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Question Number 92838 by i jagooll last updated on 09/May/20

$$(\mathrm{x}+\mathrm{y})\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{x}}^2+\mathrm{xy}+\mathrm{x}+1$$

Commented by john santu last updated on 09/May/20

$$(\mathrm{x}+\mathrm{y})\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}(\mathrm{x}+\mathrm{y})+\mathrm{x}+1$$$$\mathrm{set}\mathrm{x}+\mathrm{y}=\mathrm{v}\Rightarrow 1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}-1$$$$\Rightarrow \mathrm{v}[\frac{\mathrm{dv}}{\mathrm{dx}}-1]=\mathrm{xv}+\mathrm{x}+1$$$$\mathrm{v}\frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}(\mathrm{x}+1)+\mathrm{x}+1$$$$\mathrm{v}\frac{\mathrm{dv}}{\mathrm{dx}}=(\mathrm{x}+1)(\mathrm{v}+1)$$$$\frac{\mathrm{v}\mathrm{dv}}{\mathrm{v}+1}=(\mathrm{x}+1)\mathrm{dx}$$$$\int \frac{\mathrm{v}+1-1}{\mathrm{v}+1}\mathrm{dv}=\frac{1}{2}{(\mathrm{x}+1)}^2+\mathrm{c}$$$$\mathrm{v}-\ln (\mathrm{v}+1)=\frac{1}{2}{(\mathrm{x}+1)}^2+\mathrm{c}$$$$\mathrm{x}+\mathrm{y}-\ln (\mathrm{x}+\mathrm{y}+1)=\frac{1}{2}{(\mathrm{x}+1)}^2+\mathrm{c}$$$$\mathrm{y}-\ln (\mathrm{x}+\mathrm{y}+1)=\frac{1}{2}{(\mathrm{x}+1)}^2-\mathrm{x}+\mathrm{c}$$

Commented by i jagooll last updated on 09/May/20

cool man sir ��

Commented by niroj last updated on 09/May/20

great dear����

Commented by john santu last updated on 09/May/20

ha ha ha ������

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