Question

If y = 3 cos (log x) + 4 sin (log x), show that x²y₂ + xy₁ + y = 0.

Answer 1

Given, y = 3 cos (log x) + 4 sin (log x)  ...(1)

Differentiating both sides with respect to x,

`dy/dx = 3 d/dx cos (log x) + 4 d/dx sin (log x)`

= `3 [- sin (log x)] d/dx (log x) + 4 cos (log x) d/dx (log x)`

= `-3 sin (log x) xx 1/x + 4 cos (log x) xx 1/x`

Multiplying both sides by x,

`x dy/dx` = −3 sin (log x) + 4 cos (log x)

Differentiating both sides again with respect to x,

`x d/dx (dy/dx) + dy/dx * d/dx (x) = - 3 cos (log x) d/dx (log x) - 4 sin (log x) d/dx (log x)`

`x (d^2 y)/dx^2 + 1 * dy/dx = - 3 cos (log x) 1/x - 4 sin (log x) * 1/x`

Multiplying both sides by x,

`x^2 (d^2 y)/dx^2 + x dy/dx` = −[3 cos (log x) + 4 sin (log x)]

`x^2 (d^2 y)/dx^2 + x dy/dx` = −y  ...[From equation (1)]

`=> x^2 (d^2 y)/dx^2 + x dy/dx + y = 0`

Or, x²y₂ + xy₁ + y = 0