Question

If  y `sqrt(x^2 + 1) = log sqrt(x^2 + 1) - x`, show that `(x^2 + 1)(dy)/(dx) + xy + 1 = 0.`

Answer 1

Given:

y `sqrt(x^2 + 1) = log (sqrt(x^2 + 1) - x)`

Differentiate the Left-Hand Side:

Using the product rule (uv)′ = u′v + uv′:

Let u = y and v = `sqrt(x^2 + 1)`

`d/dx (y sqrt(x^2 + 1)) = (dy)/(dx) . sqrt(x^2 + 1) + y . d/dx (sqrt(x^2 + 1))`

= `sqrt(x^2 + 1) (dy)/(dx)  + y . (1/(2sqrt(x^2 + 1)) . 2x)`

= `sqrt(x^2 + 1) (dy)/(dx) + (xy)/sqrt(x^2 + 1)`   ...(i)

Differentiate the Right-Hand Side:

Using the chain rule for log(u):

`d/dx [log (sqrt(x^2 + 1) - x)] = 1/(sqrt(x^2 + 1) - x) . d/dx (sqrt(x^2 + 1) - x)`

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

Take the LCM in the bracket:

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

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

= `-1/(sqrt(x^2 + 1)`   ...(ii)

Equate LHS and RHS

`sqrt(x^2 + 1) (dy)/(dx) + (xy)/sqrt(x^2 + 1) = -1/(sqrt(x^2 + 1)`

Multiply the entire equation by `sqrt(x^2 + 1)` to clear the denominators:

`(sqrt(x^2 + 1) . sqrt(x^2 + 1)) (dy)/(dx) + xy = -1`

`(x^2 + 1) (dy)/(dx) + xy = -1`

`(x^2 + 1) (dy)/(dx) + xy + 1 = 0`

Hence proved