First Order ODEs: Variable Separable, Homogeneous & Exact Equations with Integrating Factors
Idea in plain words
Differential Equation of First Order & First Degree: M(x, y) dx + N(x, y) dy = 0.
Exam tip
High-yield exam topic (Very High).
Explanation
First-order ordinary differential equations describe instantaneous rates of change in dynamical systems. An exact differential equation corresponds to the level curves of a potential scalar function U(x, y) = C such that dU = (∂U/∂x)dx + (∂U/∂y)dy = 0. When ∂M/∂y ≠ ∂N/∂x, the differential form is non-conservative; multiplying by an integrating factor μ(x, y) restores exactness by warping the geometry into a conservative potential field.
Key points
- Variable Separable: f(x) dx + g(y) dy = 0 => ∫ f(x) dx + ∫ g(y) dy = C.
- Homogeneous Equations: dy/dx = f(y/x). Substitute y = vx => dy/dx = v + x (dv/dx) to separate variables in v and x.
- Exact Differential Equations: M dx + N dy = 0 is exact if and only if ∂M/∂y = ∂N/∂x.
- Solution of Exact ODE: ∫_{y=const} M dx + ∫ (terms of N free from x) dy = C.
- Integrating Factors (IF) Rules:
- Rule 1 (Homogeneous): If M dx + N dy = 0 is homogeneous and Mx + Ny ≠ 0, then IF = 1 / (Mx + Ny).
- Rule 2 (f₁(xy)y dx + f₂(xy)x dy = 0): If Mx - Ny ≠ 0, then IF = 1 / (Mx - Ny).
- Rule 3 (Function of x alone): If (1/N)(∂M/∂y - ∂N/∂x) = f(x), then IF = e^{∫ f(x) dx}.
- Rule 4 (Function of y alone): If (1/M)(∂N/∂x - ∂M/∂y) = g(y), then IF = e^{∫ g(y) dy}.
Example
Example
Homogeneous ODE (x² - y²) dx + 2xy dy = 0: Let y = vx => dy = v dx + x dv. (x² - v²x²)dx + 2x(vx)(v dx + x dv) = 0 => (1 + v²)dx + 2xv dv = 0 => dx/x + 2v dv/(1 + v²) = 0. Integrating: ln|x| + ln(1 + v²) = ln C => x(1 + y²/x²) = C => x² + y² = Cx.
Code examples
// Complete Analytical Solution of Non-Exact ODE via Integrating Factor:
// Problem: (x^2 + y^2 + 2x) dx + 2y dy = 0
// Step 1: Identify M and N:
// M = x^2 + y^2 + 2x
// N = 2y
// Step 2: Test for exactness:
// ∂M/∂y = 2y
// ∂N/∂x = 0
// ∂M/∂y ≠ ∂N/∂x => NOT EXACT!
// Step 3: Check Rule 3 (Function of x alone):
// (1/N) * (∂M/∂y - ∂N/∂x) = [1 / (2y)] * (2y - 0) = 2y / 2y = 1 = f(x) (constant)
// Step 4: Calculate Integrating Factor (IF):
// IF = e^{∫ 1 dx} = e^x
// Step 5: Multiply entire ODE by IF = e^x:
// e^x (x^2 + y^2 + 2x) dx + 2y e^x dy = 0
// Now: M' = e^x (x^2 + y^2 + 2x), N' = 2y e^x
// ∂M'/∂y = 2y e^x, ∂N'/∂x = 2y e^x (EXACT!)
// Step 6: Integrate to find potential function U(x, y):
// U = ∫_{y=const} e^x (x^2 + y^2 + 2x) dx + ∫ (terms of N' free from x) dy
// Notice: d/dx [ e^x (x^2 + y^2) ] = e^x (x^2 + y^2) + e^x (2x) = e^x (x^2 + y^2 + 2x)
// Terms of N' = 2y e^x free from x: NONE (0 terms)
// Therefore: e^x (x^2 + y^2) = C (General Solution!)
Exam questions
- [8 Marks] Solve the differential equation: (x⁴ e^x - 2m y² x) dx + 2m x² y dy = 0.
- [8 Marks] Solve: (x² y - 2x y²) dx - (x³ - 3x² y) dy = 0 using an integrating factor.
- [7 Marks] Solve the homogeneous differential equation: [x - y cos(y/x)] dx + x cos(y/x) dy = 0.