Differential Equations Quiz: Repeated Roots

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Repeated RootsQuestion 1 of 6

The differential equation $y'' + ay' + by = 0$ has characteristic equation with a repeated root at $r = -3$ . If $u(x) = y(x)e^{3x}$ where $y(x)$ is a solution to the original equation, which differential equation does $u(x)$ satisfy?

u'' + 6u' + 9u = 0

u'' = 0

u'' - 6u' + 9u = 0

u'' + 9u = 0

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Differential Equations Quiz

Differential Equations Quiz: Repeated Roots

Practice Repeated Roots in Differential Equations with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Repeated Roots, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$u'' + 6u' + 9u = 0$
$u'' = 0$ (correct answer)
$u'' - 6u' + 9u = 0$
$u'' + 9u = 0$

Explanation: Since the characteristic equation has repeated root

r = -3

, we have

(r + 3)^2 = r^2 + 6r + 9 = 0

, so

a = 6

and

b = 9

. The original equation is

y'' + 6y' + 9y = 0

. The general solution is

y = (c_1 + c_2 x)e^{-3x}

. If

u(x) = y(x)e^{3x}

, then

y(x) = u(x)e^{-3x}

. Substituting:

y' = u'e^{-3x} - 3ue^{-3x}

and

y'' = u''e^{-3x} - 6u'e^{-3x} + 9ue^{-3x}

. Substituting into the original equation:

(u''e^{-3x} - 6u'e^{-3x} + 9ue^{-3x}) + 6(u'e^{-3x} - 3ue^{-3x}) + 9(ue^{-3x}) = 0

. Factoring out

e^{-3x}

u'' - 6u' + 9u + 6u' - 18u + 9u = 0

, which simplifies to

u'' = 0

Question 2

A mass-spring system is governed by the differential equation $m\frac{d^2x}{dt^2} + 2\gamma m\frac{dx}{dt} + kx = 0$ where $m > 0$ , $\gamma > 0$ , and $k > 0$ . If the system is critically damped (repeated roots), and the solution with initial conditions $x(0) = x_0$ and $x'(0) = v_0$ can be written as $x(t) = (A + Bt)e^{-\omega t}$ , which relationship must hold between the parameters?

$\gamma^2 = \frac{k}{m}$ and $\omega = \gamma$
$\gamma^2 = \frac{k}{4m}$ and $\omega = \sqrt{\frac{k}{m}}$
$\gamma^2 = \frac{k}{m}$ and $\omega = \sqrt{\frac{k}{m}}$ (correct answer)
$4\gamma^2 = \frac{k}{m}$ and $\omega = \gamma$

Explanation: Dividing the equation by

m

\frac{d^2x}{dt^2} + 2\gamma\frac{dx}{dt} + \frac{k}{m}x = 0

. The characteristic equation is

r^2 + 2\gamma r + \frac{k}{m} = 0

. For critical damping (repeated roots), the discriminant must be zero:

(2\gamma)^2 - 4 \cdot 1 \cdot \frac{k}{m} = 0

, which gives

4\gamma^2 = \frac{4k}{m}

, or

\gamma^2 = \frac{k}{m}

. The repeated root is

r = -\gamma

, so

\omega = \gamma = \sqrt{\frac{k}{m}}

Question 3

Consider the differential equation $y'' + 4y' + 4y = 0$ . If $z(x) = \int_0^x y(t) dt$ where $y(x)$ is the solution satisfying $y(0) = 0$ and $y'(0) = 1$ , what differential equation does $z(x)$ satisfy?

$z''' + 4z'' + 4z' = 0$ with $z(0) = z'(0) = 0$ , $z''(0) = 1$ (correct answer)
$z''' + 4z'' + 4z' = 1$ with $z(0) = z'(0) = z''(0) = 0$
$z''' + 4z'' + 4z' = 0$ with $z(0) = 0$ , $z'(0) = 1$ , $z''(0) = 4$
$z'' + 4z' + 4z = y(x)$ with $z(0) = z'(0) = 0$

Explanation: Since

z(x) = \int_0^x y(t) dt

, we have

z'(x) = y(x)

and

z''(x) = y'(x)

. From the original equation

y'' + 4y' + 4y = 0

, substituting:

z''' + 4z'' + 4z' = 0

. For initial conditions:

z(0) = \int_0^0 y(t) dt = 0

z'(0) = y(0) = 0

, and

z''(0) = y'(0) = 1

Question 4

Consider the equation $y'' - 6y' + 9y = 0$ with general solution $y = (c_1 + c_2 x)e^{3x}$ . If $f(x) = y(x) \cdot g(x)$ where $g(x)$ is a twice-differentiable function, and we want $f(x)$ to satisfy $f'' - 6f' + 9f = h(x)$ for some function $h(x)$ , what must be true about $g(x)$ ?

