# Using Laplace Transform , solve the following differential equation {y}text{}-{4}{y}={e}^{{-{3}{t}}},{y}{left({0}right)}={0},{y}'{left({0}right)}={2}

Using Laplace Transform , solve the following differential equation
$y-4y={e}^{-3t},y\left(0\right)=0,{y}^{\prime }\left(0\right)=2$
a) $\frac{14}{20}{e}^{2t}-\frac{5}{30}{e}^{-2t}-\frac{9}{30}{e}^{-6t}$
b) $\frac{11}{20}{e}^{2t}-\frac{51}{20}{e}^{-2t}-\frac{4}{20}{e}^{-3t}$
c) $\frac{14}{15}{e}^{2t}-\frac{5}{10}{e}^{-2t}-\frac{9}{20}{e}^{-3t}$
d) $\frac{14}{20}{e}^{2t}+\frac{5}{20}{e}^{-2t}-\frac{9}{20}{e}^{-3t}$
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Step 1
As per guidelines, only typed answers are acceptable. so we will give only typed solution.
Given initial value problem is
$y-4y={e}^{-3t},y\left(0\right)=0,{y}^{\prime }\left(0\right)=2$
Step 2
Applying Laplace transform on both sides
$L\left\{y-4y\right\}=L\left\{{e}^{-3t}\right\}$
$⇒L\left\{y\right\}-4L\left\{y\right\}=\frac{1}{s+3}$
$⇒{s}^{2}L\left\{y\right\}-sy\left(0\right)-{y}^{\prime }\left(0\right)-4L\left\{y\right\}=\frac{1}{s+3}$
$⇒\left({s}^{2}-4\right)L\left\{y\right\}-2=\frac{1}{s+3}$
$⇒L\left\{y\right\}=\frac{1}{\left(s+3\right)\left({s}^{2}-4\right)}+\frac{2}{{s}^{2}-4}$
$=\left(\frac{1}{5\left(s+3\right)}+\frac{1}{20\left(s-2\right)}-\frac{1}{4\left(s+2\right)}\right)+\left(-\frac{1}{2\left(s+2\right)}+\frac{1}{2\left(s-2\right)}\right)$
$=\frac{1}{5\left(s+3\right)}+\frac{11}{20\left(s-2\right)}-\frac{3}{4\left(s+2\right)}$
Taking inverse Laplace transform on both sides, we get
$y={L}^{-1}\left\{\frac{1}{5\left(s+3\right)}+\frac{11}{20\left(s-2\right)}-\frac{3}{4\left(s+2\right)}\right\}$
$={L}^{-1}\left\{\frac{1}{5\left(s+3\right)}\right\}+{L}^{-1}\left\{\frac{11}{20\left(s-2\right)}\right\}-{L}^{-1}\left\{\frac{3}{4\left(s+2\right)}\right\}$
$=\frac{1}{5}{e}^{-3t}+\frac{11}{20}{e}^{2t}-\frac{3}{4}{e}^{-2t}$
Hence, solution is

Comparing the answer with options, none of the option is correct.
The option given are all wrong.