solve the initial value problem with using laplace Y'+3Y=6e^{-3t}cos 6t+24 Y(0)=0

Question
Laplace transform
solve the initial value problem with using laplace
$$Y'+3Y=6e^{-3t}\cos 6t+24$$
$$Y(0)=0$$

2020-12-17
Step 1
The given IVP is as follows.
$$Y'+3Y=6e^{-3t}\cos 6t+24$$
$$Y(0)=0$$
Apply Laplace transform on both sides of the given differential equation as follows.
$$L\left\{Y'+3Y\right\}=L\left\{6e^{-3t}\cos 6t+24\right\}$$
$$L\left\{Y'\right\}+\left\{3Y\right\}=L\left\{6e^{-3t}\cos 6t\right\}+L\left\{24\right\}$$
$$sL\left\{Y\right\}-Y(0)+3L\left\{Y\right\}=6L\left\{e^{-3t}\cos 6t\right\}+L\left\{24\right\}$$
$$(s+3)L\left\{Y\right\}=6\frac{s+3}{(s+3)^{2}+6^{2}}+\frac{24}{s}$$
$$L\left\{Y\right\}=\frac{6}{(s+3)^{2}+6^{2}}+\frac{24}{s(s+3)}$$
$$L\left\{Y\right\}=\frac{6}{(s+3)^{2}+6^{2}}+\frac{8}{s}-\frac{8}{(s+3)}$$
Step 2
Apply inverse Laplace transform on both sides as follows.
$$L^{-1}\left\{L\left\{Y\right\}\right\}=L^{-1}\left\{\frac{6}{(s+3)^{2}+6^{2}}+\frac{8}{s}-\frac{8}{(s+3)}\right\}$$
$$Y(t)=L^{-1}\left\{\frac{6}{(s+3)^{2}+6^{2}}\right\}+8L^{-1}\left\{\frac{1}{s}\right\}-8L^{-1}\left\{\frac{1}{(s+3)}\right\}$$
$$Y(t)=e^{-3t}\sin 6t+8-8L^{-1}\left\{\frac{1}{(s+3)}\right\}$$

Relevant Questions

Solve the initial value problem below using the method of Laplace transforms
$$y"-35y=144t-36^{-6t}$$
$$y(0)=0$$
$$y'(0)=47$$
Solve the third-order initial value problem below using the method of Laplace transforms
$$y'''+3y''−18y'−40y=−120$$
$$y(0)=6$$
$$y'(0)=45$$
$$y"(0)=-21$$
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 $$\displaystyle{\left\lbrace\begin{matrix}{y}\text{}+{16}{y}= \cos{{\left({4}{t}\right)}}\\{y}{\left({0}\right)}={1}\\{y}'{\left({0}\right)}={1}\end{matrix}\right.}$$ using the Laplace transform.
With the aid of Laplace Transform, solve the Initial Value Problem
$${y}\text{}{\left({t}\right)}-{y}'{\left({t}\right)}={e}^{t} \cos{{\left({t}\right)}}+ \sin{{\left({t}\right)}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={0}$$
Use laplace transform to solve the given system
$$a) \frac{dx}{dt} -2x- \frac{dx}{dt}-y =6e^{3t}$$
$$2\left(\frac{dx}{dt}\right)-3x+\frac{dx}{dt}-3y=6e^{3t} x(0)=3 y(0)=0$$
use the Laplace transform to solve the given initial-value problem.
$$y"-3y'+2y=4 \ , \ y(0)=0 \ , \ y'(0)=1$$
use the Laplace transform to solve the initial value problem.
$$y"-3y'+2y=\begin{cases}0&0\leq t<1\\1&1\leq t<2\\ -1&t\geq2\end{cases}$$
$$y(0)=-3$$
$$y'(0)=1$$
$$y"-4y'=-4te^{2t}, y_0=0, y'_0 =1$$
$$y"+2y'+10y=-6e^{-t}\sin3t, y_0=0, y'_0=1$$
$$a) \frac{d^2y}{dt^2}+4\frac{dy}{dt}+3y=1,$$
$$y(0=0) , y'(0)=0$$
$$b) \frac{d^2y}{dt^2}+4\frac{dy}{dt}=1,$$
$$y(0=0) , y'(0)=0$$
$$c) 2\frac{d^2y}{dt^2}+3\frac{dy}{dt}-2y=te^{-2t},$$
$$y(0=0) , y'(0)=-2$$