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Question
If `y = (x + sqrt(a^2 + x^2))^m`, prove that `(a^2 + x^2)(d^2y)/(dx^2) + xdy/dx - m^2y = 0`
Solution 1
`y = [x + sqrt(a^2 + x^2)]^m`
`dy/dx = (m[x + sqrt(a^2 + x^2)]^m)/(x + sqrt(a^2 + x^2))*[1 + (1*2x)/(2sqrt(a^2 + x^2))]`
= `my 1/sqrt(a^2 + x^2)`
`sqrt(a^2 + x^2)dy/dx = my`
`sqrt(a^2 + x^2)(d^2y)/(dx^2) + dy/dx * 1/2 * (2x)/sqrt(a^2 + x^2) = mdy/dx`
`(a^2 + x^2)(d^2y)/(dx^2) + xdy/dx - msqrt(a^2 + x^2)dy/dx = 0`
`(a^2 + x^2)(d^2y)/(dx^2) + xdy/dx - m^2y = 0`
Solution 2
`y = [x + sqrt(a^2 + x^2)]^m`
`dy/dx = (m[x + sqrt(a^2 + x^2)]^m)/(x + sqrt(a^2 + x^2)]*[1 + (1.2x)/(2sqrt(a^2 + x^2))]`
= `my 1/sqrt(a^2 + x^2)`
`sqrt(a^2 + x^2)dy/dx = my`
Squaring both sides we get,
`(a^2 + x^2)(dy/dx)^2 = m^2y^2`
Differentiating w.r.t ‘ЁЭСе’,
`2x(dy/dx)^2 + (a^2 + x^2)2dy/dx*(d^2y)/(dx^2) = 2m^2ydy/dx`
`\implies (a^2 + x^2)(d^2y)/(dx^2) + xdy/dx - m^2y = 0`
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