Question

Solve the following equation with Laplace Transform Method (Inverse Laplace the equation to find the solution) y"-3y'-4y=3e^{2x} y(0)=1 , y'(0)=0

Laplace transform
ANSWERED
asked 2021-02-06
Solve the following equation with Laplace Transform Method (Inverse Laplace the equation to find the solution)
\(y"-3y'-4y=3e^{2x}\)
\(y(0)=1 , y'(0)=0\)

Answers (1)

2021-02-07
\(\text{Step 1}\)
\(\text{Given differential equation,}\)
\(y"-3y'-4y=3e^{2x}\)
\(y(0)=1 , y'(0)=0\)
\(\text{Solve this differential equation by Laplace Transform method.}\)
\(\text{Step 2}\)
\(y"-3y'-4y=3e^{2x}\)
\(\text{Take Laplace Transform of both sides,}\)
\(L\left[y"-3y'-4y\right]=L\left[{3e^{2x}}\right]\)
\(L\left[y"\right]-3L\left[y'\right]-4L\left[y\right]=3L\left[e^{2x}\right]\)
\(\text{Use the formula such that }\)
\(L\left[y"\right]=s2L\left[y\right]-sy(0)-y'(0)\)
\(L\left[y'\right]=sL\left[y\right]-y(0)\)
\(L\left[e^{ax}\right]=\frac{1}{s-a}\)
\(\text{Then from (1),}\)
\(\text{Step 3}\)
\(s^2L\left[y\right]-sy(0)-y'(0)-3\left[sL\left[y\right]-y(0)\right]-4L\left[y\right]=3\times \frac{1}{s-2}\)
\(s^2L\left[y\right]-s\times1-0-3\left[sL\left[y\right]-1\right]-4L\left[y\right]=\frac{3}{s-2}\)
\(\left\{s^2-3s-4\right\}L\left[y\right]=\frac{3}{s-2}+s-3\)
\(L\left[y\right]=\frac{3}{(s-2)(s^2-3s-4)}+\frac{s-3}{s^2-3s-4}\)
\(y=L^{-1}\left[\frac{3}{(s-2)(s^2-3s-4)}+\frac{s-3}{s^2-3s-4}\right]\)
\(y=3L^{-1}\left[\frac{1}{(s-2)(s^2-3s-4)}\right]+L^{-1}\left[\frac{s-3}{s^2-3s-4}\right] ...(2)\)
\(\text{Step 4}\)
\(\text{Now}\)
\(\frac{1}{(s-2)(s^2-3s-4)}=-\frac{1}{6(s-2)}+\frac{1}{15(s+1)}+\frac{1}{10(s-4)}\)
\(\Rightarrow L^{-1}\left[\frac{1}{(s-2)(s^2-3s-4)}\right]=L^{-1}\left[-\frac{1}{6(s-2)}\right]+L^{-1}\left[\frac{1}{15(s+1)}\right]+L^{-1}\left[\frac{1}{10(s-4)}\right]\)
\(\Rightarrow L^{-1}\left[\frac{1}{(s-2)(s^2-3s-4)}\right]=-\frac{1}{6}L^{-1}\left[\frac{1}{(s-2)}\right]+\frac{1}{15}L^{-1}\left[\frac{1}{(s+1)}\right]+\frac{1}{10}L^{-1}\left[\frac{1}{(s-4)}\right]\)
\(=-\frac{1}{6}\times e^{2x}+\frac{1}{15}\times e^{-x}+\frac{1}{10}\times e^{4x}\)
\(\text{and}\)
\(\frac{s-3}{s^2-3s-4}=\frac{4}{5(s+1)}+\frac{1}{5(s-4)}\)
\(\Rightarrow L^{-1}\left[\frac{s-3}{s^2-3s-4}\right]=L^{-1}\left[\frac{4}{5(s+1)}\right]+L^{-1}\left[\frac{1}{5(s-4)}\right]\)
\(\Rightarrow L^{-1}\left[\frac{s-3}{s^2-3s-4}\right]=\frac{4}{5}L^{-1}\left[\frac{1}{(s+1)}\right]+\frac{1}{5}L^{-1}\left[\frac{1}{(s-4)}\right]\)
\(=\frac{4}{5}\times e^{-x}+\frac{1}{5}\times e^{4x}\)
\(\text{Step 5}\)
\(\text{From (2), }\)
\(y=3L^{-1}\left[\frac{1}{(s-2)(s^2-3s-4)}\right]+L^{-1}\left[\frac{s-3}{s^2-3s-4}\right]\)
\(y=3\left[-\frac{1}{6}\times e^{2x}+\frac{1}{15}\times e^{-x}+\frac{1}{10}\times e^{4x}\right]+\frac{4}{5}\times e^{-x}+\frac{1}{5}\times e^{4x}\)
\(y=-\frac{1}{2}\times e^{2x}+\frac{1}{5}\times e^{-x}+\frac{3}{10}\times e^{4x}+\frac{4}{5}\times e^{-x}+\frac{1}{5}\times e^{4x}\)
\(y=-\frac{1}{2} e^{2x}+ e^{-x}+\frac{1}{2} e^{4x}\)
\(\text{Step 6}\)
\(\text{Hence, the solution of the given differential equation is}\)
\(y(x)=-\frac{1}{2} e^{2x}+ e^{-x}+\frac{1}{2} e^{4x}\)
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