Tahmid Knox

2020-11-10

use the Laplace transform to solve the given initial-value problem.

Arnold Odonnell

$\text{Given differential equation is}$

$\text{Step 2}$
$\text{Laplace transform is denoted as}$
$Y\left(s\right)=L\left(y\left(t\right)\right)$

$L\left({y}^{n}\left(t\right)\right)={s}^{n}Y\left(s\right)-{s}^{n-1}y\left(0\right)-{s}^{n-2}{y}^{\prime }\left(0\right)-\cdots -{y}^{n-1}\left(0\right)$
$\text{Let's take the Laplace on both sides, of given differential equation}$
$L\left(y"-3{y}^{\prime }+2y\right)=L\left(4\right)$
${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)-3\left[sY\left(s\right)-y\left(0\right)\right]+2Y\left(s\right)=\frac{4}{s}$
$Y\left(s\right)\left({s}^{2}-3s+2\right)+y\left(0\right)\left(3-s\right)-{y}^{\prime }\left(0\right)=\frac{4}{s}$

$Y\left(s\right)\left({s}^{2}-3s+2\right)-1-\frac{4}{s}=0$
$Y\left(s\right)=\frac{\frac{4+s}{s}}{{s}^{2}-3s+2}$
$=\frac{4+s}{s\left(s-2\right)\left(s-1\right)}$
$\text{By partial fraction}$
$Y\left(s\right)=\frac{2}{s}-\frac{5}{s-1}+\frac{3}{s-2}$

$y\left(t\right)=2-5{e}^{t}+{3}^{2t}$
$\text{This is the required general solution.}$

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