To solve the given differential equation $(y + xy^2)dx - xdy = 0$ by finding the integrating factor, we follow these steps:

Step 1: Write the equation in the standard form

The given differential equation is: $(y + xy^2)dx - xdy = 0$ This is of the form $M(x, y)dx + N(x, y)dy = 0$, where: $M(x, y) = y + xy^2$ $N(x, y) = -x$

Step 2: Check if the equation is exact

For a differential equation to be exact, the following condition must hold: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

First, we compute the partial derivatives:

$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(y + xy^2) = 1 + 2xy$

$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(-x) = -1$

Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact.

Step 3: Find the integrating factor

Since the equation is not exact, we need an integrating factor to make it exact. Often, integrating factors depend on $x$ or $y$. Let's try to find an integrating factor that depends on $x$.

If the integrating factor $\mu(x)$ depends only on $x$, the new equation will become: $\mu(x)(M(x, y)dx + N(x, y)dy) = 0$ To make the equation exact, we need the modified version of the equation to satisfy: $\frac{\partial}{\partial y}(\mu(x)M(x, y)) = \frac{\partial}{\partial x}(\mu(x)N(x, y))$

Now, compute the derivatives of $\mu(x)M(x, y)$ and $\mu(x)N(x, y)$.

$\frac{\partial}{\partial y}(\mu(x)M(x, y)) = \mu(x)(1 + 2xy)$ $\frac{\partial}{\partial x}(\mu(x)N(x, y)) = \frac{\partial}{\partial x}(-\mu(x)x) = -\mu'(x)x - \mu(x)$

Equating these two expressions: $\mu(x)(1 + 2xy) = -\mu'(x)x - \mu(x)$

We now solve for $\mu(x)$ from this equation. To simplify, let's focus on terms involving $x$ only and find a suitable function for $\mu(x)$.

Assuming $\mu(x) = \frac{1}{x}$, we substitute it back into the equation.

Step 4: Apply the integrating factor

Multiply the entire equation by $\frac{1}{x}$: $\frac{1}{x}((y + xy^2)dx - xdy) = 0$ Simplifying: $\left(\frac{y}{x} + y^2\right)dx - dy = 0$

Step 5: Check if the new equation is exact

Now, let's check if this new equation is exact.

Let: $M(x, y) = \frac{y}{x} + y^2$ $N(x, y) = -1$

Now, compute the partial derivatives:

$\frac{\partial M}{\partial y} = \frac{1}{x} + 2y$ $\frac{\partial N}{\partial x} = 0$

Since $\frac{\partial M}{\partial y} = \frac{1}{x} + 2y$ and $\frac{\partial N}{\partial x} = 0$, the equation is now exact.

Step 6: Solve the exact equation

To solve the exact equation, we need to find a potential function $\Psi(x, y)$ such that: $\frac{\partial \Psi}{\partial x} = M(x, y) = \frac{y}{x} + y^2$ $\frac{\partial \Psi}{\partial y} = N(x, y) = -1$

Integrating $M(x, y)$ with respect to $x$: $\Psi(x, y) = \int \left( \frac{y}{x} + y^2 \right) dx = y \ln|x| + xy^2 + g(y)$ where $g(y)$ is an arbitrary function of $y$.

Now, differentiate $\Psi(x, y)$ with respect to $y$: $\frac{\partial \Psi}{\partial y} = \ln|x| + 2xy + g'(y)$ Set this equal to $N(x, y) = -1$: $\ln|x| + 2xy + g'(y) = -1$ This gives us: $g'(y) = -1 - \ln|x| - 2xy$ Now