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Question Number 98808 by bemath last updated on 16/Jun/20

Answered by john santu last updated on 16/Jun/20

$$y=wx\Rightarrow \frac{dy}{dx}=w+x\frac{dw}{dx}$$$$\iff w+x\frac{dw}{dx}=\frac{{wx}^2-w^2{x}^2}{x^2+{wx}^2}$$$$x\frac{dw}{dx}=\frac{w-w^2}{1+w}-w=\frac{-2w^2}{1+w}$$$$\frac{(1+w)dw}{w^2}=-2\frac{dx}{x}$$$$\int (w^{-2}+w^{-1})dw=-2\int \frac{dx}{x}$$$$-\frac{1}{w}+\ln \mid w\mid =-2\ln \mid x\mid +\mathrm{c}$$$$\ln \mid \frac{w}{x^2}\mid =\frac{1}{w}+c$$$$\frac{y}{x^3}=e^{\frac{1}{x}+c}\iff \mathrm{y}=x^3\left({Ce}^{\frac{1}{x}}\right)$$

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