This set of Linear Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “System of Simultaneous Linear D.E with Constant Coefficients”.

1. Which of the following is obtained by evaluating \(\frac {dx}{dt}\) – y = t and \(\frac {dy}{dx}\) + x = t?

a) x = C_{1}cost + C_{2}sint + t^{2} – 1 and y = C_{2}cost – C_{1}sint + t

b) x = C_{1}cost + C_{2}sint and y = C_{2}cost – C_{1}sint + t

c) x = C_{1}cost + C_{2}sint + t^{2} and y = C_{2}cost – C_{1}sint + t

d) x = C_{1}cost + C_{2}sint + t^{2} – 1 and y = C_{2}cost

View Answer

Explanation:

The given equations are

\(\frac {dx}{dt}\) – y = t….(1)

\(\frac {dy}{dx}\) + x = t….(2)

First we eliminate y.

Differentiating w.r.t. t,

\(\frac {d^2 x}{dt^2} – \frac {dy}{dt}\) = 1

From (2), \(\frac {dy}{dt}\) = t

^{2}– x

\(\frac {d^2 x}{dt^2}\) – (t

^{2}– x) = 1

\(\frac {d^2 x}{dt^2}\) + x = 1 + t

^{2}

D

^{2}x + x = 1 + t

^{2}, where D = \(\frac {d}{dt}\)

(D

^{2}+ 1)x = 1 + t

^{2}

This is a second order linear differential equation with constant coefficients in x.

To find the complementary function, solve (D

^{2}+ 1)x = 0

Auxiliary equation is m

^{2}+ 1 = 0 → m = ± i

C.F = C

_{1}cost + C

_{2}sint [α = 0; β = 1]

P.I = \(\frac {1}{D^2 + 1}\)(1 + t

^{2})

= (1 + D

^{2})

^{-1}(1 + t

^{2})

= (1 – D

^{2}+ D

^{4}– …)(1 + t

^{2})

= 1 + t

^{2}– 2 = t

^{2}– 1

x = C.F + P.I

x = C

_{1}cost + C

_{2}sint + t

^{2}– 1 and y = C

_{2}cost – C

_{1}sint + t

2. Which of the following is obtained by evaluating \(\frac {dx}{dt}\) + y = sint; \(\frac {dy}{dt} \) + x = cost if x = 2, y = 0 at t = 0 ?

a) x = e^{-t} + e^{t} and y = e^{-t} – e^{t} + sint

b) x = e^{t} and y = e^{-t} – e^{t} + sint

c) x = e^{t} + e^{-t} and y = sint

d) x = e^{t} + e^{-t} and y = e^{-t} – e^{t}

View Answer

Explanation:

The given equations are

\(\frac {dx}{dt}\) + y = sint…..(1)

\(\frac {dy}{dt} \) + x = cost…..(2)

First we eliminate y.

Differentiating w.r.t. t,

\(\frac {d^2 x}{dt^2} + \frac {dy}{dt}\) = cost

\(\frac {dy}{dt} \) = cost – x

From (2), \(\frac {d^2 x}{dt^2} \) + cost – x = cost

\(\frac {d^2 x}{dt^2} \) – x = 0

(D

^{2}– 1)x = 0, where D = \(\frac {d}{dt} \) …..(3)

This is a second order linear differential equation with constant coefficients in x.

Auxiliary equation is m

^{2}– 1 = 0 → m = ±1

x = C

_{1}e

^{t}+ C

_{2}e

^{-t}

y = – \(\frac {dx}{dt}\) + sint

= C

_{2}e

^{-t}– C

_{1}e

^{t}+ sint

Given, when t = 0, x = 2, y = 0

∴ C

_{1}+ C

_{2}= 2

C

_{2}– C

_{1}= 0 → C

_{1}= C

_{2}

2C

_{2}= 2 → C

_{2}= C

_{1}= 1

The solution is x = e

^{t}+ e

^{-t}and y = e

^{-t}– e

^{t}+ sint.

3. Which of the following is obtained by evaluating \(\frac {dx}{dt}\) + 2y = – sint; \(\frac {dy}{dt} \) – 2x = cost?

A) x = C_{1}cos2t + C_{2}sin2t – cost and y = C_{1}sin2t – C_{2}cos2t – sint

b) x = C_{1} and y = C_{1}sin2t – C_{2}cos2t – sint

c) x = C_{1}cos2t + C_{2}sin2t – cost and y = C_{1}sin2t – C_{2}cos2t

d) x = C_{1}cos2t + C_{2}sin2t and y = C_{1}sin2t – C_{2}cos2t – sint

View Answer

Explanation:

The given equations are

\(\frac {dx}{dt}\) + 2y = – sint; \(\frac {dy}{dt} \) – 2x = cost

First we eliminate y.

