Unit 5: Introduction to Differential Equations
        
        Introduction and Definitions
        
            A differential equation (DE) is an equation that involves an unknown function and one or more of its derivatives.
        
        Differential equations are used to model real-world phenomena where a quantity's rate of change depends on the quantity itself (e.g., population growth, radioactive decay, circuits).
        
        
            - Ordinary Differential Equation (ODE): Contains derivatives with respect to only one independent variable. (e.g., `dy/dx`, `d²y/dt²`). This unit focuses *only* on ODEs.
- Partial Differential Equation (PDE): Contains partial derivatives with respect to two or more independent variables. (e.g., `∂u/∂t`, `∂²u/∂x²`).
Order and Degree of a DE
        
        Order
        
            The order of a differential equation is the order of the highest-order derivative present in the equation.
        
        
        Degree
        
            The degree of a differential equation is the power (exponent) of the highest-order derivative, *after* the equation has been cleared of any fractions or radicals in its derivatives.
        
        Examples:
        
            - (d²y/dx²)³ + 5(dy/dx)⁷ + y = 0
                
                    - Highest derivative is `d²y/dx²`, so Order = 2.
- The power of this highest derivative is 3, so Degree = 3.
 
- d³y/dx³ + 2x = √(1 + dy/dx)
                
                    - Highest derivative is `d³y/dx³`, so Order = 3.
- The power of this derivative is 1, so Degree = 1. (The radical on `dy/dx` is irrelevant as it's a lower-order term).
 
- y'' = (1 + (y')²)³ᐟ²
                
                    - First, clear the radical/fractional power: Square both sides.
- (y'')² = (1 + (y')²)³
- Highest derivative is `y''`, so Order = 2.
- The power of `y''` is 2, so Degree = 2.
 
General and Particular Solutions
        
        General Solution
        A general solution to a DE is a solution that contains one or more arbitrary constants (like `C`, `A`, `B`). The number of arbitrary constants is equal to the order of the DE.
        Example: For `dy/dx = 2x`, the general solution is `y = x² + C`.
        
        Particular Solution
        A particular solution is a solution obtained from the general solution by assigning specific values to the arbitrary constants. This is usually done using "initial conditions" or "boundary conditions."
        Example: For `y = x² + C`, if we are given the initial condition `y(0) = 5`, we can find `C`:
        `5 = (0)² + C` ⇒ `C = 5`.
        
The particular solution is `y = x² + 5`.
        
        Formation of a Differential Equation
        This is the reverse process: given a general solution (a family of curves), find the DE they satisfy by eliminating the arbitrary constants.
        
        Method:
        
            - Count the number of arbitrary constants (say, `n`). The order of your final DE will be `n`.
- Differentiate the given equation `n` times.
- You now have `n+1` equations (the original plus the `n` derivatives).
- Use algebra (substitution, elimination) to eliminate all `n` constants from these equations.
Example: Form the DE for the family of curves `y = A cos(x) + B sin(x)` (n=2 constants: A, B)
        
            - Original: `y = A cos(x) + B sin(x)` --- (1)
- Differentiate 1: `y' = -A sin(x) + B cos(x)` --- (2)
- Differentiate 2: `y'' = -A cos(x) - B sin(x)` --- (3)
- 
                Eliminate: Notice that equation (3) is the negative of equation (1).
                
 `y'' = -(A cos(x) + B sin(x))`
 Substitute (1): `y'' = -y`
The resulting DE is y'' + y = 0 or d²y/dx² + y = 0.
        Solving DEs: Variable Separable
        This is the simplest method for solving first-order, first-degree DEs.
        
            A DE is variable separable if it can be algebraically rearranged into the form:
            
            f(y) dy = g(x) dx
            
            (All `y` terms on one side, all `x` terms on the other).
        
        Solution: Integrate both sides and add a constant of integration `C`.
        ∫ f(y) dy = ∫ g(x) dx + C
        
        Example: Solve `dy/dx = xy²`
        
            - Separate variables: `dy / y² = x dx`
- Integrate both sides: `∫ y⁻² dy = ∫ x dx + C`
- `-y⁻¹ = (x²/2) + C`
- -1/y = x²/2 + C (This is the general solution).
Solving DEs: Homogeneous Equation
        
            A first-order DE dy/dx = F(x, y) is homogeneous if `F(x, y)` is a homogeneous function of degree 0.
            
            This means `F(tx, ty) = F(x, y)`, or more simply, it can be written as a function of `y/x`.
            
            Example: `dy/dx = (x² + y²) / (xy)`
        
        Solution Method:
        
            - Use the substitution: y = vx
- Differentiate with respect to `x` (using product rule): dy/dx = v(1) + x(dv/dx)
- Substitute both `y = vx` and `dy/dx = v + x(dv/dx)` into the original DE.
- The `x` variables will cancel, leaving a new DE involving only `v` and `x`.
- This new DE will always be variable separable. Solve it for `v`.
- Substitute v = y/x back in to get the final solution.
Solving DEs: Linear Differential Equation
        This is one of the most important types of DEs.
        
            A first-order DE is linear if it can be written in the standard form:
            
            dy/dx + P(x)y = Q(x)
            
            where P(x) and Q(x) are functions of `x` only (or constants).
        
        
        Solution Method (3 Steps):
        
            - 
                Step 1: Identify `P(x)` and `Q(x)` from the standard form.
                
 (Make sure the coefficient of `dy/dx` is 1).
- 
                Step 2: Find the Integrating Factor (I.F.).
                
 I.F. = e^(∫ P(x) dx)
- 
                Step 3: The general solution is given by the formula:
                
 y · (I.F.) = ∫ [ Q(x) · (I.F.) ] dx + C
            Exam Tip: Memorize this 3-step process.
            
            
Example: Solve `dy/dx + (1/x)y = x²`
            
                - Standard form is matched. `P(x) = 1/x`, `Q(x) = x²`.
- I.F. = e^(∫ (1/x) dx) = e^(ln x) = x. (So, I.F. = x).
- Solution: `y · (x) = ∫ [ x² · (x) ] dx + C`
                    
 `xy = ∫ x³ dx + C`
 `xy = (x⁴ / 4) + C`
 y = x³/4 + C/x (This is the general solution).