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प्रश्न
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
y = x sin x : xy' = `y + x sqrt (x^2 - y^2)` (x ≠ 0 and x > y or x < -y)
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उत्तर
y = x sin x …(i)
y’ = x cos x + sin x
and cos x + `sqrt(1 - sin^2 x) = sqrt((1 - y^2)/x^2)`
`= sqrt(x^2 - y^2)/x`
Hence, y' = x. `sqrt(x^2 - y^2)/x +y/x`
`xy' = x sqrt(x^2 - y^2) + y`
The given function is the solution of the given differential equation.
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