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Question Number 114978 by srijachakraborty last updated on 22/Sep/20

$$$$$$Solve$$$$x^{2}dy+y(x+y)dx=0$$

Answered by soumyasaha last updated on 22/Sep/20

Answered by Dwaipayan Shikari last updated on 22/Sep/20

$$x^2\frac{dy}{dx}+y(x+y)=0$$$$x^2\frac{dy}{dx}+vx(x+vx)=0y=vx\frac{dy}{dx}=v+x\frac{dv}{dx}$$$$\frac{dy}{dx}=-v(1+v)$$$$x\frac{dv}{dx}=-2v-v^2$$$$\int \frac{dv}{-2v-v^2}=\int \frac{dx}{x}$$$$-\int \frac{dv}{v(v+2)}=\int \frac{dx}{x}$$$$-\frac{1}{2}\int \frac{1}{v}-\frac{1}{v+2}dv=log\left(x\right)+C$$$$\frac{1}{2}log\left(\frac{v+2}{v}\right)=log\left(Cx\right)$$$$1+\frac{2x}{y}={Cx}^2$$$$y=\frac{2x}{1-{Cx}^2}$$

Commented by srijachakraborty last updated on 22/Sep/20

thank you sir

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