# Linear Differential Equations Questions and Answers – Linear Independence and Dependence

This set of Linear Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “Linear Independence and Dependence”.

1. If m (t) and n (t) be any two linear differential equations and c and d be any two nonzero constants, then which of the following option is applicable to make these two equations linearly dependent?
a) c m(t) + d n(t) = c + d
b) c m(t) + d n(t) = 0
c) c m(t) + d n(t) = c – d
d) c m(t) + d n(t) ≠ 0

Explanation: According to linear algebra two vectors A and B are called linear dependent if there are nonzero constants u and v with u A + v B = 0. Here we can compare differential functions as vectors. So for any two constants c and d and any two linear differential equations m (t) and n (t), the necessary condition to make these equations linearly dependent is c m(t) + d n(t) = 0. In any other case these equations are called linearly independent.

2. If a (t) and b (t) be any two linear differential equations and x and y be any two nonzero constants, then which of the following option is applicable to make these two equations linearly independent?
a) x a(t) + y b(t) = 0
b) x a(t) – y b(t) = 0
c) x b(t) + y a(t) = 0
d) x a(t) + y b(t) ≠ 0

Explanation: If there are two vectors A and B and two nonzero constants u and v with u A + v B ≠ 0, then according to linear algebra these two vectors are called linearly independent. Here we can compare differential functions as vectors. So for any two constants x and y and any two linear differential equations a (t) and b (t), the necessary condition to make these equations linearly independent is x a (t) + y b (t) ≠ 0.

3. Let two linear differential equations x (t) = sin2t and y (t) = (cos2t – 1). Find out the two constant values c and d for which these equations will be linearly dependent.
a) c = 2 and d = 3
b) c = 3 and d = 2
c) c = 2 and d = 2
d) c = 1 and d = 2

Explanation: For any two constants c and d and any two linear differential equations x (t) and y (t), the necessary condition to make these equations linearly dependent is c x(t) + d y(t) = 0.
So, if c = 2 and d = 2,
c x(t) + d y(t)
= 2 sin2t + 2 (cos2t – 1)
= 2 (sin2t + cos2t) – 2
= (2 – 2), since (sin2t + cos2t) = 1
= 0.

4. Let two linear differential equations x (t) = (sec2t – 1) and y (t) = tan2t and two constant values c = 4 and d = -4. Given two differential equations are linearly dependent. Find out this statement is True or False for the constants c and d.
a) True
b) False

Explanation: If c and d are any two constants and x (t) and y (t) are any two linear differential equations, then the necessary condition to make these equations linearly dependent is c x(t) + d y(t) = 0.
So, if c = 4 and d = 4,
c x(t) + d y(t)
= 4 (sec2t – 1) + (-4) tan2t
= 4 (sec2t – tan2t) – 4
= (4 – 4), since (sec2t – tan2t) = 1
= 0.
So the above statement is True.

5. Let two linear differential equations x (t) = (cosec2t – 1) and y (t) = cot2t and two constant values c = 2 and d = -3. Given two differential equations are linearly dependent. Find out this statement is True or False for the given constant values.
a) True
b) False

Explanation: Suppose any two constants c and d and any two linear differential equations x (t) and y (t), then the necessary condition to make these equations linearly dependent is c x(t) + d y(t) = 0.
So, if c = 2 and d = -3,
c x(t) + d y(t)
= 2 (cosec2t – 1) + (-3) cot2t
= 2 (cosec2t – cot2t) – cot2t – 2
= (2 – cot2t – 2), since (cosec2t – cot2t) = 1
= (-cot2t)
≠ 0.
So the above statement is False.
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6. Let two linear differential equations x (t) = (2t + 8) and y (t) = (3t + 12). Find out the two constant values c and d for which these equations will be linearly dependent.
a) c = 3 and d = -2
b) c = -2 and d = 3
c) c = 2 and d = -2
d) c = -3 and d = 2

Explanation: The necessary condition to make x(t) and y(t) linearly dependent is c x(t) + d y(t) = 0, where c and d are any two constants and x (t) and y (t) are any two linear differential equations.
So, if c = 3 and d = -2,
c x(t) + d y(t)
= 3 (2t + 8) + (-2) (3t + 12)
= (6t + 24 – 6t -24)
= 0.

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