# Linear Differential Equations Questions and Answers – System of Simultaneous Linear D.E with Constant Coefficients

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

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 + et and y = e-t – et + sin⁡t
b) x = et and y = e-t – et + sin⁡t
c) x = et + e-t and y = sin⁡t
d) x = et + e-t and y = e-t – et

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 et + C2 e-t
y = – $$\frac {dx}{dt}$$ + sin⁡t
= C2 e-t – C1 et + 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 = et + e-t and y = e-t – et + 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

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 = 5et; $$\frac {dy}{dt}$$ – 2x = 5et if x = – 1 and y = 3?
a) x = – et and y = 3et
b) x = et and y = 3et
C) x = – e-t and y = 3et
d) x = – et and y = – 3et

Explanation:
The given equations are
$$\frac {dx}{dt}$$ + 2y = 5et; $$\frac {dy}{dt}$$ – 2x = 5et
First we eliminate y.
Differentiating w.r.t. t,
$$\frac {d^2 x}{dt^2}$$ + 2$$\frac {dy}{dt}$$ = 5et
$$\frac {d^2 x}{dt^2}$$ + 2[2x + 5et] = 5et
(D2 + 4)x = – 5et
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 = $$\frac {1}{D^2 + 4}$$(-5et)
= – et
x = C.F + P.I
x = C1cos⁡2t + C2sin⁡2t – et
The general solution is x = -et and y = 3et

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)]

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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