Questions: Second-Order Linear Homogeneous Differential Equations

Question 1

The characteristic equation for y'' − 5y' + 6y = 0 has roots r = 2 and r = 3. What is the general solution?

• A. y = e^(5x)
• B. y = c₁e^(2x) + c₂e^(3x)
• C. y = c₁cos(2x) + c₂sin(3x)
• D. y = (c₁ + c₂x)e^(5x)

Answer: (B)

For two distinct real roots r₁ and r₂, the general solution is y = c₁e^(r₁x) + c₂e^(r₂x). With r₁ = 2 and r₂ = 3, this gives c₁e^(2x) + c₂e^(3x). The sinusoidal option applies only when roots are complex conjugates; the repeated-root option (c₁ + c₂x)eʳˣ applies only when r₁ = r₂.

Question 2

A second-order linear homogeneous ODE can have three or more linearly independent solutions.

• T. True
• F. False

Answer: False

The solution space of a second-order linear homogeneous ODE is exactly two-dimensional — it always has a basis of exactly two linearly independent solutions. Any solution is a linear combination c₁y₁ + c₂y₂ of those two. This mirrors how a second-order equation requires two initial conditions to determine a unique solution.

Question 3

For a constant-coefficient second-order ODE, why do we try y = eʳˣ as the solution form?

Model answer: Because differentiation of eʳˣ reproduces the same exponential: (eʳˣ)' = r·eʳˣ and (eʳˣ)'' = r²·eʳˣ. Substituting y = eʳˣ into the ODE therefore converts it into the algebraic equation r² + pr + q = 0 (the characteristic equation), which can be solved directly for r.

The exponential eʳˣ is an eigenfunction of the differentiation operator — differentiating it just multiplies by r. This property is what makes the substitution work. No other elementary function has this property in general, which is why the guess y = eʳˣ is so productive for constant-coefficient equations.