a) y=x
b) y+x=2
c) y=-x
d) y-x=2
Particular solution of the differential equation \(\frac{dy}{dx} = \frac{y^2-2xy-x^2}{y^2+2xy-x^2}\) given y=-1 at x=1.
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answerhappygod
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Particular solution of the differential equation \(\frac{dy}{dx} = \frac{y^2-2xy-x^2}{y^2+2xy-x^2}\) given y=-1 at x=1.
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