Method of Variation of Parameters & Cauchy-Euler Homogeneous Linear Equations

Explanation

The Method of Variation of Parameters replaces the arbitrary constants c₁, c₂ of the homogeneous solution with function variables u(x), v(x), imposing the condition u'y₁ + v'y₂ = 0 to eliminate second derivatives of the unknown functions. Cauchy-Euler equations represent scale-invariant differential systems where powers of independent variable x balance corresponding derivative orders.

Key points

• Wronskian Determinant: W = y₁ y₂' - y₁' y₂ ≠ 0, where y₁ and y₂ are independent solutions of the homogeneous equation.
• Particular Integral Formula: y_p = u(x) y₁(x) + v(x) y₂(x) where: u(x) = -∫ [y₂ R / W] dx and v(x) = ∫ [y₁ R / W] dx.
• Cauchy-Euler Equation: a_n x^n (d^n y/dx^n) + ... + a₁ x (dy/dx) + a₀ y = X(x).
• Transformation to Constant Coefficients: Let x = e^z (z = ln x).
• Operator Substitutions: x (d/dx) = D_z, x² (d²/dx²) = D_z(D_z - 1), x³ (d³/dx³) = D_z(D_z - 1)(D_z - 2).
• Legendre's Linear Equation: (ax + b)^n (d^n y/dx^n) + ... = X(x). Substitute ax + b = e^z.

Example

Code examples

// Detailed Solution: Cauchy-Euler Differential Equation
// Problem: x^2 (d^2y/dx^2) - x (dy/dx) + 4y = cos(ln x) + x sin(ln x)
// Step 1: Transformation:
//    Let x = e^z => z = ln x. Let D = d/dz.
//    x (dy/dx) = D y
//    x^2 (d^2y/dx^2) = D(D - 1) y
// Step 2: Substitute into ODE:
//    [ D(D - 1) - D + 4 ] y = cos z + e^z sin z
//    (D^2 - 2D + 4) y = cos z + e^z sin z
// Step 3: Complementary Function:
//    Auxiliary Equation: m^2 - 2m + 4 = 0
//    Roots: m = [2 ± √(4 - 16)] / 2 = 1 ± i √3
//    y_c = e^z [ c1 cos(√3 z) + c2 sin(√3 z) ]
//        = x [ c1 cos(√3 ln x) + c2 sin(√3 ln x) ]
// Step 4: Particular Integral y_p = y_p1 + y_p2:
//    y_p1 = [1 / (D^2 - 2D + 4)] cos z = [1 / (-1 - 2D + 4)] cos z
//         = [1 / (3 - 2D)] cos z = (3 + 2D) / (9 - 4D^2) cos z
//         = (3 cos z - 2 sin z) / (9 + 4) = (3 cos z - 2 sin z) / 13
//    y_p2 = [1 / (D^2 - 2D + 4)] e^z sin z = e^z [1 / ((D+1)^2 - 2(D+1) + 4)] sin z
//         = e^z [1 / (D^2 + 3)] sin z = e^z [1 / (-1 + 3)] sin z = (e^z sin z) / 2
// Step 5: Back-substitute z = ln x, e^z = x:
//    y_p = [3 cos(ln x) - 2 sin(ln x)] / 13 + [x sin(ln x)] / 2
// General Solution: y = y_c + y_p

Exam questions

• [8 Marks] Solve by variation of parameters: d²y/dx² + y = cosec x.
• [8 Marks] Solve the Cauchy-Euler equation: x² (d²y/dx²) - 2x (dy/dx) - 4y = x⁴.
• [7 Marks] Solve Legendre's linear differential equation: (2x + 3)² (d²y/dx²) - 2(2x + 3) (dy/dx) - 12y = 6x.