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The question asks you to determine which of the provided differential equations are exact, and then solve them.

To check if a differential equation is exact, you follow these steps:

1. For an equation in the form \( M(x, y)dx + N(x, y)dy = 0 \), calculate the partial derivatives:
   - \( M_y = \frac{\partial M}{\partial y} \)
   - \( N_x = \frac{\partial N}{\partial x} \)
   
2. If \( M_y = N_x \), then the differential equation is exact. If not, it is not exact.

Let’s go through each equation to see which one(s) are exact:

### a. \( (x^2 - y)dx - xdy = 0 \)
- Here, \( M(x, y) = x^2 - y \) and \( N(x, y) = -x \).
- \( M_y = \frac{\partial}{\partial y}(x^2 - y) = -1 \)
- \( N_x = \frac{\partial}{\partial x}(-x) = -1 \)
- Since \( M_y = N_x \), **this equation is exact**.

### b. \( y(x - 2y)dx - x^2dy = 0 \)
- Here, \( M(x, y) = y(x - 2y) = xy - 2y^2 \) and \( N(x, y) = -x^2 \).
- \( M_y = \frac{\partial}{\partial y}(xy - 2y^2) = x - 4y \)
- \( N_x = \frac{\partial}{\partial x}(-x^2) = -2x \)
- Since \( M_y \neq N_x \), **this equation is not exact**.

### c. \( dx - \left(a^2 - x^2\right)^{\frac{1}{2}}dy = 0 \)
- Here, \( M(x, y) = 1 \) and \( N(x, y) = -\left(a^2 - x^2\right)^{\frac{1}{2}} \).
- \( M_y = \frac{\partial}{\partial y}(1) = 0 \)
- \( N_x = \frac{\partial}{\partial x}\left[-\left(a^2 - x^2\right)^{\frac{1}{2}}\right] = \frac{x}{\left(a^2 - x^2\right)^{\frac{1}{2}}} \)
- Since \( M_y \neq N_x \), **this equation is not exact**.

### d. \( (x^2 + y^2)dx + xydy = 0 \)
- Here, \( M(x, y) = x^2 + y^2 \) and \( N(x, y) = xy \).
- \( M_y = \frac{\partial}{\partial y}(x^2 + y^2) = 2y \)
- \( N_x = \frac{\partial}{\partial x}(xy) = y \)
- Since \( M_y \neq N_x \), **this equation is not exact**.

### e. \( \left(4x^3y^3 + \frac{1}{x}\right)dx + \left(3x^4y^2 - \frac{1}{y}\right)dy = 0 \)
- Here, \( M(x, y) = 4x^3y^3 + \frac{1}{x} \) and \( N(x, y) = 3x^4y^2 - \frac{1}{y} \).
- \( M_y = \frac{\partial}{\partial y}(4x^3y^3 + \frac{1}{x}) = 12x^3y^2 \)
- \( N_x = \frac{\partial}{\partial x}(3x^4y^2 - \frac{1}{y}) = 12x^3y^2 \)
- Since \( M_y = N_x \), **this equation is exact**.

### Conclusion:
- The exact equations are **a** and **e**.
  
Next, you would solve these exact equations by finding a potential function \( \psi(x, y) \) such that \( \psi_x = M(x, y) \) and \( \psi_y = N(x, y) \).

Would you like to see the solutions for these exact equations or any further details? 

Here are 5 related questions:
1. How do you solve an exact differential equation once you know it's exact?
2. Can an equation be made exact using an integrating factor? How?
3. What are some applications of exact differential equations in physics or engineering?
4. How do you find an integrating factor for a non-exact equation?
5. Can there be multiple integrating factors for a given differential equation?

**Tip:** For exact differential equations, always check if the mixed second derivatives of the potential function match for consistency in the solution process.

Choose from the following exact equations and solve:
 a. (x^2 - y)dx - xdy = 0
 b. y(x - 2y)dx - x^2dy = 0
 c. dx - (a^2 - x^2)^(1/2)dy = 0
 d. (x^2 + y^2)dx + xydy = 0
 e. (4x^3y^3 + 1/x)dx + (3x^4y^2 - 1/y)dy = 0

Learn how to identify and solve exact differential equations. This problem covers the method of checking exactness (M_y = N_x) and solving through potential functions. Suitable for advanced mathematics students.

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