The solution of the differential equation $(x^2 + xy)dy = (x^2 + y^2)dx$ is

$\frac{y}{x} + \log_e\left|\frac{(y-x)^2}{x}\right|= c$, ($c$ is an arbitrary constant)

The correct answer is Option (2) → $\frac{y}{x} + \log_e\left|\frac{(y-x)^2}{x}\right|= c$, ($c$ is an arbitrary constant)

Given differential equation: $(x^2 + xy)dy = (x^2 + y^2)dx$

Rewriting:

$\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy}$

Divide numerator and denominator by $x^2$:

$= \frac{1 + \left(\frac{y}{x}\right)^2}{1 + \frac{y}{x}}$

Put $v = \frac{y}{x} \Rightarrow y = vx$

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

Now substitute:

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

$\Rightarrow x\frac{dv}{dx} = \frac{1 + v^2}{1 + v} - v = \frac{1 + v^2 - v(1 + v)}{1 + v}$

$= \frac{1 + v^2 - v - v^2}{1 + v} = \frac{1 - v}{1 + v}$

Now separate and integrate:

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

Integrate both sides:

$\int \frac{1 + v}{1 - v} dv = \int \frac{dx}{x}$

Use: $\frac{1 + v}{1 - v} = \frac{(1 - v) + 2v}{1 - v} = 1 + \frac{2v}{1 - v}$

Then:

$\int \left(1 + \frac{2v}{1 - v} \right) dv = \ln|x| + c$

$= v - 2\ln|1 - v| = \ln|x| + c$

Substitute $v = \frac{y}{x}$:

$\frac{y}{x} - 2\ln\left|1 - \frac{y}{x}\right| = \ln|x| + c$

Multiply both sides by $1$ and adjust into answer form:

$\frac{y}{x} + \ln\left|\frac{(y - x)^2}{x}\right| = c$