Question

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

Answer 1

Given that y = 3cos(log x) + 4sin(log x) 

Differentiating both sides w.r.t. x, we get

dy/dx = y₁ = -3sin(logx)(d/dx)(logx) + 4cos(logx)(d/dx)(logx)

= - 3sin(logx)1/x + 4cos(logx)1/x = 1/x[-3sin(logx) + 4cos(logx)]

⇒ xy₁ = - 3sin(log x) 4cos(logx) 

Again, Differentiating both sides w.r.t. x, we get