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Linear Differential Equations, Bernoulli's Form & Clairaut's Singular Solutions

AI draft, not yet reviewedUnit 2: Differential Equations of the First OrderAbout 3 minutes

Idea in plain words

Leibniz's Linear First Order ODE: dy/dx + P(x) y = Q(x).

Exam tip

High-yield exam topic (Very High).

Explanation

Linear differential equations satisfy the principle of superposition. When an equation is non-linear solely due to a power y^n on the right side, Bernoulli's substitution v = y^{1-n} collapses the non-linear curve into a linear differential manifold. Clairaut's equations represent a remarkable class where the singular solution forms the envelope of the family of straight line solutions, representing states unreachable through any choice of the arbitrary constant C.

Key points

  • Integrating Factor: IF = e^{∫ P(x) dx}. General solution: y · (IF) = ∫ [Q(x) · (IF)] dx + C.
  • Alternative Form: dx/dy + P(y) x = Q(y) with IF = e^{∫ P(y) dy} and solution x · (IF) = ∫ [Q(y) · (IF)] dy + C.
  • Bernoulli's Equation: dy/dx + P(x) y = Q(x) y^n. Divide by y^n: y^{-n} (dy/dx) + P(x) y^{1-n} = Q(x). Substitute v = y^{1-n} to obtain linear ODE: (1/(1-n)) (dv/dx) + P(x) v = Q(x).
  • Clairaut's Equation: y = p x + f(p) where p = dy/dx.
  • Clairaut's Solutions: Differentiating with respect to x gives p = p + [x + f'(p)] (dp/dx) => (dp/dx) [x + f'(p)] = 0.
  • (1) General Solution: Set dp/dx = 0 => p = c => y = c x + f(c) (family of straight lines).
  • (2) Singular Solution: Set x + f'(p) = 0 and eliminate p between x + f'(p) = 0 and y = px + f(p) (envelope of the family of lines).

Example

Example

Solve dy/dx + y tan x = sec x: P = tan x, Q = sec x. IF = e^{∫ tan x dx} = e^{ln sec x} = sec x. Solution: y · sec x = ∫ sec x · sec x dx = ∫ sec² x dx = tan x + C => y = sin x + C cos x.

Code examples

// Solution of Clairaut's Equation & Singular Solution Extraction:
// Problem: y = p x + a / p   (where p = dy/dx)
// Step 1: Differentiate with respect to x:
//    dy/dx = p = p + x (dp/dx) - (a / p^2) (dp/dx)
//    0 = (dp/dx) [ x - a / p^2 ]
// Step 2: Case 1: dp/dx = 0 => p = c
//    Substitute p = c into original equation:
//    y = c x + a / c  <== GENERAL SOLUTION (Family of tangent lines!)
//
// Step 3: Case 2: x - a / p^2 = 0 => p^2 = a / x => p = √(a / x)
//    Substitute p back into original equation:
//    y = [√(a / x)] * x + a / [√(a / x)]
//      = √(a x) + √(a x) = 2 √(a x)
//    Squaring both sides:
//    y^2 = 4 a x  <== SINGULAR SOLUTION (Parabola envelope!)
// Notice: The parabola y^2 = 4ax cannot be obtained from y = cx + a/c for any choice of c!

Exam questions

  • [8 Marks] Solve the differential equation: (1 + y²) dx = (tan⁻¹ y - x) dy.
  • [8 Marks] Solve the Bernoulli equation: dy/dx + (xy / (1 - x²)) = x √y.
  • [7 Marks] Find the general and singular solutions of Clairaut's equation: y = p x + √(a² p² + b²).

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Unit 1: Multiple Integrals

Unit 2: Differential Equations of the First Order

Unit 3: Linear Differential Equations with Constant and Variable Coefficients

Unit 4: Fourier Series and Integrals

Unit 5: Functions of a Complex Variable

Unit 6: Complex Series, Residues and Poles

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