Differentiating w.r.t. t,

\(\frac {d^2 x}{dt^2} \) + 2\(\frac {dy}{dt} \) = – cost

\(\frac {dy}{dt} \) = cost + 2x

\(\frac {d^2 x}{dt^2} \) + 2(cost + 2x) = – cost

D

^{2}x + 4x = – 3cost, where D = \(\frac {d}{dt} \)

(D

^{2}+ 4)x = – 3cost

This is a second order linear differential equation with constant coefficients in x.

Auxiliary equation is m

^{2}+ 4 = 0 → m = ±2i

The roots are complex numbers with α = 0 and β = 1

C.F = C

_{1}cos2t + C

_{2}sin2t

P.I. = \(\frac {1}{D^2 + 4}\)(-3cost)

= -3.\(\frac {cost}{(-1)^2 + 4}\)

= – cost

x = C.F + P.I

= C

_{1}cos2t + C

_{2}sin2t – cost

The solution is x = C

_{1}cos2t + C

_{2}sin2t – cost

y = C

_{1}sin2t – C

_{2}cos2t – sint

4. Which of the following is obtained by evaluating \(\frac {dx}{dt}\) + 2y = 5e^{t}; \(\frac {dy}{dt} \) – 2x = 5e^{t} if x = – 1 and y = 3?

a) x = – e^{t} and y = 3e^{t}

b) x = e^{t} and y = 3e^{t}

C) x = – e^{-t} and y = 3e^{t}

d) x = – e^{t} and y = – 3e^{t}

View Answer

Explanation:

The given equations are

\(\frac {dx}{dt}\) + 2y = 5e

^{t}; \(\frac {dy}{dt} \) – 2x = 5e

^{t}

First we eliminate y.

Differentiating w.r.t. t,

\(\frac {d^2 x}{dt^2} \) + 2\(\frac {dy}{dt} \) = 5e

^{t}

\(\frac {d^2 x}{dt^2} \) + 2[2x + 5e

^{t}] = 5e

^{t}

(D

^{2}+ 4)x = – 5e

^{t}

This is a second order linear differential equation with constant coefficients in x.

Auxiliary equation is m

^{2}+ 4 = 0 → m = ±2i

The roots are real and different.

C.F = C

_{1}cos2t + C

_{2}sin2t

P.I = \(\frac {1}{D^2 + 4}\)(-5e

^{t})

= – e

^{t}

x = C.F + P.I

x = C

_{1}cos2t + C

_{2}sin2t – e

^{t}

The general solution is x = -e

^{t}and y = 3e

^{t}

5. Which of the following is obtained by evaluating \(\frac {dx}{dt}\) + 5x – 2y = t; \(\frac {dy}{dt} \) + 2x + y = 0 if x = y = 0?

a) x = \(\frac {1}{27} \) [1 + 3t – (1 + 6t) e^{-3t}] and y = \(\frac {2}{27} \)[2 – 3t – e^{-3t}(3t + 2)]

b) x = [1 + 3t – (1 + 6t) e^{-3t}] and y = \(\frac {2}{27} \)[2 – 3t – e^{-3t}(3t + 2)]

c) x = \(\frac {1}{27} \) [1 + 3t – (1 + 6t) e^{-3t}] and y = [2 – 3t – e^{-3t}(3t + 2)]

d) x = \(\frac {1}{27} \) [3t – (1 + 6t) e^{-3t}] and y = \(\frac {2}{27} \)[2 – 3t – e^{-3t}(3t + 2)]

View Answer

Explanation:

The given equations are

\(\frac {dx}{dt}\) + 5x – 2y = t; \(\frac {dy}{dt} \) + 2x + y = 0

Let D = \(\frac {d}{dt} \) then

(D + 5)x – 2y = t

2x + (D + 1)y = 0

First we eliminate y.

Differentiating w.r.t. t,

(D

^{2}+ 6D + 9)x = 1 + t

This is a second order linear differential equation with constant coefficients in x.

Auxiliary equation is m

^{2}+ 6m + 9 = 0 → m = m = -3, -3

The roots are real and different.

C.F = (C

_{1}+ C

_{2}t)e

^{-3t}

P.I = \(\frac {1}{D^2 + 6D + 9}\)(1 + t)

= \(\frac {1}{9}.\frac {1}{ \big [1 + \frac {(6D + D^2)}{9} \big ]}\)(1 + t)

= \(\frac {1}{27} \)(3t + 1)

x = C.F + P.I

x = (C

_{1}+ C

_{2}t) e

^{-3t}+ \(\frac {1}{27} \)(3t + 1)

The general solution is

x = \(\frac {1}{27} \)[1 + 3t – (1 + 6t) e

^{-3t}]

y = \(\frac {2}{27} \)[2 – 3t – e

^{-3t}(3t + 2)]

**Sanfoundry Global Education & Learning Series – Linear Differential Equations.**

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