# Solve the differential equation using Laplace transform of y''-3y'+2y=e^{3t} when y(0)=0 and y'(0)=0

Solve the differential equation using Laplace transform of
${y}^{″}-3{y}^{\prime }+2y={e}^{3t}$
when y(0)=0 and y'(0)=0
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Khribechy
Step 1
To Solve:
${y}^{″}-3{y}^{\prime }+2y={e}^{3t}$
when y(0)=0 and y'(0)=0
Concept:
Let $L\left(y\left(t\right)\right)=Y\left(s\right)$ be any given Laplace Transform
Then laplace transform of the heigher order derivative is given by the following equation.
$L\left({y}^{\prime }\right)=\left(sY\left(s\right)-y\left(0\right)\right)$
$L\left({y}^{″}\right)=\left({s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right)$
$L\left({e}^{at}\right)=\frac{1}{\left(s-a\right)}$
Step 2
Explanation:
We have${y}^{″}-3{y}^{\prime }+2y={e}^{3t}$
taking the laplace transform of each term, we get
$L\left\{{y}^{″}\right\}-3L\left\{{y}^{\prime }\right\}+2L\left\{y\right\}=L\left\{{e}^{3t}\right\}$
$⇒{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)-3sY\left(s\right)-y\left(0\right)+2Y\left(s\right)=\frac{1}{\left(s-3\right)}$
$⇒{s}^{2}Y\left(s\right)-3sY\left(s\right)+2Y\left(s\right)=\frac{1}{\left(s-3\right)}\dots \left(\because y\left(0\right)={y}^{\prime }\left(0\right)=0\right)$
$⇒Y\left(s\right)\left[{s}^{2}-3s+2\right]=\frac{1}{\left(s-3\right)\left[{s}^{2}-3s+2\right]}$
$⇒Y\left(s\right)=\frac{1}{2\left(s-1\right)}-\frac{1}{\left(s-2\right)}+\frac{1}{2\left(s-3\right)}$
Now, taking the inverse of the laplace, we get
$y\left(t\right)={L}^{-1}\left\{Y\left(s\right)\right\}=\frac{1}{2}{L}^{-1}\left\{\frac{1}{\left(s-1\right)}\right\}-{L}^{-1}\left\{\frac{1}{\left(s-2\right)}\right\}+\frac{1}{2}{L}^{-1}\left\{\frac{1}{\left(s-3\right)}\right\}$
$⇒y\left(t\right)=\frac{1}{2}{e}^{t}-{e}^{2t}+\frac{1}{2}{e}^{3t}$
Therefore, the solution of the given Initial value problem is
$y\left(t\right)=\frac{1}{2}{e}^{t}-{e}^{2t}+\frac{1}{2}{e}^{3t}$
Step 3
$y\left(t\right)=\frac{1}{2}{e}^{t}-{e}^{2t}+\frac{1}{2}{e}^{3t}$