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