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Question Number 147621 by mathdanisur last updated on 22/Jul/21

$$Findthegeneralsolutionfor:$$$$\frac{dy}{dx}={(3x+2y+1)}^2$$

Answered by gsk2684 last updated on 22/Jul/21

$$putu=3x+2y+1\Rightarrow \frac{du}{dx}=3+2\frac{dy}{dx}$$$$\frac{du}{dx}=3+2u^2$$$$\frac{du}{{\left(\sqrt{2}u\right)}^2+{\left(\sqrt{3}\right)}^2}=dx$$$$\int \frac{du}{{\left(\sqrt{2}u\right)}^2+{\left(\sqrt{3}\right)}^2}=\int dx+c$$$$\frac{1}{\sqrt{2}}\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{\sqrt{2}u}{\sqrt{3}}=x+c$$$$\frac{1}{\sqrt{6}}{\tan }^{-1}\left(\sqrt{\frac{2}{3}}(3x+2y+1)\right)=x+c$$

Commented by puissant last updated on 22/Jul/21

$$\mathrm{jolie}\mathrm{prof}..\mathrm{Nice}$$

Commented by mathdanisur last updated on 22/Jul/21

$$ThankyouSircool$$

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