Q.Solve the following differential equation $(x^2 - yx^2)\,dy + (y^2 + xy^2)\,dx = 0$.
Separating the variables, the given equation can be written as:
$\square \, dy + \square \, dx = 0$
∴ $\left(y^{-2} - \frac{1}{y}\right)dy + \left(x^{-2} + \frac{1}{x}\right)dx = 0$
$\square \, dy - \frac{1}{y} \, dy + x^{-2}\,dx + \square \, dx = 0$
Integrating, we get
$\int y^{-2}\,dy - \int \frac{1}{y}\,dy + \int x^{-2}\,dx + \int \frac{1}{x}\,dx = 0$
∴ $\frac{y^{-1}}{-1} - \square + \frac{x^{-1}}{-1} + \square = c$
$-\frac{1}{y} - \frac{1}{x} + \log x - \log y = c$
$\log x - \log y = \square + c$
is the required solution.

Factor the equation, divide by $x^2y^2$ to separate the variables, integrate each term using standard forms, and collect the log terms to get $\log\dfrac{x}{y} = \dfrac{1}{x}+\dfrac{1}{y}+c$.

The given equation is

$(x^2 - yx^2)\,dy + (y^2 + xy^2)\,dx = 0.$

Step 1 — factor each coefficient. Take out common factors: $x^2 - yx^2 = x^2(1-y)$ and $y^2 + xy^2 = y^2(1+x)$, so

$x^2(1-y)\,dy + y^2(1+x)\,dx = 0.$

Step 2 — separate the variables. Divide throughout by $x^2 y^2$ (valid for $x\neq 0,\ y\neq 0$):

$\frac{1-y}{y^2}\,dy + \frac{1+x}{x^2}\,dx = 0.$

Splitting each fraction, $\dfrac{1-y}{y^2}=\dfrac{1}{y^2}-\dfrac{1}{y}=y^{-2}-\dfrac{1}{y}$ and $\dfrac{1+x}{x^2}=\dfrac{1}{x^2}+\dfrac{1}{x}=x^{-2}+\dfrac{1}{x}$. Hence

$\left(y^{-2}-\frac{1}{y}\right)dy + \left(x^{-2}+\frac{1}{x}\right)dx = 0.$

Step 3 — integrate term by term.

$\int y^{-2}\,dy - \int \frac{1}{y}\,dy + \int x^{-2}\,dx + \int \frac{1}{x}\,dx = 0.$ …