Question

If y = `log(sqrtx + 1/sqrtx)^2`, then show that x(x + 1)² y₂ + (x + 1)² y₁ = 2.

Answer 1

y = `log(sqrtx + 1/sqrtx)^2`

= `2 log((x + 1)/sqrtx)`

= `2[log(x + 1) - 1/2 log x]`

Differentiating on both sides w.r.t., x

`y_1 = 2[1/(x + 1) - 1/(2x)]`

= `2[(2x -x - 1)/(2x(x + 1))]`

`y_1 = (x - 1)/(x(x + 1))`   ...(1)

 

`y_2 = ((x^2 + x) (1)- (x - 1)(2x + 1))/(x^2(x + 1)^2)`

= `(x^2 + x - (2x^2 + x - 2x - 1))/(x^2(x + 1)^2)`

= `(-x^2 + 2x + 1)/(x^2(x + 1)^2)`   ...(2)

Multiply equation (2) by x(x + 1)²

`x(x + 1)^2 y_2 = (-x^2 + 2x + 1)/x`

`x(x + 1)^2 y_2 = 2x/x - (x^2 - 1)/x`

`x(x + 1)^2 y_2 = 2 - ((x + 1)(x - 1))/x`   ...(3)

Multiply equation (1) by (x + 1)²

`y_1 = (x - 1)/(x(x + 1))`

`x(x + 1)^2 y_1 = ((x + 1)^2 (x - 1))/(x(x + 1))`

= `((x + 1)(x - 1))/x`   ...(4)

Substitute equation (4) into equation (3):

x(x + 1)² y₂ = 2 − (x + 1)² y₁

So, x(x + 1)²y₂ + (x + 1)²y₁ = 2