Find the integrating factor of the following differential equation which will

make it exact. Hence solve the equation.

(y + xy

2

)dx − xdy = 0.

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

Find the integrating factor of the following differential equation which will make it exact. Hence solve the equation. (y + xy^2)dx − xdy = 0.

Learn how to solve the differential equation (y + xy^2)dx - xdy = 0 by finding an integrating factor to make the equation exact. This problem involves concepts from differential equations, including integrating factors and exact equations. Suitable for college-level calculus students.

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