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^xI 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

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

1.	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^xI 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
Answer» 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.

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