Question

a) In a circuit containing inductance 𝐿, resistance 𝑅 and voltage 𝐸, the current 𝐼 is

given by 𝐸 = 𝑅 𝐼 + 𝐿

𝑑𝐼

𝑑𝑡

. Given 𝐿 = 640 ℎ , 𝑅 = 250 𝛺 and 𝐸 =

500 𝑣𝑜𝑙𝑡 . 𝐼 being 0 when 𝑡 = 0. Find the time 𝑡 that elapses, before the current

𝐼 reaches 90% of its maximum value.

[5]

b) Solve the system:

𝑑𝑥

𝑑𝑡

+ 𝑥 − 𝑦 = 𝑡𝑒

𝑡

, 2𝑦 −

𝑑𝑥

𝑑𝑡

+

𝑑𝑦

𝑑𝑡

= 𝑒

Answer 1

a) Differential equation of a LR circuit 

$$L\dfrac{dI}{dt}+RI=E$$ $$\dfrac{dI}{dt}+\dfrac{R}{L}I=\dfrac{E}{L}$$

Integrating factor

$$\mu(t)=e^{\int(R/L)dt}=e^{(R/L)t}$$

$$e^{(R/L)t}(\dfrac{dI}{dt}+\dfrac{R}{L}I)=e^{(R/L)t}\dfrac{E}{L}$$

$$d(e^{(R/L)t}I)=e^{(R/L)t}\dfrac{E}{L}dt$$

Integrate

$$\int d(e^{(R/L)t}I) =\int e^{(R/L)t}\dfrac{E}{L}dt$$

$$e^{(R/L)t}I=\dfrac{E}{R}e^{(R/L)t}+C$$

$$I(t)=\dfrac{E}{R}+Ce^{-(R/L)t}$$

$I(0)=0$

$$0=\dfrac{E}{R}+C$$

$$C=-\dfrac{E}{R}$$

$$I(t)=\dfrac{E}{R}(1-e^{-(R/L)t})$$

Given $E=500V, R=250\ Ohm, L=640\ H$

Substitute

$$I(t)=\dfrac{500}{250}(1-e^{-(250/640)t})$$

$$I(t)=2(1-e^{-(25/64)t})$$

Find the time 𝑡 that elapses, before the current $I$ reaches 90% of its maximum value

$$I(t_{1})=2(1-e^{-(25/64)t_1})=2(0.9)$$

$$e^{-(25/64)t_1}=0.1$$

$$\dfrac{25}{64}t_1=\ln (10)$$

$$t_1=2.56\ln(10) \ s$$

$$t_1\approx5.8946\ s$$

b)

$$\dfrac{dx}{dt}+x-y=te^t$$

$$2x-\dfrac{dx}{dt}+\dfrac{dy}{dt}=e^t$$

Differentiate the first equation with respect to $t$

$$\dfrac{d^2x}{dt^2}+\dfrac{dx}{dt}-\dfrac{dy}{dt}=te^t+e^t$$

Substitute

$$\dfrac{dy}{dt}=\dfrac{dx}{dt}-2x+e^t$$

$$\dfrac{d^2x}{dt^2}+\dfrac{dx}{dt}-\dfrac{dx}{dt}+2x-e^t=te^t+e^t$$

$$\dfrac{d^2x}{dt^2}+2x=te^t+2e^t$$

Corresponding homogeneous differential equation

$$\dfrac{d^2x}{dt^2}+2x=0$$

Auxiliary equation

$$r^2+2=0$$

$$r=\pm\sqrt{2}i$$

The general solution of the homogeneous differential equation is

$$x_h=c_1\cos(\sqrt{2}t)+c_2\sin(\sqrt{2}t)$$

Find the partial solution of the nonhomogeneous differential equation

$$\tilde{x}=Ate^t+Be^t$$

$$\tilde{x}'=Ate^t+Ae^t+Be^t$$

$$\tilde{x}''=Ate^t+2Ae^t+Be^t$$

Substitute

$$Ate^t+2Ae^t+Be^t+2Ate^t+2Be^t$$

$$=te^t+2e^t$$

$$A=\dfrac{1}{3}, B=\dfrac{4a}{9}$$

The general solution of the nonhomogeneous differential equation is

$$x(t)=c_1\cos(\sqrt{2}t)+c_2\sin(\sqrt{2}t)+\dfrac{1}{3}te^t +\dfrac{4}{9}e^t$$ $$\dfrac{dx}{dt}=-c_1\sqrt{2}\sin(\sqrt{2}t)+c_2\sqrt{2}\cos(\sqrt{2}t)$$ $$+\dfrac{1}{3}te^t +\dfrac{1}{3}e^t +\dfrac{4}{9}e^t$$

$$y= \dfrac{dx}{dt}+x-te^t$$

$$y=-c_1\sqrt{2}\sin(\sqrt{2}t)+c_2\sqrt{2}\cos(\sqrt{2}t)$$

$$+\dfrac{1}{3}te^t +\dfrac{7}{9}e^t+c_1\cos(\sqrt{2}t)+c_2\sin(\sqrt{2}t)$$

$$+\dfrac{1}{3}te^t +\dfrac{4}{9}e^t-te^t$$

$$x(t)=c_1\cos(\sqrt{2}t)+c_2\sin(\sqrt{2}t)+\dfrac{1}{3}te^t +\dfrac{4}{9}e^t$$

$$y(t)=(c_1+c_2\sqrt{2})\cos(\sqrt{2}t)+(-c_1\sqrt{2}+c_2)\sin(\sqrt{2}t)$$

$$-\dfrac{1}{3}te^t +\dfrac{11}{9}e^t$$