If y = 3cos(logx) + 4sin(logx), show that x²d²y/dx² + xdy/dx + y = 0

Given, y = 3cos(logx) + 4sin(logx)
Differentiating w.r.t to x, we have
dy/dx = -3sin(logx)/x + 4cos(logx)/x
=> y₁ = (1/x)[-3sin(logx) + 4cos(logx)]
Again differentiating w.r.t to x, we have
d²y/dx² = x[-3cos(logx)/x - 4sin(logx)/x] - [-3sin(logx) + 4cos(logx)]/x²
= -3cos(logx) - 4sin(logx) + 3sin(logx) - 4cos(logx)/x²
d²y/dx² = -[3cos(logx) + 4sin(logx)]/x²
=> y₂ = -[3cos(logx) + 4sin(logx)]/x²
Now, L.H.S = x²y₂ + xy₁ + y
= x²(-[3cos(logx) + 4sin(logx)]/x²) + x × (1/x)[-3sin(logx) + 4cos(logx)] + 3cos(logx) + 4sin(logx)
= -[3cos(logx) + 4sin(logx)] + [-3sin(logx) + 4cos(logx)] + 3cos(logx) + 4sin(logx)
= 0
= R.H.S
Hence proved.