If y = (x + √(1 + x^2))^m, then show that (1 + x^2) y2 + xy1 – m^2y = 0

Question

If y = \((x+\sqrt{1+x^2})^m\), then show that (1 + x²) y₂ + xy₁ – m²y = 0

Answer 1

y = \((x+\sqrt{1+x^2})^m\)

y₁ = \(\frac{my}{\sqrt{1+x^2}}\)

Squaring both sides we get,

y₁² = \(\frac{m^2y^2}{{1+x^2}}\)

(1 + x²) (y₁²) = m² y² 

Differentiating with respect to x, we get 

(1 + x²).2(y₁) (y₂) + (y₁)² (2x) = 2m² yy₁ 

Dividing both sides by 2y₁ we get,

(1 + x²) y₂ + xy₁ = m² y 

⇒ (1 + x²) y₂ + xy₁ – m² y = 0