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"`

`y_1 = "m"(x + sqrt(1 + x^2))^("m"-1) {1 + (2x)/(2sqrt(1 + x^2))}`

`= "m" (x + sqrt(1 + x^2))^("m" - 1){(sqrt (1+ x^2) + x)/(sqrt(1 + x^2))} = ("m"(x + sqrt(1 + x^2))^"m")/(sqrt(1 + x^2))`

`y_1 = "my"/sqrt(1 + x^2)`

Squaring both sides we get,

`y_1^2 = ("m"^2y"^2)/(1 + x^2)`

(1 + x²) `(y_1^2) = m<sup>2</sup>y<sup>2</sup>`

Differentiating with respect to x, we get

(1 + x²) . 2(y₁) (y₂) + (y₁)² (2x) = 2m²yy₁

Dividing both sides by 2y₁ we get,

(1 + x²) y₂ + xy₁ = m²y

⇒ (1 + x²) y₂ + xy₁ – m²y = 0