Answer to Question #272865 in Differential Equations for Lads

Question

Tank A initially contains 200, litres of brine containing 225 N of salt. Eight litres of fresh water per A and the mixture, assumed uniform, passes from A to b. initially .containing 200 litres of fresh water, at 8 litres per minute. The resulting mixture, also kept uniform, leaves B at the rate of 8 litres/min. Find the amount of minute enter salt in tank B after one hour.

Answer 1

Let $A(t) =$ amount, in N of salt in tank A at time $t.$ Then we have

$\dfrac{dA}{dt}=$(rate of salt into tank A) − (rate of salt out of tank A)

$$\dfrac{dA}{dt}=0-\dfrac{8A}{200}$$

So we get the differential equation

$$\dfrac{dA}{A}=-\dfrac{dt}{25}$$

Integrate

$$A(t)=Ce^{-t/25}$$

Given $A(0)=225.$ Then

$$A(t)=225e^{-t/25}$$

Let $B(t) =$ amount, in N of salt in tank B at time $t.$ Then we have

$\dfrac{dB}{dt}=$(rate of salt into tank B) − (rate of salt out of tank B)

$$\dfrac{dB}{dt}=\dfrac{8}{200}(225e^{-t/25})-\dfrac{8}{200}B$$

$$\dfrac{dB}{dt}+0.04B=e^{-0.04t}$$

$$\dfrac{dB}{dt}+0.04B=0$$

$$\dfrac{dB}{B}=-0.04dt$$

$$B=c_1e^{-0.04t}$$

$$\dfrac{dB}{dt}=\dfrac{dc_1}{dt}e^{-0.04t}-0.04c_1e^{-0.04t}$$

$$\dfrac{dc_1}{dt}e^{-0.04t}-0.04c_1e^{-0.04t}+0.04c_1e^{-0.04t}$$

$$=e^{-0.04t}$$

$$\dfrac{dc_1}{dt}=1$$

$$c_1=t+c_2$$

$$B(t)=(t+c_2)e^{-0.04t}$$

Given $B(0)=0.$ Then

$$B(t)=te^{-0.04t}$$

Hence

$$B(60)=60e^{-0.04(60)}\approx5.443 \ N$$