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

Step 1 We need to solve the given differential equations using Laplace transformation.
$a)\frac{d^{2}y}{d{t}^2}+4\frac{dy}{dt}+3y=1,$
$y(0=0),y^{\prime }(0)=0$
Now , taking Laplace transformation both of the sides.
$L\{\frac{d^{2}y}{d{t}^2}+4\frac{dy}{dt}+3y\}=L\left\{1\right\}$
$\Rightarrow s^{2}Y(s)-sy(0)-y^{\prime }(0)+4(sY(s)-y(0))+3Y(s)=\frac{1}{s}$
$\Rightarrow (s^2+4s+3)Y(s)=\frac{1}{s}(\text{ Using }y(0)=0,y^{\prime }(0)=0)$
$\Rightarrow Y(s)=\frac{1}{s^2+4s+3}$
$\Rightarrow Y(s)=\frac{1}{s(s+1)(s+3)}$
$\Rightarrow Y(s)=\frac{1}{3s}-\frac{1}{2(s+1)}+\frac{1}{6(s+3)}$ (By the method of partial fraction)
Step 2
Now, taking inverse Laplace transformation to get the final answer.
$\Rightarrow y(t)=L^{-1}\{\frac{1}{3s}-\frac{1}{2(s+1)}+\frac{1}{6(s+3)}\}$
$\Rightarrow y(t)=\frac{1}{3}{L}^{-1}\left\{\frac{1}{s}\right\}-\frac{1}{2}{L}^{-1}\left\{\frac{1}{s+1}\right\}+\frac{1}{6}{L}^{-1}\left\{\frac{1}{s+3}\right\}$
$\Rightarrow y(t)=\frac{1}{3}\cdot 1-\frac{1}{2}\cdot e^{-t}+\frac{1}{6}\cdot e^{-3t}$
$\Rightarrow y(t)=\frac{1}{3}-\frac{e^{-t}}{2}+\frac{e^{-3t}}{6}$
Answer $y(t)=\frac{1}{3}-\frac{e^{-t}}{2}+\frac{e^{-3t}}{6}$
Step 3
$b)\frac{d^{2}y}{d{t}^2}+4\frac{dy}{dt}=1,$
$y(0=0),y^{\prime }(0)=0$
Now , taking Laplace transformation both of the sides. $L\{\frac{d^{2}y}{d{t}^2}+4\frac{dy}{dt}\}=L\left\{1\right\}$
$\Rightarrow s^{2}Y(s)-sy(0)-y^{\prime }(0)+4(sY(s)-y(0))=\frac{1}{s}$
$\Rightarrow (s^2+4s)Y(s)=\frac{1}{s}(\text{Using }y(0)=0,y^{\prime }(0)=0)$
$\Rightarrow Y(s)=\frac{1}{s(s^2+4s)}$
$\Rightarrow Y(s)=\frac{1}{s(s+1)(s+3)}$
$\Rightarrow Y(s)=\frac{1}{4s^2}+\frac{1}{16(s+4)}-\frac{1}{16s}$ (By the method of partial fraction)
Step 4
Now, taking inverse Laplace transformation to get the final answer.
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