Question

# Solve the initial value problem below using the method of Laplace transforms. y"-4y'+40y=225 e^{5t} y(0)=5 y'(0)=31

Laplace transform
Solve the initial value problem below using the method of Laplace transforms.
$$y"-4y'+40y=225 e^{5t}$$
$$y(0)=5$$
$$y'(0)=31$$

2021-02-01
Step 1
Taking Laplace transform in the both the direction:
$$y"-4y'+40y=225 e^{5t}$$
$$y(0)=5$$
$$y'(0)=31$$
$$L\left\{y"-4y'+40y\right\}=L\left\{Y(s)\right\}$$
$$L\left\{y"\right\}-4L\left\{y'\right\}+40L\left\{y\right\}=225L\left\{Y(s)\right\}$$
$$L\left\{y"\right\}=s^2Y(s)-sy(0)-y'(0)$$
$$L(y')=sY(s)-y(0)$$
$$L(y)=Y(s) , L\left\{e^{at}\right\}=\frac{1}{(s-a)}$$
$$\left\{s^2Y(s)-sy(0)-y'(0)\right\}-4\left\{sY(s)-y(0)\right\}+40\left\{Y(s)\right\}=225\left\{\frac{1}{(s-5)}\right\}$$
$$\left\{s^2-4s+40\right\}Y(s)+\left\{-5s-11\right\}=225\left\{\frac{1}{(s-5)}\right\}$$
$$\left\{s^2-4s+40\right\}Y(s)+\left\{-5s-11\right\}=225\left\{\frac{1}{(s-5)}\right\}$$
$$\left\{s^2-4s+40\right\}Y(s)=225\left\{\frac{1}{(s-5)}\right\}+5s+11$$
$$Y(s)=\frac{225\left\{\frac{1}{(s-5)}\right\}+5s+11}{s^2-4s+40}$$
Step 2
Find the factor of the function:
$$Y(s)=\frac{225\left\{\frac{1}{(s-5)}\right\}+5s+11}{s^2-4s+40}=$$
$$=\frac{225}{(s-5)(s^2-4s+40)}+\frac{5s}{(s^2-4s+40)}+\frac{11}{(s^2-4s+40)}$$
$$=225\left\{\frac{-5s-5}{s^2-4s+40}+\frac{5}{(s-5)}\right\}+5 \cdot \frac{(s-2)}{(s-2)^2+36}+ 10 \cdot \frac{1}{(s-2)^2+36} +\left\{11 \cdot \frac{1}{(s-2)^2+36}\right\}$$
Step 3 Taking inverse Laplace transform in both direction:
$$L^{-1}\left\{Y(s)\right\}=L^{-1}\left\{\left\{\frac{(-5s-5)}{(s^2-4s+40)}+\frac{5}{(s-5)}\right\}+5 \cdot \frac{(s-2)}{(s-2)^2+36}+ 10 \cdot \frac{1}{(s-2)^2+36} +\left\{11 \cdot \frac{1}{(s-2)^2+36}\right\}\right\}$$
$$=L^{-1}{Y(s)}=L^{-1}\left\{\left\{\frac{(-5s-5)}{(s^2-4s+40)}+\frac{5}{(s-5)}\right\}+5 \cdot \frac{(s-2)}{(s-2)^2+36}+ 10 \cdot \frac{1}{(s-2)^2+36} +\left\{11 \cdot \frac{1}{(s-2)^2+36}\right\}\right\}$$
$$L^{-1}\left\{\frac{(-5s-5)}{(s^2-4s+40)}+\frac{5}{(s-5)}\right\}=L^{-1}\left\{-5 \cdot \frac{(s-2)}{(s-2)^2+36}- 15 \cdot \frac{1}{(s-2)^2+36}+ \frac{5}{(s-5)}\right\}$$
$$=-5L^{-1}\left\{\frac{(s-2)}{(s-2)^2+36}\right\}-15L^{-1}\left\{\frac{1}{(s-2)^2+36}\right\}+L^{-1}\left\{\frac{5}{(s-5)}\right\}$$
$$\left\{L^{-1}\left\{\frac{s-2}{(s-2)^2+36}\right\}=e^{2t} \cos(6t) , L^{-1}\left\{\frac{1}{(s-2)^2+36}\right\}=\frac{1}{6}e^{2t} \sin(6t) , L^{-1}\left\{\frac{5}{(s-5)}\right\}=5e^{5t}\right\}$$
$$=-5e^{2t} \cos (6t)-15e^{2t}\frac{1}{6}\sin(6t)+5e^{5t}$$
Step 4
Further solve:
$$L^{-1}\left\{\frac{5s}{(s-2)^2+36}\right\}=5L^{-1}\left\{\frac{(s-2)}{(s-2)^2+36}\right\}+10L^{-1}\left\{\frac{1}{(s-2)^2+36}\right\}$$
$$=5e^{2t}\cos(6t)+10e^{2t}\frac{1}{6}\sin(6t)$$
$$L^{-1}\left\{\frac{11}{s^2-4s+40}\right\}=11L^{-1}\left\{\frac{1}{(s-2)^2+36}\right\}$$
$$=\frac{11}{6}e^{2t}sin(6t)$$
$$L^{-1}(Y(s))=-5e^{2t}\cos(6t)-15e^{2t}\frac{1}{6}\sin(6t)+5e^{5t}+5e^{2t}\cos(6t)+10e^{2t}\frac{1}{6}\sin(6t)+\frac{11}{6}e^{2t}\sin(6t)$$
$$y=e^{2t}\sin(6t)+5e^{5t}$$
Step 5
Hence the solution of the given Laplace equation is given as:
$$y=e^{2t}\sin(6t)+5e^{5t}$$