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

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

Question
Laplace transform
asked 2021-02-09
Solve the initial value problem below using the method of Laplace transforms
\(2ty''-3ty'+3y=6,y(0)=2, y'(0)=-3\)

Answers (1)

2021-02-10
Step 1
Given initial value problem,
\(2ty''-3ty'+3y=6,y(0)=2, y'(0)=-3\)
Step 2
Take Laplace transform of both sides of the given initial value problem,
\(L[2ty''-3ty'+3y]=L[6]\)
\(2L[ty'']-3L[ty']+3L[y]=6L[1]\)
Use the formula such that
\(L[ty'']=(-1)\frac{d}{(ds)}L[y'']\)
\(L[y'']=s^2L[y]-sy(0)-y'(0)\)
\(L[y']=sL[y]-y(0)\)
\(L[1]=\frac{1}{s}\)
Step 3
Then,
\(2\left[s^2L[y]-sy(0)-y'(0)\right]-3\left[sL[y]−y(0)\right]+3L[y]=6 \cdot \frac{1}{s}\)
\(2\left[s^2L[y]-2s+3\right]-3\left[sL[y]-2\right]+3L[y]=\frac{6}{s}\)
\(2s2L[y]-4s+6-3sL[y]+6+3L[y]=\frac{6}{s}\)
\((2s^2-3s+3)L[y]-4s+12=\frac{6}{s}\)
\((2s^2-3s+3)L[y]=\frac{6}{s}+4s-12\)
\(L[y]=\frac{6+4s^2-12s}{2s^2-3s+3}\)
\(L[y]=\frac{6+4s^2-12s}{2s^2-3s+3}\)
Step 4
Taking inverse Laplace transform,
\(y=L^{-1}\left[\frac{6+4s^2-12s}{2s^2-3s+3}\right] \dots (1)\)
Now
\(\frac{6+4s^2-12s}{2s^2-3s+3}=2-\frac{6s}{2s^2-3s+3}\)
\(=2-\frac{3s}{s^2-\frac{3}{2s}+\frac{3}{2}}\)
\(=2-\frac{3(s-\frac{3}{4}+\frac{3}{4})}{(s-\frac{3}{4})^2+\frac{15}{16}}\)
\(=2-\frac{3(s-\frac{3}{4})}{(s-\frac{3}{4})^2+\frac{15}{16}}-\frac{9}{4}\frac{1}{(s-\frac{3}{4})^2+\frac{15}{16}}\)
Step 5
Then,
\(L^{-1}\left[\frac{6+4s^2-12s}{2s^2-3s+3}\right]=L^{-1}\left[2-\frac{3(s-\frac{3}{4})}{(s-\frac{3}{4})^2+\frac{15}{16}}-\frac{9}{4}\frac{1}{(s-\frac{3}{4})^2+\frac{15}{16}}\right]\)
\(=2L^{-1}[1]-3L^-1\left[\frac{s-\frac{3}{4}}{(s-\frac{3}{4})^2+\frac{15}{16}}\right]-\frac{9}{4}L^{-1}\left[\frac{1}{(s-\frac{3}{4})^2+\frac{15}{16}}\right]\)
\(=2L^{-1}[1]-3L^{-1}\left[\frac{s-\frac{3}{4}}{(s-\frac{3}{4})^2+(\sqrt{\frac{15}{16}})^2}\right]-\frac{9}{\sqrt15} L^{-1}\left[\frac{\sqrt{\frac{15}{16}}}{(s-\frac{3}{4})^2+(\sqrt{\frac{15}{16}})^2}\right]\)
\(=2\delta(t)-3e^{\frac{3t}{4}}\cos\left(\sqrt{\frac{15}{4t}}\right)-\frac{9}{\sqrt{15}}e^{\frac{3t}{4}}\sin\left(\frac{\sqrt15}{4t}\right)\)
Step 6
From(1) ,
\(y=2\delta(t)-3e^{\frac{3t}{4}}\cos\left(\sqrt{\frac{15}{4t}}\right)-\frac{9}{\sqrt{15}}e^{\frac{3t}{4}}\sin\left(\frac{\sqrt15}{4t}\right)\)
\(y=2\delta(t)-3e^{\frac{3t}{4}}\cos\left(\sqrt{\frac{15}{4t}}\right)-\frac{3\sqrt3}{\sqrt{15}}e^{\frac{3t}{4}}\sin\left(\sqrt{\frac{15}{4t}}\right)\)
Hence, the solution of the given initial value problem is
\(y(t)=2\delta(t)-3e^{\frac{3t}{4}}\cos\left(\sqrt{\frac{15}{4t}}\right)-\frac{3\sqrt3}{\sqrt{15}}e^{\frac{3t}{4}}\sin\left(\sqrt{\frac{15}{4t}}\right)\)
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