Question

Solve the initial value problem y′+y=f(t),y(0)=0 where f(t)={1,−1, if t<4 if t≥4 Use h(t−a) for the Heaviside function shifted a units horizontally.

Answer 1

Looks like the function on the right hand side is

$f(t)=\begin{cases}1&\text{for }t<4\\-1&\text{for }t\ge4\end{cases}$

We can write it in terms of the Heaviside function,

$h(t-a)=\begin{cases}1&\text{for }t\ge a\\0&\text{for }t>a\end{cases}$

as

$f(t)=h(t)-2h(t-4)$

Now for the ODE: take the Laplace transform of both sides:

$y'(t)+y(t)=f(t)$

$\implies s Y(s)-y(0)+Y(s)=\dfrac{1-2e^{-4s}}s$

Solve for Y(s), then take the inverse transform to solve for y(t):

$(s+1)Y(s)=\dfrac{1-e^{-4s}}s$

$Y(s)=\dfrac{1-e^{-4s}}{s(s+1)}$

$Y(s)=(1-e^{-4s})\left(\dfrac1s-\dfrac1{s+1}\right)$

$Y(s)=\dfrac1s-\dfrac{e^{-4s}}s-\dfrac1{s+1}+\dfrac{e^{-4s}}{s+1}$

$\implies y(t)=1-h(t-4)-e^{-t}+e^{-(t-4)}h(t-4)$

$\boxed{y(t)=1-e^{-t}-h(t-4)(1-e^{-(t-4)})}$