The solution of the differential equation d^2y/dx^3 - 8d^2y/dx^3 = 0 satisfying y(0) = 1/8, y1(0) = 0 and y2(0) = 1 is

Question

The solution of the differential equation \(\frac{d^3y}{dx^3} - 8\frac{d^2y}{dx^2} =0\) satisfying y(0) = \(\frac {1}8\), y₁(0) = 0 and y₂(0) = 1 is \(y_1 = \frac{e^{8x} - 8x + 7}\lambda\), then the numerical value of λ is

(a) 8

(b) \(\frac 7{64}\)

(c) 64

(d) \(\frac {-1}8\)

Answer 1

Correct option is (c) 64