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प्रश्न
If y=2 cos(logx)+3 sin(logx), prove that `x^2(d^2y)/(dx2)+x dy/dx+y=0`
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उत्तर
y=2 cos(logx)+3 sin(logx)
Differentiating both sides with respect to x, we get
`dy/dx=2xxd/dx cos(logx)+3xx d/dxsin(log x)`
`=-2sin(logx)xx1/x+3 cos(logx)xx1/x`
`=>x dy/dx=-2 sin(logx)+3 cos(logx)`
Again, differentiating both sides with respect to x, we get
`x (d^2y)/(dx^2)+dy/dx=-2cos(logx)xx1/x-3 sin(logx)xx1/x`
`x^2 (d^2y)/(dx^2)+xdy/dx=-[2 cos(logx)+3sin(logx)]`
`x^2 (d^2y)/(dx^2)+xdy/dx=-y`
`x^2 (d^2y)/(dx^2)+xdy/dx+y=0`
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