Questions: Characteristic Equation Method for Linear ODEs

Question 1

The characteristic equation of a 2nd-order ODE has a repeated root r = -3. What is the general solution?

• A. y = c₁e^(-3x) + c₂e^(-3x), since both roots are -3
• B. y = (c₁ + c₂x)e^(-3x), since repeated roots require an x factor in the second solution
• C. y = e^(-3x)(c₁cos(3x) + c₂sin(3x)), since repeated roots produce oscillatory behavior
• D. y = c₁e^(-3x) alone, since the two identical roots yield only one independent solution

Answer: (B)

Two copies of e^(-3x) are linearly dependent — they are the same function and cannot span the solution space. The fix is to multiply one by x: xe^(-3x) is linearly independent from e^(-3x) and still satisfies the ODE. Option A writes two identical terms, which is just c₁+c₂ times one solution. Option C incorrectly applies the complex-root formula. Option D is incomplete — a 2nd-order ODE always needs two independent solutions.

Question 2

A characteristic equation has roots r = 2 ± 3i. What will the solution look like for large x?

• A. It will oscillate with constant amplitude, like a pure undamped sine wave
• B. It will decay to zero, since complex roots always produce damped oscillations
• C. It will oscillate with exponentially growing amplitude
• D. It will approach a constant steady-state value

Answer: (C)

Complex roots α ± βi give solutions e^(αx)(c₁cos(βx) + c₂sin(βx)). Here α = 2 > 0, so the amplitude grows as e^(2x) — without bound. Constant amplitude (option A) requires α = 0; decay (option B) requires α < 0. The imaginary part β = 3 controls oscillation frequency only.

Question 3

The characteristic equation method works by substituting y = e^(rx) into the ODE, which allows e^(rx) to be divided out, leaving a polynomial equation in r alone.

• T. True
• F. False

Answer: True

Substituting y = e^(rx) gives (r² + pr + q)e^(rx) = 0. Since e^(rx) is never zero, we divide both sides by it, reducing the differential equation to the algebraic characteristic equation r² + pr + q = 0. This is how the method converts a calculus problem into an algebra problem solvable with the quadratic formula.

Question 4

If the characteristic equation of a 2nd-order ODE with real coefficients has complex roots, the ODE has no real-valued solutions.

• T. True
• F. False

Answer: False

Complex roots always come in conjugate pairs (α ± βi) when the ODE has real coefficients. Euler's formula lets us combine the complex exponentials into real-valued solutions: e^(αx)cos(βx) and e^(αx)sin(βx). Complex roots indicate oscillatory behavior, not the absence of real solutions.

Question 5

What is the key insight that allows the characteristic equation method to turn a differential equation into an algebra problem?

Model answer: The exponential function e^(rx) is the only function whose derivatives are scalar multiples of itself: y′ = re^(rx), y″ = r²e^(rx). Substituting collapses all differentiation into scalar multiplication, so e^(rx) factors out completely, leaving a polynomial in r.

This is what makes e^(rx) the right ansatz. Any other function guess — a polynomial, sin, etc. — would not collapse the equation so cleanly. Because exponents compose multiplicatively under differentiation, the characteristic equation method reduces the hardest part (solving a differential equation) to something you already know: the quadratic formula.