Question

Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 if 3≤t<50 if 5≤t<[infinity],y(0)=4. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of y(t)y(t) by Y(s)Y(s). Do not move any terms from one side of the equation to the other (until you get to part (b) below).

Answer 1

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$.

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$:

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$, then $1=5a\implies a=\frac15$.

If $s=-5$, then $1=-5b\implies b=-\frac15$.

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$. We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$