Question

# Solve the following differential equations using the Laplace transform frac{(dy)}{(dt)}-4y=4-2t , y(0)=0

Laplace transform
Solve the following differential equations using the Laplace transform
$$\frac{(dy)}{(dt)}-4y=4-2t ,$$
$$y(0)=0$$

2020-12-25
Step 1
Given:
$$\frac{(dy)}{(dt)}-4y=4-2t$$
$$y(0)=0$$
$$\frac{(dy)}{(dt)}-4y=4-2t$$
Replace $$\frac{(dy)}{(dt)}$$ as y'
$$y'-4y=4-2t$$
Applying Laplace transform on both sides,
$$L\left\{y'-4y\right\}=L\left\{4-2t\right\}$$
$$L\left\{y'-4y\right\}=sL\left\{y\right\}-y(0)-4L\left\{y\right\}$$
$$L\left\{4-2t\right\}=\frac{4}{s}-\frac{2}{s^2}$$
Step 2
$$sL\left\{y\right\}-y(0)-4L\left\{y\right\}=\frac{4}{s}-\frac{2}{s^2}$$
$$sL\left\{y\right\}-0-4L\left\{y\right\}=\frac{4}{s}-\frac{2}{s^2}$$
$$sL\left\{y\right\}−4L\left\{y\right\}=\frac{4}{s}-\frac{2}{s^2}$$
$$L\left\{y\right\}=\frac{4s-2}{s^2(s-4)}$$
Take the inverse Laplace transform
$$y=L^{-1}\left(\frac{4s-2}{s^2(s-4)}\right)$$
$$L^{-1}\left(\frac{4s-2}{s^2(s-4)}\right)=-\frac{7}{8}+\frac{t}{2}+\frac{7}{8}e^{4t}$$
$$y=-\frac{7}{8}+\frac{t}{2}+\frac{7}{8}e^{4t}$$