 # 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 sanuluy 2021-01-10 Answered
Solve the following initial value problems using Laplace Transforms:
$a\right)\frac{{d}^{2}y}{d{t}^{2}}+4\frac{dy}{dt}+3y=1,$
$y\left(0=0\right),{y}^{\prime }\left(0\right)=0$
$b\right)\frac{{d}^{2}y}{d{t}^{2}}+4\frac{dy}{dt}=1,$
$y\left(0=0\right),{y}^{\prime }\left(0\right)=0$
$c\right)2\frac{{d}^{2}y}{d{t}^{2}}+3\frac{dy}{dt}-2y=t{e}^{-2t},$
$y\left(0=0\right),{y}^{\prime }\left(0\right)=-2$
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Step 1 We need to solve the given differential equations using Laplace transformation.
$a\right)\frac{{d}^{2}y}{d{t}^{2}}+4\frac{dy}{dt}+3y=1,$
$y\left(0=0\right),{y}^{\prime }\left(0\right)=0$
Now , taking Laplace transformation both of the sides.
$L\left\{\frac{{d}^{2}y}{d{t}^{2}}+4\frac{dy}{dt}+3y\right\}=L\left\{1\right\}$
$⇒{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+4\left(sY\left(s\right)-y\left(0\right)\right)+3Y\left(s\right)=\frac{1}{s}$

$⇒Y\left(s\right)=\frac{1}{{s}^{2}+4s+3}$
$⇒Y\left(s\right)=\frac{1}{s\left(s+1\right)\left(s+3\right)}$
$⇒Y\left(s\right)=\frac{1}{3s}-\frac{1}{2\left(s+1\right)}+\frac{1}{6\left(s+3\right)}$ (By the method of partial fraction)
Step 2
Now, taking inverse Laplace transformation to get the final answer.
$⇒y\left(t\right)={L}^{-1}\left\{\frac{1}{3s}-\frac{1}{2\left(s+1\right)}+\frac{1}{6\left(s+3\right)}\right\}$
$⇒y\left(t\right)=\frac{1}{3}{L}^{-1}\left\{\frac{1}{s}\right\}-\frac{1}{2}{L}^{-1}\left\{\frac{1}{s+1}\right\}+\frac{1}{6}{L}^{-1}\left\{\frac{1}{s+3}\right\}$
$⇒y\left(t\right)=\frac{1}{3}\cdot 1-\frac{1}{2}\cdot {e}^{-t}+\frac{1}{6}\cdot {e}^{-3t}$
$⇒y\left(t\right)=\frac{1}{3}-\frac{{e}^{-t}}{2}+\frac{{e}^{-3t}}{6}$
Answer $y\left(t\right)=\frac{1}{3}-\frac{{e}^{-t}}{2}+\frac{{e}^{-3t}}{6}$
Step 3
$b\right)\frac{{d}^{2}y}{d{t}^{2}}+4\frac{dy}{dt}=1,$
$y\left(0=0\right),{y}^{\prime }\left(0\right)=0$
Now , taking Laplace transformation both of the sides. $L\left\{\frac{{d}^{2}y}{d{t}^{2}}+4\frac{dy}{dt}\right\}=L\left\{1\right\}$
$⇒{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+4\left(sY\left(s\right)-y\left(0\right)\right)=\frac{1}{s}$

$⇒Y\left(s\right)=\frac{1}{s\left({s}^{2}+4s\right)}$
$⇒Y\left(s\right)=\frac{1}{s\left(s+1\right)\left(s+3\right)}$
$⇒Y\left(s\right)=\frac{1}{4{s}^{2}}+\frac{1}{16\left(s+4\right)}-\frac{1}{16s}$ (By the method of partial fraction)
Step 4
Now, taking inverse Laplace transformation to get the final answer.