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The general solution of the differential equation x2dydx+y2=xyx^2\dfrac{dy}{dx} + y^2 = xy is: (take log⁡ex=log⁡x\log_e x = \log x)

Solution

✅ Correct Option: 1

The equation dydx=xy−y2x2\frac{dy}{dx} = \frac{xy-y^2}{x^2} is homogeneous. Putting y=vxy = vx gives xdvdx=−v2x\frac{dv}{dx} = -v^2, so dvv2=−dxx\frac{dv}{v^2} = -\frac{dx}{x} and 1v=log⁡x+C\frac{1}{v} = \log x + C. Since 1v=xy\frac{1}{v} = \frac{x}{y}, the solution is xy=log⁡x+C\frac{x}{y} = \log x + C.

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