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Question Number 87504 by M±th+et£s last updated on 04/Apr/20

$$solve$$$$\frac{dy}{dx}=2{\left(\frac{2+y}{1+x+y}\right)}^2$$

Answered by TANMAY PANACEA. last updated on 04/Apr/20

$$ithink$$$$\frac{dy}{dx}=2{\left(\frac{x+y}{1+x+y}\right)}^2$$$$t=x+y\to \frac{dt}{dx}=1+\frac{dy}{dx}$$$$\frac{dt}{dx}-1=2{\left(\frac{t}{1+t}\right)}^2$$$$\frac{dt}{dx}=1+\frac{2t^2}{1+2t+t^2}$$$$\frac{1+2t+t^2}{3t^2+2t+1}dt=dx$$$$\frac{1}{3}\int \frac{3t^2+2t+1+4t+2}{3t^2+2t+1}dt=\int dx$$$$\int dt+\frac{2}{3}\int \frac{6t+2+1}{3t^2+2t+1}dt=3\int dx$$$$\int dt+\frac{2}{3}\int \frac{d(3t^2+2t+1)}{3t^2+2t+1}+\frac{2}{9}\int \frac{dt}{t^2+2\times t\times \frac{1}{3}+\frac{1}{9}+\frac{1}{3}-\frac{1}{9}}=3\int dx$$$$\int dt+\frac{2}{3}\int \frac{d(3t^2+2t+1)}{3t^2+2t+1}+\frac{2}{9}\int \frac{dt}{{(t+\frac{1}{3})}^2+{\left(\frac{\sqrt{2}}{3}\right)}^2}=3\int dx$$$$t+\frac{2}{3}ln(3t^2+2t+1)+\frac{2}{9}\times \frac{3}{\sqrt{2}}{tan}^{-1}\left(\frac{t+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right)=3x+c$$$$nowplsputt=x+y$$$$$$$$$$

Commented by M±th+et£s last updated on 04/Apr/20

$$thankyousirbut2\left(\frac{2+y}{1+x+y}\right)$$

Commented by TANMAY PANACEA. last updated on 04/Apr/20

$$okishalltry...$$

Commented by M±th+et£s last updated on 04/Apr/20

$$thankyousir$$

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