c),
$$2\frac{d^2y}{dt^2}+3\frac{dy}{dt}-2y=te^{-2t}$$
$$y(0=0) , y'(0)=-2$$
Now , taking Laplace transformation both of the sides.
$$L\left\{2\frac{d^2y}{dt^2}+3\frac{dy}{dt}-2y\right\}=L\left\{te^(-2t)\right\}$$
$$\Rightarrow 2(s^2Y(s)-sy(0)-y'(0))+3(sY(s)-y(0))-2Y(s)=\frac{1}{(s+2)^2}$$
$$\Rightarrow (2s^2+3s-2)Y(s)+4=\frac{1}{(s+2)^2} (\text{ Using } y(0)=0 , y'(0)=-2)$$
$$\Rightarrow (2s^2+3s-2)Y(s)=\frac{1}{(s+2)^2-4}$$
$$\Rightarrow Y(s)=\frac{\frac{1}{(s+2)^2}-4}{2s^2+3s-2}$$
$$\Rightarrow Y(s)=\frac{-4s^2-16s-15}{(s+2)^2(2s^2+3s-2)}$$
$$\Rightarrow Y(s)=-\frac{192}{125(2s-1)}+\frac{96}{125(s+2)}-\frac{2}{25(s+2)^2}-\frac{1}{5(s+2)^3}$$
(By the method of partial fraction)
Step 6
Now, taking inverse Laplace transformation to get the final answer.
$$\Rightarrow y(t)=L^{-1}\left\{-\frac{192}{250(s-\frac{1}{2})}+\frac{96}{125(s+2)}-\frac{2}{25(s+2)^2}-\frac{1}{5(s+2)^3}\right\}$$
$$\Rightarrow y(t)=-\frac{192}{250}L^{-1}\left\{\frac{1}{s-\frac{1}{2}}\right\}+\frac{96}{125}L^{-1}\left\{\frac{1}{s+2}\right\}-\frac{2}{25}L^{-1}\left\{\frac{1}{(s+2)^2}\right\}-\frac{1}{5}L^{-1}\left\{\frac{1}{(s+2)^3}\right\}$$
$$\Rightarrow y(t)=-\frac{192}{250}\cdot e^{\frac{1}{2t}}+\frac{96}{125}\cdot e^{-2t}-\frac{2}{25}\cdot te^{-2t}-\frac{1}{5}\cdot t^2e^{-2t}$$
$$\Rightarrow y(t)=-\frac{192}{250}e^{\frac{t}{2}}+\frac{96}{125}e^{-2t}-\frac{2}{25}te^{-2t}-\frac{1}{5}t^2e^{-2t}$$
Answer: $$y(t)=-\frac{192}{250}e^{\frac{t}{2}}+\frac{96}{125}e^{-2t}-\frac{2}{25}te^{-2t}-\frac{1}{5}t^2e^{-2t}$$

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