Create a differential equation and solve for an initial value

Question

Water flows continuously at the rate of 1 liter per hour into a bucket, but flows out continuously at the rate of 5% of the total amount per hour. Let $W(t)$ be the number of liters in the bucket at time $t$. Solve for $W(t)=0$

We begin by determining the first-order differential equation that models this scenario.

Strategy: (net rate of W) = (rate of flow in) - (rate of flow out). $$\frac {dW}{dt}=1-0.05W$$

This differential equation is both separable and linear; using separability gives: $$\int\frac {dW}{1-0.05W}=\int dt$$ $$\int\frac {dW}{1-\frac {W}{20}}=\int dt$$ Let $u={1-\frac {W}{20}}$

then $du =-20dW$ and $\frac {du}{20} = dW$

So: $$-20\ln|W-20|=t+C$$ And solving is a matter of simplification.

However, can this differential equation be solved using an integration factor rather than through use of separability? If so, what does it look like?

Answer 1

your equation has the following form

$$W'=1-aW$$ When $a=0.05$. So $$ (W'(t)+aW(t))e^{at}=e^{at}$$ $$\dfrac{d}{dt} \left( W(t)e^{at} \right)=e^{at}$$

$$ W(t)e^{at}= \frac{1}{a}e^{at}+c$$ $$ W(t)= ce^{-at} +\frac{1}{a}$$

Answer 2

You can also handle this as a linear equation with constant coefficients.

The general solution of the homogeneous equation

$$W'+aW=0$$

is

$$W=Ce^{-at}$$

as the single root of the characteristic equation is $-a$.

Now a particular solution of the non-homogeneous equation is a constant

$$W=\frac1a,$$ as this corresponds to $W'=0$.

Combining both,

$$W(t)=\frac1a+Ce^{-at}$$ and $$C=W_0-\frac1a.$$

Finally,

$$W(t)=0=\frac1a+\left(W_0-\frac1a\right)e^{-at}$$

for $$t=-\frac1a\log(1-aW_0).$$

Notice that a solution only exists when $aW_0<1$.