The characteristic equation for y'' − 4y' + 4y = 0 has a repeated root r = 2. A student writes the general solution as y = c₁e^(2x) + c₂e^(2x). What is wrong with this answer?
AThe student should use r = ±2 as two separate roots, giving y = c₁e^(2x) + c₂e^(−2x)
BThe two terms are not linearly independent — both are multiples of e^(2x), so this reduces to y = Ce^(2x) with only one free constant, which cannot be a general solution to a second-order ODE
CThe exponent should be 4x, not 2x, because the characteristic root must be squared
DThis ODE actually has complex roots, not repeated real roots, because the discriminant is negative
A general solution to a second-order ODE must have two linearly independent solutions and two free constants. Writing c₁e^(2x) + c₂e^(2x) = (c₁ + c₂)e^(2x) = Ce^(2x) is really just one free constant disguised as two — the terms are proportional and therefore linearly dependent. The characteristic equation only gives one independent solution for a repeated root; the second independent solution xe^(2x) must be found by reduction of order, giving the correct general solution y = (c₁ + c₂x)e^(2x).
Question 2 Multiple Choice
In the reduction-of-order substitution y₂ = v(x)·e^(rx), after substituting into the ODE and differentiating with the product rule, the terms containing v(x) itself cancel. Why does this cancellation occur?
ABecause e^(rx) becomes zero when differentiated twice and substituted back
BBecause e^(rx) is already a solution to the ODE — when the coefficient of v(x) is collected, it equals exactly the left-hand side of the ODE evaluated at y₁ = e^(rx), which is zero by definition
CBecause v(x) is assumed to be a constant throughout the reduction-of-order method
DBecause the product rule always eliminates all terms involving the original undifferentiated factor
This cancellation is the elegant core of the method. When you substitute y₂ = v(x)·y₁ into the ODE and expand using the product rule, you collect terms in v, v', and v''. The coefficient of v turns out to be exactly y₁'' + py₁' + qy₁ — which equals zero because y₁ is a solution to the ODE. This leaves an equation only in v' and v'', which is a first-order ODE for w = v'. The technique 'reduces the order' by one, making it solvable.
Question 3 True / False
If the characteristic equation of a second-order linear ODE has two distinct real roots, reduction of order is still needed to find the second independent solution.
TTrue
FFalse
Answer: False
Reduction of order is only needed when the characteristic equation fails to yield two independent solutions — specifically, when there is a repeated root. For two distinct real roots r₁ ≠ r₂, the characteristic equation directly produces two solutions e^(r₁x) and e^(r₂x), which are automatically linearly independent (neither is a constant multiple of the other). The general solution c₁e^(r₁x) + c₂e^(r₂x) is complete without any further work.
Question 4 True / False
The general solution to a second-order ODE with repeated root r is y = (c₁ + c₂x)e^(rx), and the two basis solutions e^(rx) and xe^(rx) are linearly independent.
TTrue
FFalse
Answer: True
Linear independence means neither solution is a constant multiple of the other. Although both contain the factor e^(rx), the extra factor of x in xe^(rx) prevents proportionality — there is no constant C such that xe^(rx) = C·e^(rx) for all x. This can be confirmed by the Wronskian: W(e^(rx), xe^(rx)) = e^(rx)·(e^(rx) + rxe^(rx)) − rxe^(rx)·e^(rx) = e^(2rx) ≠ 0, confirming independence. The general solution therefore has two genuinely free constants as required.
Question 5 Short Answer
Explain why a repeated root in the characteristic equation produces only one solution, and what the reduction-of-order technique does to find the missing second solution.
Think about your answer, then reveal below.
Model answer: The characteristic equation method tries y = e^(rx) and produces a quadratic in r. A repeated root r means the quadratic has only one value of r, yielding one solution e^(rx). No other exponential form e^(sx) works because s would have to equal r. To find the second solution, reduction of order tries y₂ = v(x)·e^(rx) — multiplying the known solution by an unknown function. After substituting and differentiating, the v(x) terms cancel (because e^(rx) already satisfies the ODE), leaving a simpler equation for v'(x). Solving gives v(x) = c₁ + c₂x, so the second independent solution is y₂ = xe^(rx).
The reduction-of-order technique is more general than just repeated roots — it works for any second-order linear ODE given one known solution, including variable-coefficient equations. The key insight is that knowing one solution lets you 'factor out' its structure and reduce the remaining problem by one order of difficulty.