Find the general solution of the differential equation: (xy - x^2) dy…

The correct answer is Option (1) → $\frac{y}{x} - \ln|y| = C$ ##

Given differential equation is

$\frac{dy}{dx} = \frac{y^2}{xy - x^2} \quad \dots(i)$

Let $F(x, y) = \frac{y^2}{xy - x^2}$

Now, on replacing $x$ by $\lambda x$ and $y$ by $\lambda y$, we get

$F(\lambda x, \lambda y) = \frac{\lambda^2 y^2}{\lambda^2(xy - x^2)} = \lambda^0 \frac{y^2}{xy - x^2} = \lambda^0 F(x, y)$

Thus, the given differential equation is a homogeneous differential equation.

Now, to solve it, put $y = vx$

or $\frac{dy}{dx} = v + x \frac{dv}{dx}$

From Eq. (i), we get

$v + x \frac{dv}{dx} = \frac{v^2 x^2}{vx^2 - x^2} = \frac{v^2}{v - 1}$

$\text{or } x \frac{dv}{dx} = \frac{v^2}{v - 1} - v = \frac{v^2 - v^2 + v}{v - 1}$

$\text{or } x \frac{dv}{dx} = \frac{v}{v - 1}$

$\text{or } \frac{v - 1}{v} dv = \frac{dx}{x}$

On integrating both sides, we get

$\int \left( 1 - \frac{1}{v} \right) dv = \int \frac{dx}{x}$

$\text{or } v - \log |v| = \log |x| + C$

$\text{or } \frac{y}{x} - \log \left| \frac{y}{x} \right| = \log |x| + C \quad \left[ \text{put } v = \frac{y}{x} \right]$

$\text{or } \frac{y}{x} - \log |y| + \log |x| = \log |x| + C \quad \left[ ∵\log \left( \frac{m}{n} \right) = \log m - \log n \right]$

$∴\frac{y}{x} - \log |y| = C$ which is the required solution.