Answer to Question #268219 in Differential Equations for jay

Question

1. An RL circuit has emf of 24t volts, a resistance of 10 k-ohms, an inductance of 5H with an initial current of 1 mA. What is the current after 0.05 seconds? Ans. 118.8µA
2. Initially a tank holds 75 gal of a brine containing 12 lb of salt. At t=0, fresh water is poured into the tank at the rate of 3 gal/min, while the well stirred mixture leaves the tank at the rate of 2 gal/min. what is the amount of salt in the tank after 1 hour? Ans. 3.70 lb

Answer 1

1.

$$Li'+Ri=V$$ $$5i'+10\cdot10^{3}i=24t$$

$$5i'=24t-10000i$$

Substitute

$$u=24t-10000i$$

$$u'=24-10000i'$$

$$i'=0.0024-0.0001u'$$

$$0.0005(24-u')=u$$

$$u'=-2000(u-0.012)$$

$$\dfrac{du}{u-0.012}=-2000dt$$

Integrate

$$u=0.012+c_1e^{-2000t}$$

$$24t-10000i=0.012+c_1e^{-2000t}$$

$$i=c_2e^{-2000t}+0.0024t-0.0000012$$

Given $i(0)=10^{-3}\ A$

$$0.001=c_2-0.0000012$$

$$c_2=-0.0009988e^{-2000t}+0.0024t-0.0000012$$

$$i(t)=-0.0009988e^{-2000t}+0.0024t-0.0000012$$

$$i(0.05)=-0.0009988e^{-2000(0.05)}+0.0024(0.05)$$

$$-0.0000012=0.0001188(A)$$

$118.8\ \mu A$

2.

Let $s(t) =$ amount, in lb of salt at time $t.$ Then we have

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

$$\dfrac{ds}{dt}=0-\dfrac{2s}{75+(3-2)t}$$

So we get the differential equation

$$\dfrac{ds}{dt}=-\dfrac{2s}{75+t}$$

$$\int \dfrac{ds}{s}=-\int\dfrac{2dt}{75+t}$$

$$s(t)=c_1(75+t)^{-2}$$

$$s(0)=c_1(75)^{-2}=12$$

$$c_1=67500$$

$$s(t)=67500(75+t)^{-2}$$

$$s(60)=67500(75+60)^{-2}$$

$$s(60)=3.7037\ lb$$

$3.70\ lb$