What will be the required solution of d^2y/dx^2 – 3dy/dx + 4y = 0?

(a) Ae^-4x + Be^-x

(b) Ae^4x – Be^-x

(c) Ae^4x + Be^-x

(d) Ae^4x + Be^x

I had been asked this question by my school principal while I was bunking the class.

This key question is from Linear Second Order Differential Equations topic in division Differential Equations of Mathematics – Class 12

NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options

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The correct choice is (c) Ae^4x + Be^-x

To elaborate: d^2y/dx^2 – 3dy/dx + 4y = 0 …..(1)

Let, y = e^mx be a trial solution of (1), then,

=> dy/dx = me^mx and d^2y/dx^2 = m^2e^mx

Clearly, y = e^mx will satisfy equation (1). Hence, we have,

m^2e^mx – 3m * e^mx – 4e^mx = 0

=>m^2 – 3m – 4 = 0 (as e^mx ≠ 0) …….(2)

=> m^2 – 4m + m – 4 = 0

=> m(m – 4) + 1(m – 4) = 0

Or, (m – 4)(m + 1) = 0

Thus, m = 4 or m = -1

Clearly, the roots of the auxiliary equation (2) are real and unequal.

Therefore, the required general solution of (1) is

y = Ae^4x + Be^-x where A and B are constants.