# Solve the following differential equations using the Laplace transform and the unit step function y"+4y=g(t) y(0)=-1 y'(0)=0 , text{ where } g(t)=begin{cases}t &, tleq 25 & ,t > 2end{cases} y"-y=g(t) y(0)=1 y'(0)=2 , text{ where } g(t)=begin{cases}1 &, tleq 3t & ,t > 3end{cases}

Question
Laplace transform
Solve the following differential equations using the Laplace transform and the unit step function
$$y"+4y=g(t)$$
$$y(0)=-1$$
$$y'(0)=0 , \text{ where } g(t)=\begin{cases}t &, t\leq 2\\5 & ,t > 2\end{cases}$$
$$y"-y=g(t)$$
$$y(0)=1$$
$$y'(0)=2 , \text{ where } g(t)=\begin{cases}1 &, t\leq 3\\t & ,t > 3\end{cases}$$

2021-02-03
Step 1 Given :The differential equations using the Laplace Transform and the unit step function
$$y"+4y=g(t)$$
$$y(0)=-1$$
$$y'(0)=0 , \text{ where } g(t)=\begin{cases}t &, t\leq 2\\5 & ,t > 2\end{cases}$$
Step 2
The given initial value problem is
$$y"+4y=g(t)$$
$$y(0)=-1$$
$$y'(0)=0 , \text{ where } g(t)=\begin{cases}t &, t\leq 2\\5 & ,t > 2\end{cases}$$
Write the step using the Heaviside step functions.
If f(t) is a function of t for $$t\geq0$$ whose Laplace Transform F(s) exists , then for any constant $$a\geq0$$ , the function
$$g(t)=f(t-a)u(t-a)=\begin{cases}0&\text{ if } ta \end{cases}$$
has the Laplace Transform $$e^{-as}F(s)$$
$$g(t)=(2-u(t-2))+5u(t-2)=2-u(t-2)+5u(t-2)$$
$$g(t)=2+4u(t-2)$$ So, the initial value problem can be written as
$$y"+4y=2+4u(t-2) , y(0)=-1 , y'(0)=0$$
Taking the Laplace Transform of both sides , we get
$$s^2\bar y - sy(0)-y'(0)+4 \bar y =\frac{2}{s}+4e^{-2s}$$
$$s^2\bar y - s(-1)+4 \bar y =\frac{2}{s}+4e^{-2s}$$
$$s^2\bar y +s+4 \bar y =\frac{2}{s}+4e^{-2s}$$
$$(s^2+4)\bar y =\frac{2}{s}+4e^{-2s}-s$$
$$\bar y =\frac{2}{s(s^2+4)}+4\frac{e^{-2s}}{s^2+4}-\frac{s}{s^2+4}$$
Taking Inverse Laplace Transform , we get
$$\text{If } L^{-1} \left[F(s)\right] = f(t) \text{ then } L^{-1} \left(F(s)\right) = \int_0^t f(t) dt$$
$$L (1) =\frac{1}{s} , L^{-1}\left(\frac{s}{s^2+4}\right) = \cos at , L^{-1}\left(\frac{1}{s^2+4}\right) = \frac{1}{a} \sin at$$
$$L^-1[\bar y] = L^{-1}\left[\frac{2}{s(s^2+4)}\right]+4L^{-1}\left[\frac{e^{-2s}}{s^2+4}\right]-L^{-1}\left[\frac {s}{s^2+4}\right]$$
$$y=\frac{2}{2} \int_0^t \sin2t dt + \frac{4e^{-2s}}{2} \sin 2t - \cos 2 t$$
$$y=-\left[\frac{\cos 2t}{2}\right]_0^t+2u(t-2)\sin2t-\cos2t$$
$$y=-\frac{1}{2}\left[\cos 2t-1\right]+2u(t-2)\sin2t-\cos2t$$
Hence, the differential equation is
$$y=-\frac{1}{2}\left[\cos 2t-1\right]+2u(t-2)\sin2t-\cos2t$$

### Relevant Questions

Given the function $$\begin{cases}e^{-t}& \text{if } 0\leq t<2\\ 0&\text{if } 2\leq t\end{cases}$$
Express f(t) in terms of the shifted unit step function u(t -a)
F(t) - ?
Now find the Laplace transform F(s) of f(t)
F(s) - ?
Let x(t) be the solution of the initial-value problem
(a) Find the Laplace transform F(s) of the forcing f(t).
(b) Find the Laplace transform X(s) of the solution x(t).
$$x"+8x'+20x=f(t)$$
$$x(0)=-3$$
$$x'(0)=5$$
$$\text{where the forcing } f(t) \text{ is given by }$$
$$f(t) = \begin{cases} t^2 & \quad \text{for } 0\leq t<2 ,\\ 4e^{2-t} & \quad \text{for } 2\leq t < \infty . \end{cases}$$
Use a Laplace transform to determine the solution of the following systems with differential equations
a) $$x' +4x+3y=0 \text{ with } x(0)=0$$
$$y'+3x+4y=2e^t , y(0)=0$$
By using Laplace transforms, solve the following differential equations subject to the given initial conditions.
$$y"-4y'=-4te^{2t}, y_0=0, y'_0 =1$$
$$y"+2y'+10y=-6e^{-t}\sin3t, y_0=0, y'_0=1$$
Using Laplace Transform , solve the following differential equation
$${y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}$$
a) $$\frac{14}{{20}}{e}^{{{2}{t}}}-\frac{5}{{30}}{e}^{{-{2}{t}}}-\frac{9}{{30}}{e}^{{-{6}{t}}}$$
b) $$\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{51}{{20}}{e}^{{-{2}{t}}}-\frac{4}{{20}}{e}^{{-{3}{t}}}$$
c) $$\frac{14}{{15}}{e}^{{{2}{t}}}-\frac{5}{{10}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
d) $$\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
Solve the IVP with Laplace Transform:
$$\begin{cases} y"+4y'+4y=(3+t)e^{-2t} \\ y(0)=2 \\ y'(0)=5 \end{cases}$$
Solve the following differential equations using the Laplace transform
$$\frac{(dy)}{(dt)}-4y=4-2t ,$$
$$y(0)=0$$
Solve $$y"+4y=f(t) \text{ subject to } y(0)=0 , y'(0)=2 \text{ and }$$
$$f(t)=\begin{cases}0 & t< \pi\\1 & t\geq\pi\end{cases}$$
How to solve for third order differential equation of $$y"'-7y'+6y =2 \sin (t)$$ using Method of Laplace Transform when $$y(0)=0, y'(0)=0, y"(0)=0$$?
$$\text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by }$$
$$F(s)=\int_0^\infty e^{-st} f(t)dt \(\text{where we assume s is a positive real number. For example, to find the Laplace transform of } f(t)=e^{-t} \text{ , the following improper integral is evaluated using integration by parts:} \(F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{s+1}$$
$$\text{ Verify the following Laplace transforms, where u is a real number. }$$
$$f(t)=t \rightarrow F(s)=\frac{1}{s^2}$$