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Question
If y = xx, prove that `(d^2y)/(dx^2)−1/y(dy/dx)^2−y/x=0.`
Solution
y = xx
Applying logarithm,
log y = x log x
`1/y dy/dx=logx+xxx1/x=1+logx`
`dy/dx=x^x[1+logx]`
`(d^2y)/dx^2=(d(x^2))/dx(1+logx)+x^x[d/dx(1+logx)]`
`=x^x(1+logx)(1+logx)+x^x[1/x]`
`=x^x(1+logx)^2+x^(x-1)`
`(d^2y)/dx^2-1/y(dy/dx)^2-y/x=x^x(1+logx)^2+x^(x-1)-1/(x^2)(x^x(1+logx)^2)-x^2/x`
= xx(1+log x)2 + xx−1 −xx(1+log x)2 − xx−1
= 0
Hence proved.
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\[\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \] ?
If logy = tan–1 x, then show that `(1+x^2) (d^2y)/(dx^2) + (2x - 1) dy/dx = 0 .`
