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Question Number 139426 by I want to learn more last updated on 26/Apr/21

Answered by physicstutes last updated on 26/Apr/21

$$1.x\frac{dy}{dx}=x^2+y$$$$y=x^2+cx\Rightarrow \frac{dy}{dx}=2x+c$$$$\mathrm{hence}x(2x+c)=x^2+(x^2+cx)$$$$2x^2+c=2x^2+c\mathrm{hence}!$$$$2.(x+1)\frac{dy}{dx}=x(y^2+1)$$$$\Rightarrow \frac{dy}{dx}=\frac{x(y^2+1)}{x+1}\mathrm{or}\frac{dy}{y^2+1}=\frac{x}{x+1}dx$$$${\tan }^{-1}y=x-\ln \mid x+1\mid +c$$$$\Rightarrow y=\tan (x-\ln \mid x+1\mid +c)$$

Commented by I want to learn more last updated on 28/Apr/21

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.$$

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