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प्रश्न
If y = 3 cos(log x) + 4 sin(log x), show that `x^2 (d^2y)/(dx^2) + x dy/dx + y = 0`
योग
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उत्तर
y = 3 cos (log x) + 4 sin (log x)
On differentiating both sides w.r.t. x, we get
`dy/dx = -3sin (log x) xx 1/x + 4 cos(log x) xx 1/x`
`x dy/dx = -3sin (log x) + 4 cos (log x)`
Again differentiating both sides w.r.t x, we get
`x (d^2y)/(dx^2) + (dy/dx) = -3 cos (log x) xx 1/x - 4 sin(log x) xx 1/x`
`\implies x^2 (d^2y)/(dx^2) + x(dy/dx) = -[3 cos (log x) + 4 sin (log x)]`
`\implies x^2 (d^2y)/(dx^2) + x(dy/dx) = -y`
`\implies x^2 (d^2y)/(dx^2) + x dy/dx + y = 0`
Hence proved
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