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The given equation is:

\[
(3xy + 2)dx + x^2 dy = 0 \quad \text{with} \quad y(x=1) = 1
\]

This is a first-order differential equation. We can approach this problem using exact differential equation methods since it appears in the form \( M(x, y) dx + N(x, y) dy = 0 \).

### Step 1: Check for Exactness

The equation is of the form:

\[
M(x, y) = 3xy + 2 \quad \text{and} \quad N(x, y) = x^2
\]

For the equation to be exact, the condition is:

\[
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}
\]

Now, calculate these partial derivatives:

\[
\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3xy + 2) = 3x
\]
\[
\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^2) = 2x
\]

Since \( \frac{\partial M}{\partial y} = 3x \) and \( \frac{\partial N}{\partial x} = 2x \), these are **not equal**, meaning the equation is not exact.

### Step 2: Try an Integrating Factor

Since the equation is not exact, we need an integrating factor to solve it. One common integrating factor depends on \( x \) or \( y \). Let's first check if an integrating factor of the form \( \mu(x) \) will work.

We need to multiply the entire equation by \( \mu(x) \) to attempt to make it exact:

\[
\mu(x)[(3xy + 2)dx + x^2 dy] = 0
\]

Now, find \( \mu(x) \) by solving the equation:

\[
\frac{1}{\mu(x)} \frac{d\mu(x)}{dx} = \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N}
\]

Substituting the known values:

\[
\frac{1}{\mu(x)} \frac{d\mu(x)}{dx} = \frac{3x - 2x}{x^2} = \frac{x}{x^2} = \frac{1}{x}
\]

Integrating both sides with respect to \( x \):

\[
\ln(\mu(x)) = \ln(x) \quad \Rightarrow \quad \mu(x) = x
\]

### Step 3: Multiply the Equation by \( \mu(x) = x \)

Multiplying the original equation by \( x \), we get:

\[
x(3xy + 2) dx + x(x^2) dy = 0
\]

Simplifying:

\[
(3x^2y + 2x) dx + x^3 dy = 0
\]

Now, the equation is exact because:

\[
M(x, y) = 3x^2y + 2x \quad \text{and} \quad N(x, y) = x^3
\]

The partial derivatives are:

\[
\frac{\partial M}{\partial y} = 3x^2 \quad \text{and} \quad \frac{\partial N}{\partial x} = 3x^2
\]

These are equal, confirming the equation is now exact.

### Step 4: Solve the Exact Equation

We solve the exact equation by finding the potential function \( \psi(x, y) \) such that:

\[
\frac{\partial \psi}{\partial x} = M(x, y) = 3x^2y + 2x
\]

Integrate with respect to \( x \):

\[
\psi(x, y) = \int (3x^2y + 2x) dx = x^3y + x^2 + h(y)
\]

Now, differentiate \( \psi(x, y) \) with respect to \( y \):

\[
\frac{\partial \psi}{\partial y} = x^3 + h'(y)
\]

Since \( \frac{\partial \psi}{\partial y} = N(x, y) = x^3 \), we get:

\[
x^3 + h'(y) = x^3 \quad \Rightarrow \quad h'(y) = 0
\]

Thus, \( h(y) \) is

Learn how to solve the differential equation (3xy + 2)dx + x^2dy = 0 with the initial condition y(1) = 1. The solution involves using the exact differential equation method and applying an integrating factor for non-exact equations. Suitable for undergraduate students studying calculus.

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