What will be the required solution of d2y/dx2 – 3dy/dx + 4y = 0?

What will be the required solution of d2y/dx2 – 3dy/dx + 4y = 0? Correct Answer Ae4x + Be-x

d2y/dx2 – 3dy/dx + 4y = 0 …..(1) Let, y = emx be a trial solution of (1), then, => dy/dx = memx and d2y/dx2 = m2emx Clearly, y = emx will satisfy equation (1). Hence, we have, m2emx – 3m * emx – 4emx = 0 =>m2 – 3m – 4 = 0 (as emx ≠ 0) …….(2) => m2 – 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 = Ae4x + Be-x where A and B are constants.
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