Question

If `x = sin(1/2 log y)` show that (1 − x²)y₂ − xy₁ − a²y = 0.

Answer 1

Given,

x = sin`(1/a log y)`

`(logy) = asin^-1 x`

y = `e^(asin^-1 x)`       ...(i)

To prove: `(1 - x^2)y_2 - xy_1 - a^2 = 0`

First, determine the second-order derivative, as we have noticed one in the expression that needs to be demonstrated.

Lets find `(d^2y)/(dx^2)`

As, `(d^2y)/(dx^2) = d/dx ((dy)/(dx))`

So, lets first find dy/dx

∵ `y = e^(asin^-1 x)`

Let t = `asin^-1 x => (dt)/(dx) = a/(sqrt((1 - x^2)))[d/dx sin^-1 x = 1/(sqrt((1 - x^2)))]`

And y = e^t

`(dy)/(dx) = e^t a/(sqrt((1 - x^2))) = (ae^(asin^-1 x))/sqrt((1 - x^2))`    ...(ii)

Again, differentiating with respect to x applying product rule:

`(d^2y)/(dx^2) = ae^(a sin^-1 x) d/dx (1/sqrt((1 - x^2))) + a/(sqrt((1 - x^2))) d/dx e^(asin^-1 x)`

Using chain rule and equation 2:

`(d^2y)/(dx^2) = -(ae^(asin-1 x))/(2(1 - x^2)sqrt((1 - x^2)))(-2x) + (a^2e^(asin^-1 x))/((1 - x^2)) ["Using" d/dx (x^n) = nx^(n-1) d/dx sin^-1 x = 1/(sqrt((1 - x^2)))]`

`(d^2y)/(dx^2) = (Xae^(asin^-1 x))/((1 - x^2)sqrt(1 -x^2)) + (a^2e^(asin^-1 x))/((1 - x^2))`

`(1 - x^2) (d^2y)/(dx^2) = a^2e^(asin^-1 x) + (Xae^(asin^-1 x))/(sqrt(1 - x^2))`

Using eq (i) and (ii):

`(1 - x^2) (d^2y)/(dx^2) - a^2y + x dy/dx`

∴ (1 − x²)y₂ − xy₁ − a²y = 0    ...proved