System of Simultaneous Linear D.E with Constant Coefficients Questions and Answers

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 = C₁cos⁡t + C₂sin⁡t + t² – 1 and y = C₂cos⁡t – C₁sin⁡t + t
b) x = C₁cos⁡t + C₂sint ⁡ and y = C₂cos⁡t – C₁sin⁡t + t
c) x = C₁cos⁡t + C₂sin⁡t + t² and y = C₂cos⁡t – C₁sin⁡t + t
d) x = C₁cos⁡t + C₂sin⁡t + t² – 1 and y = C₂cos⁡t
View Answer

Answer: a
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), \(\frac {dy}{dt}\) = t² – x
\(\frac {d^2 x}{dt^2}\) – (t² – x) = 1
\(\frac {d^2 x}{dt^2}\) + x = 1 + t²
D² x + x = 1 + t², where D = \(\frac {d}{dt}\)
(D² + 1)x = 1 + t²
This is a second order linear differential equation with constant coefficients in x.
To find the complementary function, solve (D² + 1)x = 0
Auxiliary equation is m² + 1 = 0 → m = ± i
C.F = C₁cos⁡t + C₂sin⁡t [α = 0; β = 1]
P.I = \(\frac {1}{D^2 + 1}\)(1 + t²)
= (1 + D²)^-1(1 + t²)
= (1 – D² + D⁴ – …)(1 + t²)
= 1 + t² – 2 = t² – 1
x = C.F + P.I
x = C₁cos⁡t + C₂sin⁡t + t² – 1 and y = C₂cos⁡t – C₁sin⁡t + t

2. Which of the following is obtained by evaluating \(\frac {dx}{dt}\) + y = sin⁡t; \(\frac {dy}{dt} \) + x = cos⁡t if x = 2, y = 0 at t = 0 ?
a) x = e^-t + e^t and y = e^-t – e^t + sin⁡t
b) x = e^t and y = e^-t – e^t + sin⁡t
c) x = e^t + e^-t and y = sin⁡t
d) x = e^t + e^-t and y = e^-t – e^t
View Answer

Answer: a
Explanation:
The given equations are
\(\frac {dx}{dt}\) + y = sin⁡t…..(1)
\(\frac {dy}{dt} \) + x = cos⁡t…..(2)
First we eliminate y.
Differentiating w.r.t. t,
\(\frac {d^2 x}{dt^2} + \frac {dy}{dt}\) = cos⁡t
\(\frac {dy}{dt} \) = cos⁡t – x
From (2), \(\frac {d^2 x}{dt^2} \) + cos⁡t – x = cos⁡t
\(\frac {d^2 x}{dt^2} \) – x = 0
(D² – 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 m² – 1 = 0 → m = ±1
x = C₁ e^t + C₂ e^-t
y = – \(\frac {dx}{dt}\) + sin⁡t
= C₂ e^-t – C₁ e^t + sin⁡t
Given, when t = 0, x = 2, y = 0
∴ C₁ + C₂ = 2
C₂ – C₁ = 0 → C₁ = C₂
2C₂ = 2 → C₂ = C₁ = 1
The solution is x = e^t + e^-t and y = e^-t – e^t + sin⁡t.

3. Which of the following is obtained by evaluating \(\frac {dx}{dt}\) + 2y = – sin⁡t; \(\frac {dy}{dt} \) – 2x = cos⁡t?
A) x = C₁cos⁡2t + C₂sin⁡2t – cos⁡t and y = C₁sin⁡2t – C₂cos⁡2t – sin⁡t
b) x = C₁ and y = C₁sin⁡2t – C₂cos⁡2t – sin⁡t
c) x = C₁cos⁡2t + C₂sin⁡2t – cos⁡t and y = C₁sin⁡2t – C₂cos⁡2t
d) x = C₁cos⁡2t + C₂sin⁡2t and y = C₁sin⁡2t – C₂cos⁡2t – sin⁡t
View Answer

Answer: a
Explanation:
The given equations are
\(\frac {dx}{dt}\) + 2y = – sin⁡t; \(\frac {dy}{dt} \) – 2x = cos⁡t
First we eliminate y.
Differentiating w.r.t. t,
\(\frac {d^2 x}{dt^2} \) + 2\(\frac {dy}{dt} \) = – cos⁡t
\(\frac {dy}{dt} \) = cos⁡t + 2x
\(\frac {d^2 x}{dt^2} \) + 2(cos⁡t + 2x) = – cos⁡t
D²x + 4x = – 3cos⁡t, where D = \(\frac {d}{dt} \)
(D² + 4)x = – 3cos⁡t
This is a second order linear differential equation with constant coefficients in x.
Auxiliary equation is m² + 4 = 0 → m = ±2i
The roots are complex numbers with α = 0 and β = 1
C.F = C₁cos⁡2t + C₂sin⁡2t
P.I. = \(\frac {1}{D^2 + 4}\)(-3cos⁡t)
= -3.\(\frac {cos⁡t}{(-1)^2 + 4}\)
= – cos⁡t
x = C.F + P.I
= C₁cos⁡2t + C₂sin⁡2t – cos⁡t
The solution is x = C₁cos⁡2t + C₂sin⁡2t – cos⁡t
y = C₁sin⁡2t – C₂cos⁡2t – sin⁡t

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

Answer: a
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
(D² + 4)x = – 5e^t
This is a second order linear differential equation with constant coefficients in x.
Auxiliary equation is m² + 4 = 0 → m = ±2i
The roots are real and different.
C.F = C₁cos⁡2t + C₂sin⁡2t
P.I = \(\frac {1}{D^2 + 4}\)(-5e^t)
= – e^t
x = C.F + P.I
x = C₁cos⁡2t + C₂sin⁡2t – 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

Answer: a
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,
(D² + 6D + 9)x = 1 + t
This is a second order linear differential equation with constant coefficients in x.
Auxiliary equation is m² + 6m + 9 = 0 → m = m = -3, -3
The roots are real and different.
C.F = (C₁ + C₂t)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 = (C₁ + C₂ 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)]

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