L15-L16: Differential Equations (Separable)

Lecture notes covering the definition and classification of differential equations, and the solution of first-order differential equations with separable variables.

1. Introduction to Differential Equations

Differential equation (DE) is an equation that contains variables $x$ and $y$, with at least one derivative of $y$ with respect to $x$.

Order and Degree

Term Definition
Order of DE The highest derivative present in the equation
Degree of DE The power of the highest derivative

Examples:

  • $\displaystyle \frac{dy}{dx} + 3y = x$ → 1st order, 1st degree
  • $\displaystyle \frac{d^2y}{dx^2} + x^2 = 5\frac{dy}{dx}$ → 2nd order, 1st degree
  • $\displaystyle \left(\frac{d^2y}{dx^2}\right)^3 - \frac{dy}{dx} = 3y$ → 2nd order, 3rd degree

Solutions of a DE

A solution to a differential equation is any function that satisfies the given equation.

Example verification: $y = \sin x$ is a solution to $y'' + y = 0$ because:

  • $y' = \cos x$
  • $y'' = -\sin x$
  • Substituting: $(-\sin x) + (\sin x) = 0 \quad \checkmark$

General Solution

Differentiating $y = x^3 + C$ with respect to $x$ gives: $$\frac{dy}{dx} = 3x^2$$

The equation $\displaystyle \frac{dy}{dx} = 3x^2$ is a 1st order and 1st degree differential equation, and: $$y = x^3 + C$$ is the general solution because it contains an arbitrary constant $C$.

Particular Solution

If further information (an initial condition) is given so that the constant can be determined, then the particular solution of the differential equation can be obtained.

2. Differential Equations of Separable Variable Type

A first-order differential equation whose variables $x$ and $y$ are separable is of the general form:

$$\frac{dy}{dx} = \frac{f(x)}{g(y)} \quad \text{or} \quad g(y)\frac{dy}{dx} = f(x)$$

Solution Method

  1. Separate variables: group all $y$ terms with $dy$ and all $x$ terms with $dx$
  2. Integrate both sides
  3. Solve for $y$ (if possible) to obtain the general solution
  4. Apply initial condition (if given) to find the particular solution
flowchart TD
    A(["Separable DE<br/>dy/dx = f(x)/g(y)"]) --> B[Separate Variables<br/>g(y) dy = f(x) dx]
    B --> C[Integrate Both Sides<br/>∫ g(y) dy = ∫ f(x) dx]
    C --> D([General Solution<br/>with constant C])
    D --> E{Initial Condition<br/>Given?}
    E -->|Yes| F[Substitute x₀, y₀<br/>to find C]
    F --> G([Particular Solution])
    E -->|No| H[Leave as<br/>General Solution]

3. Worked Examples

Example 1 — General Solution

Solve: $$y\frac{dy}{dx} = 3x^2$$

Solution: Separate variables: $$y,dy = 3x^2,dx$$

Integrate both sides: $$\int y,dy = \int 3x^2,dx$$

$$\frac{y^2}{2} = \frac{3x^3}{3} + C = x^3 + C$$

$$y^2 = 2x^3 + 2C$$

Let $A = 2C$: $$\boxed{y^2 = 2x^3 + A}$$

This is the general solution.

Example 2 — General Solution with Logarithms

Solve: $$x\frac{dy}{dx} = 2y$$

Solution: Separate variables: $$\frac{1}{2y},dy = \frac{1}{x},dx$$

Integrate both sides: $$\frac{1}{2}\int \frac{1}{y},dy = \int \frac{1}{x},dx$$

$$\frac{1}{2}\ln y = \ln x + C$$

$$\ln y = 2\ln x + 2C = \ln x^2 + 2C$$

Exponentiate: $$y = e^{\ln x^2 + 2C} = x^2 \cdot e^{2C}$$

Let $A = e^{2C}$: $$\boxed{y = Ax^2}$$

This is the general solution.

Example 3 — Particular Solution

Solve the differential equation: $$\frac{dy}{dx} = \frac{2y}{x^2 - 1}, \quad \text{given that } y = 1 \text{ when } x = 2$$

Solution: Separate variables: $$\frac{dy}{2y} = \frac{dx}{x^2 - 1}$$

Integrate both sides: $$\int \frac{dy}{2y} = \int \frac{dx}{x^2 - 1}$$

$$\frac{1}{2}\ln y = \int \frac{dx}{(x-1)(x+1)}$$

Use partial fractions on the RHS: $$\frac{1}{(x-1)(x+1)} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right)$$

So: $$\frac{1}{2}\ln y = \frac{1}{2}\left[\ln(x-1) - \ln(x+1)\right] + C$$

Multiply by 2: $$\ln y = \ln(x-1) - \ln(x+1) + 2C = \ln\left(\frac{x-1}{x+1}\right) + 2C$$

Exponentiate: $$y = \frac{x-1}{x+1} \cdot e^{2C}$$

Let $A = e^{2C}$: $$\boxed{y = \frac{A(x-1)}{x+1}} \quad \text{(General solution)}$$

Apply initial condition $y = 1$ when $x = 2$: $$1 = \frac{A(2-1)}{2+1} = \frac{A}{3} \quad \Rightarrow \quad A = 3$$

$$\boxed{y = \frac{3(x-1)}{x+1}} \quad \text{(Particular solution)}$$

Summary

Concept Description
DE Equation involving derivatives of $y$ w.r.t. $x$
Order Highest derivative present
Degree Power of the highest derivative
General solution Contains an arbitrary constant
Particular solution Constant determined by initial conditions
Separable DE Can be written as $g(y),dy = f(x),dx$

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