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Question Number 100684 by bemath last updated on 28/Jun/20

$$({\mathrm{x}}^2+\mathrm{xy})\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{xy}+{\mathrm{y}}^2$$

Answered by bobhans last updated on 28/Jun/20

$$\mathrm{set}\mathrm{y}=\mathrm{sx}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{s}+\mathrm{x}\frac{\mathrm{ds}}{\mathrm{dx}}$$$$({\mathrm{x}}^2+{\mathrm{x}}^2\mathrm{s})(\mathrm{s}+\mathrm{x}\frac{\mathrm{ds}}{\mathrm{dx}})={\mathrm{x}}^2\mathrm{s}+{\mathrm{x}}^2{\mathrm{s}}^2$$$$(1+\mathrm{s})(\mathrm{s}+\mathrm{x}\frac{\mathrm{ds}}{\mathrm{dx}})=\mathrm{s}(\mathrm{s}+1)$$$$\mathrm{s}+\mathrm{x}\frac{\mathrm{ds}}{\mathrm{dx}}=\mathrm{s}\Rightarrow \mathrm{x}\frac{\mathrm{ds}}{\mathrm{dx}}=0$$$$\frac{\mathrm{ds}}{\mathrm{dx}}=0\Rightarrow \mathrm{s}=0+\mathrm{C};\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{C},\mathrm{x}\ne 0$$$$\mathrm{y}=\mathrm{Cx}$$

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