The given differential equation is:

$(3x^2 y^2 - x) \, dy + (2x y^3 - y) \, dx = 0$

We can attempt to solve this by checking if it is exact or if an integrating factor can be applied.

Step 1: Check if the equation is exact

A differential equation of the form $M(x, y) \, dx + N(x, y) \, dy = 0$ is exact if:

$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

Here, we identify:

• $M(x, y) = 2x y^3 - y$
• $N(x, y) = 3x^2 y^2 - x$

Let's compute the partial derivatives.

$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y} (2x y^3 - y) = 6x y^2 - 1$

$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x} (3x^2 y^2 - x) = 6x y^2 - 1$

Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact.

Step 2: Solve for the potential function $\psi(x, y)$

Since the equation is exact, there exists a potential function $\psi(x, y)$ such that:

$\frac{\partial \psi}{\partial x} = M(x, y) = 2x y^3 - y$ $\frac{\partial \psi}{\partial y} = N(x, y) = 3x^2 y^2 - x$

To find $\psi(x, y)$, integrate $M(x, y)$ with respect to $x$:

$\psi(x, y) = \int (2x y^3 - y) \, dx = x^2 y^3 - xy + h(y)$

Here, $h(y)$ is a function of $y$ that acts as the "constant" of integration.

Step 3: Determine $h(y)$

To find $h(y)$, use the condition $\frac{\partial \psi}{\partial y} = N(x, y)$. First, compute $\frac{\partial \psi}{\partial y}$:

$\frac{\partial \psi}{\partial y} = \frac{\partial}{\partial y} (x^2 y^3 - xy + h(y)) = 3x^2 y^2 - x + h'(y)$

Now, equate this with $N(x, y) = 3x^2 y^2 - x$:

$3x^2 y^2 - x + h'(y) = 3x^2 y^2 - x$

This simplifies to:

$h'(y) = 0$

So, $h(y)$ is a constant, say $C$.

Step 4: General solution

Thus, the potential function is:

$\psi(x, y) = x^2 y^3 - xy + C$

The general solution to the differential equation is:

$x^2 y^3 - xy = C$

Final solution:

The implicit solution to the given differential equation is:

$x^2 y^3 - xy = C$

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you determine if a differential equation is exact?
2. What is an integrating factor, and when is it useful?
3. How do you solve an inexact differential equation?
4. Can this method be applied to non-homogeneous equations?
5. How does the potential function $\psi(x, y)$ help in solving exact equations?

Tip: Always check for exactness first when dealing with differential equations of this form, as it simplifies the solving process significantly.