Question

Solve Using Dirac Deltla/discontinuous forcing

Answer 1

Let $A(t)$ denote the amount of salt in the tank at time $t$. We're told that $A(0)=0$.

For $0\le t\le15$, the salt flows in at a rate of (1/5 lb/gal)*(5 gal/min) = 1 lb/min. When the regulating mechanism fails, 20 lbs of salt is dumped and no more salt flows for $t>15$. We can capture this in terms of the unit step function $u(t)$ and Dirac delta function $\delta(t)$ as

$\text{rate in}=u(t)-u(t-15)+20\delta(t-15)$

(in lb/min)

The salt from the mixed solution flows out at a rate of

$\text{rate out}=\left(\dfrac{A(t)\,\mathrm{lb}}{50+(5-5)t\,\mathrm{gal}}\right)\left(5\dfrac{\rm gal}{\rm min}\right)=\dfrac A{10}\dfrac{\rm lb}{\rm min}$

Then the amount of salt in the tank at time $t$ changes according to

$\dfrac{\mathrm dA}{\mathrm dt}=u(t)-u(t-15)+20\delta(t-15)-\dfrac A{10}$

Let $\hat A(s)$ denote the Laplace transform of $A(t)$, $\hat A(s)=\mathcal L_s\{A(t)\}$. Take the transform of both sides to get

$s\hat A(s)-A(0)=\dfrac1s-\dfrac{e^{-15s}}s+20e^{-15s}-\dfrac1{10}\hat A(s)$

Solve for $\hat A(s)$, then take the inverse of both sides.

$\hat A(s)=\dfrac{\frac{10-10e^{-15s}}{s^2}+\frac{200e^{-15s}}s}{10s+1}$

$\implies\boxed{A(t)=10-10e^{-t/10}+\left(30e^{3/2-t/10}-10\right)u(t-15)}$