Initial Value, Boundary Value, and Eigenvalue Problems

Example: Find the solution for the following boundary value problem.

d²y/dx² + 6 dy/dx + 13 y = 0

y(0) = 0, y’(π) = 1 (boundary values)

You can write this d.e. in operator notation (D² + 6D + 13) y = 0

Assume y = Ae^rx for the complementary solution.

So d²y/dx² = A r² e^rx , dy/dx = A r e^rx , and y = Ae^rx

Substitute into the d.e. yields

Ae^rx ( r² + 6r + 13 ) = 0

Since Ae^rx ≠ 0, the characteristic equation for r becomes:

r² + 6r + 13 = 0 ,

with roots -3 ± 2i using the quadratic formula.

The complementary solution, y_c , is:

y_c (x) = Fe^{(-3 + 2i)x} + Ge^{(-3 - 2i)x}

where F and G are undetermined constants (need two initial conditions)

which can be expressed as follows, (equivalent complementary solution)

y(x) = y_c (x) = e^-3x ( C₁ sin 2x+ C₂ cos 2x )

Apply the initial values to find C₁ and C₂. y(0) = 0 gives 0 = C₂

So y’(x) = -3 e^-3x ( C₁ sin 2x ) + 2 e^-3x ( C₁ cos 2x )

and y’(π) = 1 = 2 e^-3π C₁ so C₁ = 1/2 e^3π

The resulting solution for the boundary value problem is y(x) = (1/2) e^3x sin 2x

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