Question

Ify=3cos(logx)+4sin(logx), show that x^2y[2]+xy[1]+y=0

Answer 1

Given, $\bf{y=3cos(logx)+4sin(logx)}$
we have to proof that $\bf{x^2y_2+xy_1+y}=0$

y = 3cos(logx) + 4sin(logx) ------(1)
differentiate y with respect to x,
dy/dx = y₁ = 3. d(coslogx)/dx + 4. d(sinlogx)/dx
= 3{-sin(logx)}1/x + 4{cos(logx)}1/x
= {-3sin(logx) + 4cos(logx)}/x
xy₁ = {-3sin(logx) + 4cos(logx)}
again differentiate with respect to x ,
d(xy₁ )/dx = -3. d{sin(logx)}/dx + 4. d{cos(logx)}/dx}
=> x. dy₁ /dx + y₁ .dx/dx = -3.cos(logx)1/x - 4sin(logx).1/x
=> xy₂ + y₁ = -[3cos(logx) + 4sin(logx)]/x
=> x²y₂ + xy₁ = -[3cos(logx) + 4sin(logx)]
=> x²y₂ + xy₁ = -y [ from equation (1),
=> x²y₂ + xy₁ + y = 0 $\textbf{\underline{hence proved}}$