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Question Number 90810 by john santu last updated on 26/Apr/20

$$\frac{dy}{dx}=\frac{2x^2+y^2}{3x^2+2xy}$$

Commented by peter frank last updated on 26/Apr/20

$$helpQn90812$$

Commented by john santu last updated on 26/Apr/20

$$Q=\frac{y}{x}\Rightarrow \frac{dQ}{dx}.x+Q=\frac{2+Q^2}{3+2Q}$$$$x.\frac{dQ}{dx}=\frac{2+Q^2-3Q-2Q^2}{3+2Q}$$$$x.\frac{dQ}{dx}=\frac{-Q^2-3Q+2}{3+2Q}$$$$\frac{(3+2Q)dQ}{-Q^2-3Q+2}=\frac{dx}{x}$$$$\int \frac{d(Q^2+3Q-2)}{(Q^2+3Q-2)}=-\ln x+C$$$$\ln ({\mathrm{Q}}^2+3Q-2)+\ln x=C$$$$\ln \left(x(\frac{y^2}{x^2}+\frac{3y}{x}-2)\right)=C$$$$\ln (\frac{y^2}{x}+3y-2x)=C$$$$\Rightarrow \frac{y^2}{x}+3y-2x=e^C=K$$$$$$

Answered by Joel578 last updated on 26/Apr/20

$$\frac{dy}{dx}=\frac{2+{\left(\frac{y}{x}\right)}^2}{3+2\left(\frac{y}{x}\right)}$$$$\mathrm{Let}u=\frac{y}{x}\Rightarrow y=ux\Rightarrow \frac{dy}{dx}=x\frac{du}{dx}+u$$$$$$$$x\frac{du}{dx}+u=\frac{2+u^2}{3+2u}$$$$\Rightarrow x\frac{du}{dx}=\frac{2-3u-u^2}{3+2u}$$$$\Rightarrow -\frac{2u+3}{u^2+3u-2}du=\frac{1}{x}dx$$$$\Rightarrow -\ln (u^2+3u-2)=\ln x+C_1$$$$\Rightarrow \ln (u^2+3u-2)=C_2-\ln x$$$$\Rightarrow u^2+3u-2=\frac{C_3}{x}$$$$\Rightarrow \frac{y^2+3xy-2x^2}{x^2}=\frac{C_3}{x}$$$$\Rightarrow y^2+3xy-2x^2=C_{3}x$$

Commented by peter frank last updated on 26/Apr/20

$$thanks$$

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