Question

# 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}

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
$$y=-\frac{1}{2}\left[\cos 2t-1\right]+2u(t-2)\sin2t-\cos2t$$