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

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