# Solve the initial value problem below using the method of Laplace transforms 2ty''-3ty'+3y=6,y(0)=2, y'(0)=-3

Emily-Jane Bray 2021-02-09 Answered
Solve the initial value problem below using the method of Laplace transforms
$2t{y}^{″}-3t{y}^{\prime }+3y=6,y\left(0\right)=2,{y}^{\prime }\left(0\right)=-3$
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Step 1
Given initial value problem,
$2t{y}^{″}-3t{y}^{\prime }+3y=6,y\left(0\right)=2,{y}^{\prime }\left(0\right)=-3$
Step 2
Take Laplace transform of both sides of the given initial value problem,
$L\left[2t{y}^{″}-3t{y}^{\prime }+3y\right]=L\left[6\right]$
$2L\left[t{y}^{″}\right]-3L\left[t{y}^{\prime }\right]+3L\left[y\right]=6L\left[1\right]$
Use the formula such that
$L\left[t{y}^{″}\right]=\left(-1\right)\frac{d}{\left(ds\right)}L\left[{y}^{″}\right]$
$L\left[{y}^{″}\right]={s}^{2}L\left[y\right]-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left[{y}^{\prime }\right]=sL\left[y\right]-y\left(0\right)$
$L\left[1\right]=\frac{1}{s}$
Step 3
Then,
$2\left[{s}^{2}L\left[y\right]-sy\left(0\right)-{y}^{\prime }\left(0\right)\right]-3\left[sL\left[y\right]-y\left(0\right)\right]+3L\left[y\right]=6\cdot \frac{1}{s}$
$2\left[{s}^{2}L\left[y\right]-2s+3\right]-3\left[sL\left[y\right]-2\right]+3L\left[y\right]=\frac{6}{s}$
$2s2L\left[y\right]-4s+6-3sL\left[y\right]+6+3L\left[y\right]=\frac{6}{s}$
$\left(2{s}^{2}-3s+3\right)L\left[y\right]-4s+12=\frac{6}{s}$
$\left(2{s}^{2}-3s+3\right)L\left[y\right]=\frac{6}{s}+4s-12$
$L\left[y\right]=\frac{6+4{s}^{2}-12s}{2{s}^{2}-3s+3}$
$L\left[y\right]=\frac{6+4{s}^{2}-12s}{2{s}^{2}-3s+3}$
Step 4
Taking inverse Laplace transform,
$y={L}^{-1}\left[\frac{6+4{s}^{2}-12s}{2{s}^{2}-3s+3}\right]\dots \left(1\right)$
Now
$\frac{6+4{s}^{2}-12s}{2{s}^{2}-3s+3}=2-\frac{6s}{2{s}^{2}-3s+3}$
$=2-\frac{3s}{{s}^{2}-\frac{3}{2s}+\frac{3}{2}}$
$=2-\frac{3\left(s-\frac{3}{4}+\frac{3}{4}\right)}{\left(s-\frac{3}{4}{\right)}^{2}+\frac{15}{16}}$
$=2-\frac{3\left(s-\frac{3}{4}\right)}{\left(s-\frac{3}{4}{\right)}^{2}+\frac{15}{16}}-\frac{9}{4}\frac{1}{\left(s-\frac{3}{4}{\right)}^{2}+\frac{15}{16}}$
Step 5
Then,
${L}^{-1}\left[\frac{6+4{s}^{2}-12s}{2{s}^{2}-3s+3}\right]={L}^{-1}\left[2-\frac{3\left(s-\frac{3}{4}\right)}{\left(s-\frac{3}{4}{\right)}^{2}+\frac{15}{16}}-\frac{9}{4}\frac{1}{\left(s-\frac{3}{4}{\right)}^{2}+\frac{15}{16}}\right]$