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Question Number 122006 by bemath last updated on 13/Nov/20

$$\frac{dy}{dx}=\frac{x-y}{x+y}?$$

Answered by bobhans last updated on 13/Nov/20

$$lety=zx\Rightarrow \frac{dy}{dx}=z+x\frac{dz}{dx}$$$$\iff z+x\frac{dz}{dx}=\frac{x-zx}{x+zx}=\frac{1-z}{1+z}$$$$\iff x\frac{dz}{dx}=\frac{1-z}{1+z}-z=\frac{1-2z-z^2}{1+z}$$$$\iff \frac{(1+z)dz}{z^2+2z-1}=-\frac{dx}{x}$$$$\iff \int \frac{d(z^2+2z-1)}{z^2+2z-1}=-2\int \frac{dx}{x}$$$$\iff \ell n\mid z^2+2z-1\mid =\ell n\mid \frac{C}{x^2}\mid$$$$\iff {(z+1)}^2=\frac{C+2x^2}{x^2}$$$$\iff {\left(\frac{y+x}{x}\right)}^2=\frac{C+2x^2}{x^2}$$$$\iff y+x=\pm \sqrt{C+2x^2}$$$$\iff y=-x\pm \sqrt{C+2x^2}.⊝$$

Answered by TANMAY PANACEA last updated on 13/Nov/20

$$xdx-ydx=xdy+ydy$$$$xdy+ydx+ydy-xdx=0=dC$$$$d\left(xy\right)+d\left(\frac{y^2}{2}\right)-d\left(\frac{x^2}{2}\right)=dC$$$$xy+\frac{y^2-x^2}{2}=C$$

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