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Question Number 127505 by bramlexs22 last updated on 30/Dec/20

$$({\mathrm{y}}^2+{2\mathrm{x}}^2\mathrm{y})\mathrm{dx}+({2\mathrm{x}}^3-\mathrm{xy})\mathrm{dy}=0$$

Answered by liberty last updated on 30/Dec/20

$$\frac{dy}{dx}=\frac{y^2+2x^{2}y}{xy-2x^3};lety={vx}^2\Rightarrow \frac{dy}{dx}=2vx+x^2\frac{dv}{dx}$$$$wefind2vx+x^2\frac{dv}{dx}=\frac{x^4(v^2+2v)}{x^3(v-2)}$$$$2vx+x^2\frac{dv}{dx}=\frac{x(v^2+2v)}{v-2}$$$$2v+x\frac{dv}{dx}=\frac{v^2+2v}{v-2};x\frac{dv}{dx}=\frac{v^2+2v-2v^2+4v}{v-2}$$$$\frac{v-2}{v^2-6v}dv+\frac{dx}{x}=0\mathrm{\_};[\frac{v-2}{v(v-6)}=\frac{Q}{v}+\frac{R}{v-6}]$$$$\int [\frac{1}{3v}+\frac{2}{3(v-6)}]dv+\int \frac{dx}{x}=C$$$$\frac{1}{3}(\ln v+\ln {(v-6)}^2)+\ln x=C$$$$\ln \left({(v-6)}^{2}v\right)=\ln {\left(\frac{C_1}{x}\right)}^3$$$${\left(\frac{y-6x^2}{x^2}\right)}^2\left(\frac{y}{x^2}\right)=\frac{\lambda }{x^4}\Rightarrow y{(y-6x^2)}^2=\lambda x^2$$$$$$

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