$g(x)$ must satisfy $g'' - 6g' + 9g = 0$
$g(x)$ must satisfy $g''(x) = 0$
$g(x)$ must be a polynomial of degree at most 1
$g(x)$ can be any twice-differentiable function (correct answer)

Explanation: This question tests your understanding of how differential operators behave when applied to products of functions. The key insight is that we're multiplying the general solution of a homogeneous equation by another function and asking what constraints this places on that function. Let's work through what happens when we substitute

f(x) = y(x) \cdot g(x)

into the differential equation

f'' - 6f' + 9f = h(x)

. Using the product rule repeatedly to find

f'

and

f''

, then substituting into our equation, we get:

[y''g + 2y'g' + yg''] - 6[y'g + yg'] + 9[yg] = h(x)

Rearranging terms:

g(y'' - 6y' + 9y) + g'(2y' - 6y) + g''y = h(x)

Since

y

satisfies

y'' - 6y' + 9y = 0

, the first term vanishes. From the given general solution

y = (c_1 + c_2x)e^{3x}

, we can verify that

2y' - 6y = 0

as well (this is a special property of repeated roots). Therefore, our equation simplifies to

g''y = h(x)

, which means

g'' = \frac{h(x)}{y}

. This shows that

g(x)

can indeed be any twice-differentiable function - we simply define

h(x) = g''(x) \cdot y(x)

and the equation is satisfied. Answer choices A, B, and C all impose unnecessary restrictions on

g(x)

. The beauty of this problem is that no constraints are needed beyond twice-differentiability. Study tip: When dealing with products in differential equations, remember that the differential operator distributes according to product rule patterns, often revealing hidden relationships between the component functions.

Question 5

The functions $y_1(x) = e^{3x}$ and $y_2(x) = xe^{3x}$ are solutions to a second-order linear homogeneous differential equation. If $W(x)$ denotes the Wronskian of these solutions, what is $\frac{W(1)}{W(0)}$ ?

$\frac{W(1)}{W(0)} = e^3$
$\frac{W(1)}{W(0)} = 1$
$\frac{W(1)}{W(0)} = e^9$
$\frac{W(1)}{W(0)} = e^6$ (correct answer)

Explanation: When you encounter Wronskian problems involving solutions to linear homogeneous differential equations, remember that the Wronskian has a special exponential behavior that often provides shortcuts to avoid tedious calculations. The Wronskian of

y_1(x) = e^{3x}

and

y_2(x) = xe^{3x}

is:

Computing this determinant:

W(x) = e^{3x}(e^{3x} + 3xe^{3x}) - xe^{3x}(3e^{3x}) = e^{6x}

Therefore:

\frac{W(1)}{W(0)} = \frac{e^6}{e^0} = e^6

Answer A gives

e^3

, which would result from incorrectly thinking the Wronskian grows like

e^{3x}

instead of

e^{6x}

. This mistake often comes from focusing only on the exponential factor in the original solutions rather than properly computing the determinant. Answer B suggests the ratio equals 1, implying the Wronskian is constant. This would only be true if the coefficient function

p(x)

in the differential equation

y'' + p(x)y' + q(x)y = 0

were zero, which isn't the case here. Answer C gives

e^9

, which might come from incorrectly squaring the

e^{3x}

factor and then cubing it, representing a computational error in determinant evaluation. Study tip: For Wronskians of exponential-type solutions, look for patterns in the exponents. The Wronskian often has an exponential form that's predictable from the original solutions' structure, and Abel's theorem can provide useful shortcuts.

Question 6

A second-order linear homogeneous differential equation has characteristic polynomial $P(r) = r^2 + pr + q$ . If the equation has a repeated root and one particular solution is $y_1(x) = xe^{-2x}$ , what is the value of $p + q$ ?

$p + q = 8$ (correct answer)
$p + q = 6$
$p + q = 4$
$p + q = 2$

Explanation: Since

y_1(x) = xe^{-2x}

is a solution and we have a repeated root, the general solution must be of the form

y = (c_1 + c_2 x)e^{-2x}

. This means the repeated root is

r = -2

. The characteristic polynomial with repeated root

r = -2

(r + 2)^2 = r^2 + 4r + 4

. Therefore

p = 4

and

q = 4

, giving

p + q = 8