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Question Number 84212 by jagoll last updated on 10/Mar/20

$${\mathrm{y}}^{\prime }={(2\mathrm{x}+3\mathrm{y}+1)}^2$$$$\mathrm{find}\mathrm{the}\mathrm{solution}$$

Commented by niroj last updated on 10/Mar/20

$$\frac{\mathrm{dy}}{\mathrm{dx}}={(2\mathrm{x}+3\mathrm{y}+1)}^2$$$$\mathrm{put},2\mathrm{x}+3\mathrm{y}+1=\mathrm{v}$$$$2+3\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}$$$$3\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}-2$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{3}(\frac{\mathrm{dv}}{\mathrm{dx}}-2)$$$$\frac{1}{3}(\frac{\mathrm{dv}}{\mathrm{dx}}-2)={\mathrm{v}}^2$$$$\frac{\mathrm{dv}}{\mathrm{dx}}-2={3\mathrm{v}}^2$$$$\frac{\mathrm{dv}}{\mathrm{dx}}={3\mathrm{v}}^2+2$$$$\frac{1}{{3\mathrm{v}}^2+2}\mathrm{dv}=\mathrm{dx}$$$$\mathrm{integrating}\mathrm{both}\mathrm{side}.$$$$\frac{1}{3}\int \frac{1}{{\left(\mathrm{v}\right)}^2+{\left(\sqrt{\frac{2}{3}}\right)}^2}\mathrm{dv}=\int \mathrm{dx}$$$$\frac{1}{3}.\frac{1}{\sqrt{\frac{2}{3}}}{\tan }^{-1}\frac{\mathrm{v}}{\sqrt{\frac{2}{3}}}=\mathrm{x}+\mathrm{c}$$$$\frac{1}{3}.\frac{\sqrt{3}}{\sqrt{2}}.{\tan }^{-1}\frac{\mathrm{v}\sqrt{3}}{\sqrt{2}}=\mathrm{x}+\mathrm{c}$$$$\frac{1}{\sqrt{6}}{\tan }^{-1}\mathrm{v}\frac{\sqrt{6}}{2}=\mathrm{x}+\mathrm{c}$$$${\tan }^{-1}\mathrm{v}.\frac{\sqrt{6}}{2}=\sqrt{6}(\mathrm{x}+\mathrm{c})$$$$\frac{\mathrm{v}\sqrt{6}}{2}=\tan \sqrt{6}(\mathrm{x}+\mathrm{c})$$$$(2\mathrm{x}+3\mathrm{y}+1)\sqrt{6}=2\tan \sqrt{6}(\mathrm{x}+\mathrm{c})$$$$2\mathrm{x}+3\mathrm{y}+1=\frac{\sqrt{6}}{3}\tan \sqrt{6}(\mathrm{x}+\mathrm{c}).$$$$$$$$$$$$$$

Answered by mr W last updated on 10/Mar/20

$$lett=2x+3y+1$$$$\frac{dt}{dx}=2+3\frac{dy}{dx}$$$$y^{\prime }=\frac{dy}{dx}=\frac{1}{3}(\frac{dt}{dx}-2)$$$$\frac{1}{3}(\frac{dt}{dx}-2)=t^2$$$$\frac{dt}{dx}=3t^2+2$$$$\frac{dt}{3t^2+2}=dx$$$$\int \frac{dt}{3t^2+2}=\int dx$$$$\frac{{\tan }^{-1}\frac{3t}{\sqrt{6}}}{\sqrt{6}}=x+C$$$$\Rightarrow t=\frac{\sqrt{6}}{3}\tan (\sqrt{6}x+C)$$$$\Rightarrow 2x+3y+1=\frac{\sqrt{6}}{3}\tan (\sqrt{6}x+C)$$$$\Rightarrow y=\frac{1}{3}[\frac{\sqrt{6}}{3}\tan (\sqrt{6}x+C)-2x-1]$$

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