The solution of frac{d y}{d x}=frac{y^2}{x y- | vedclass

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(B) Given, $\frac{d y}{d x}=\frac{y^2}{x y-x^2}$.This is a homogeneous differential equation. Let $y = vx$, then $\frac{d y}{d x} = v + x \frac{d v}{d

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$e^{y / x}=k y$

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(B) Given, $\frac{d y}{d x}=\frac{y^2}{x y-x^2}$.This is a homogeneous differential equation. Let $y = vx$, then $\frac{d y}{d x} = v + x \frac{d v}{d x}$.Substituting these into the equation:$v + x \frac{d v}{d x} = \frac{(vx)^2}{x(vx) - x^2} = \frac{v^2 x^2}{x^2(v - 1)} = \frac{v^2}{v - 1}$.$x \frac{d v}{d x} = \frac{v^2}{v - 1} - v = \frac{v^2 - v^2 + v}{v - 1} = \frac{v}{v - 1}$.Separating the variables:$\frac{v - 1}{v} d v = \frac{d x}{x}$.$(1 - \frac{1}{v}) d v = \frac{d x}{x}$.Integrating both sides:$\int (1 - \frac{1}{v}) d v = \int \frac{d x}{x}$.$v - \ln |v| = \ln |x| + C$.Substituting $v = \frac{y}{x}$:$\frac{y}{x} - \ln |\frac{y}{x}| = \ln |x| + C$.$\frac{y}{x} = \ln |\frac{y}{x}| + \ln |x| + C = \ln |y| + C$.$e^{y/x} = e^{\ln |y| + C} = e^C \cdot y = ky$ (where $k = e^C$ is a constant).Thus, $e^{y/x} = ky$.

The solution of $\frac{d y}{d x}=\frac{y^2}{x y-x^2}$ is

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