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If y = cos (lnx). Show that x2y2 + xy1 + y = 0. |
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Answer» y = cos(lnx) \(y_1 = \frac{dy}{dx} = \frac{-sin(lnx)}{x}\) \(xy_1 = - sin(lnx)\) \(y _2 = \frac{dy}{dx^2} = \frac{- cos(lnx)}{x^2}+ \frac{sin(lnx)}{x^2}\) \(\therefore x^2y^2 = sin(lnx) - cos(lnx)\) Now, \(x^2y_2 + xy_1 + y = sin(lnx) - cos(lnx) - sin(lnx) + cos(lnx)\) = 0 Hence Proved. |
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