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Proving a Differential Equation

If y = x / (sqrt (x^2 + 2))

Show that 3 y^2 dy/dx + x d2y/dx2 = 0

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y = x /{(x^2+2)^(1/2)} ---(1)

dy/dx = [ {(x^2+2)^(1/2)} - x * x/{(x^2+2)^(1/2)} ] / (x^2+2)

= 2/ {(x^2+2)^(3/2)} --- (2)

d^y/dx^2 = 2 * [ (-3/2) * {(x^2+2)^(- 5/2)} 2x ] = -6x / {(x^2+2)^(5/2)} ---(3)

Now substitute (1) thro' (3) into 3y^2*dy/dx + x*d^2y/dx^2 and we get,

3y^2*dy/dx + x*d^2y/dx^2 = 0

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Solution Summary

A differential equation is proven.

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