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Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0
y = `a + b/x`
`(dy)/(dx) = square`
`(d^2y)/(dx^2) = square`
Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`
= `x square + 2 square`
= `square`
Hence y = `a + b/x` is solution of `square`
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Solution
`x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0
y = `a + b/x`
`("d"y)/("d"x)` = `bb((-b)/x^2)`
`(d^2y)/(dx^2)` = `bb((2b)/x^3)`
Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`
= `bb(x (2b)/x^3 + 2(-2)/x^2)`
= 0
Hence y = `a + b/x` is solution of `bb(x(d^2y)/(dx^2) + 2(dy)/(dx) = 0)`
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