1. Define the following terms; ODE, PDE, primitive and total differential. 2. Show that \( \frac{1}{x^{2}} \) is an integrating factor of \( \left(y+x^{4}\right) d x-x d y=0 \) and solve the equation.

### Definitions 1. **ODE (Ordinary Differential Equation)**: An ODE is a differential equation that contains one or more functions of a single independent variable and its derivatives. The general form of an ODE is \( F(x, y, y', y'', \ldots) = 0 \), where \( y \) is a function of \( x \), and \( y', y'', \ldots \) are its derivatives. 2. **PDE (Partial Differential Equation)**: A PDE is a differential equation that involves multiple independent variables and the partial derivatives of a function with respect to those variables. The general form of a PDE is \( F(x_1, x_2, \ldots, y, \frac{\partial y}{\partial x_1}, \frac{\partial y}{\partial x_2}, \ldots) = 0 \), where \( y \) is a function of several variables \( x_1, x_2, \ldots \). 3. **Primitive**: In calculus, a primitive of a function \( f(x) \) is another function \( F(x) \) such that \( F'(x) = f(x) \). In other words, the primitive is an antiderivative of the function. 4. **Total Differential**: The total differential of a function \( z = f(x, y) \) is an expression that gives the change in \( z \) due to changes in \( x \) and \( y \). It is given by: \[ dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \] This expression captures how \( z \) changes with respect to small changes in \( x \) and \( y \). ### Problem Solution To show that \( \frac{1}{x^{2}} \) is an integrating factor of the differential equation \( (y + x^4) dx - x dy = 0 \), we first rewrite the equation in the standard form: \[ M(x, y) = y + x^4, \quad N(x, y) = -x \] We need to check if the equation is exact, which requires that: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \] Calculating these partial derivatives: \[ \frac{\partial M}{\partial y} = 1, \quad \frac{\partial N}{\partial x} = -1 \] Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the equation is not exact. Next, we will check if \( \frac{1}{x^2} \) is an integrating factor. An integrating factor \( \mu(x) \) can make the equation exact if it satisfies: \[ \frac{\partial}{\partial y} \left( \mu M \right) = \frac{\partial}{\partial x} \left( \mu N \right) \] Substituting \( \mu = \frac{1}{x^2} \): \[ \mu M = \frac{1}{x^2}(y + x^4) = \frac{y}{x^2} + x^2 \] \[ \mu N = \frac{1}{x^2}(-x) = -\frac{1}{x} \] Now, we compute the partial derivatives: \[ \frac{\partial}{\partial y} \left( \frac{y}{x^2} + x^2 \right) = \frac{1}{x^2} \] \[ \frac{\partial}{\partial x} \left( -\frac{1}{x} \right) = \frac{1}{x^2} \] Since both derivatives are equal, \( \frac{1}{x^2} \) is indeed an integrating factor. Now we can multiply the entire original equation by \( \frac{1}{x^2} \): \[ \left( \frac{y}{x^2} + x^2 \right) dx - \left( \frac{1}{x} \right) dy = 0 \] This is now an exact equation. We can find a potential function \( \Psi(x, y) \) such that: \[ \frac{\partial \Psi}{\partial x} = \frac{y}{x^2} + x^2 \] Integrating with respect to \( x \): \[ \Psi(x, y) = \int \left( \frac{y}{x^2} + x^