If y = log[x + √(x² + a²)], show that (x² + a²)d²y/dx² + x(dy/dx) = 0

y = log[x + √(x² + a²)]
y₁ = dy/dx = 1/[x + √(x² + a²)] (1 + 2x/[2√(x² + a²) ])
y₁ = 1/[x + √(x² + a²)] (√(x² + a²) + x/√(x² + a²))
⇒ y₁√(x² + a²) = 1
Again differentiate,
⇒ y₂√(x² + a²) + (1/2) . 2x . y₁/√(x² + a²) = 0
⇒ y₂(x² + a²) + xy₁ = 0