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प्रश्न
If y = 3 cos (log x) + 4 sin (log x), show that x2y2 + xy1 + y = 0.
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उत्तर
Given, y = 3 cos (log x) + 4 sin (log x) ...(1)
Differentiating both sides with respect to x,
`dy/dx = 3 d/dx cos (log x) + 4 d/dx sin (log x)`
= `3 [- sin (log x)] d/dx (log x) + 4 cos (log x) d/dx (log x)`
= `-3 sin (log x) xx 1/x + 4 cos (log x) xx 1/x`
Multiplying both sides by x,
`x dy/dx` = −3 sin (log x) + 4 cos (log x)
Differentiating both sides again with respect to x,
`x d/dx (dy/dx) + dy/dx * d/dx (x) = - 3 cos (log x) d/dx (log x) - 4 sin (log x) d/dx (log x)`
`x (d^2 y)/dx^2 + 1 * dy/dx = - 3 cos (log x) 1/x - 4 sin (log x) * 1/x`
Multiplying both sides by x,
`x^2 (d^2 y)/dx^2 + x dy/dx` = −[3 cos (log x) + 4 sin (log x)]
`x^2 (d^2 y)/dx^2 + x dy/dx` = −y ...[From equation (1)]
`=> x^2 (d^2 y)/dx^2 + x dy/dx + y = 0`
Or, x2y2 + xy1 + y = 0
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