Differential Equations Quiz: Undetermined Coefficients

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Undetermined CoefficientsQuestion 1 of 17

A differential equation has the form $y'' + ay' + by = e^{rx}$ where $a$ , $b$ , and $r$ are constants. Under which condition would the particular solution necessarily take the form $y_p = Ax^2e^{rx}$ ?

When

r

is a repeated root of the characteristic equation

s^2 + as + b = 0

When

r

is a simple root of the characteristic equation

s^2 + as + b = 0

When

r^2 + ar + b = 0

has complex conjugate roots involving

r

When

r

satisfies

r^2 + ar + b \neq 0

but

2r + a = 0

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Differential Equations Quiz

Differential Equations Quiz: Undetermined Coefficients

Practice Undetermined Coefficients 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 Undetermined Coefficients, 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

When $r$ is a repeated root of the characteristic equation $s^2 + as + b = 0$ (correct answer)
When $r$ is a simple root of the characteristic equation $s^2 + as + b = 0$
When $r^2 + ar + b = 0$ has complex conjugate roots involving $r$
When $r$ satisfies $r^2 + ar + b \neq 0$ but $2r + a = 0$

Explanation: When solving non-homogeneous linear differential equations with exponential forcing functions like

e^{rx}

, the form of your particular solution depends critically on whether the exponent

r

relates to the roots of the characteristic equation

s^2 + as + b = 0

. The correct answer is A. When

r

is a repeated root of the characteristic equation, it means

r

satisfies

r^2 + ar + b = 0

and appears twice in the homogeneous solution. Since both

e^{rx}

and

xe^{rx}

are already part of the homogeneous solution, you must multiply by

x^2

to ensure your particular solution is linearly independent. This gives

y_p = Ax^2e^{rx}

. Choice B is incorrect because when

r

is a simple root, only

e^{rx}

appears in the homogeneous solution, so

y_p = Axe^{rx}

(multiply by

x

once) would be sufficient. Choice C misunderstands the relationship entirely. Complex conjugate roots don't directly determine the particular solution form for an

e^{rx}

forcing function. Choice D describes an impossible scenario. If

r^2 + ar + b \neq 0

, then

r

isn't a root, so the condition

2r + a = 0

(which would make

r = -a/2

) is irrelevant to determining the particular solution form. Study tip: Always check if your forcing function's exponent matches roots of the characteristic equation. No match = try

Ae^{rx}

; simple root = try

Axe^{rx}

; repeated root = try

Ax^2e^{rx}

Question 2

For the differential equation $y'' - 2y' + 2y = 5te^t$ , an appropriate guess for the particular solution is $Y_p(t) = (At+B)e^t$ . After substituting $Y_p(t)$ into the equation, one finds that $Ae^t = 5te^t$ . What conclusion can be drawn?

The guess for $Y_p(t)$ is incorrect and must be multiplied by $t$ .
The guess for $Y_p(t)$ is correct, and the conclusion is that $A=5t$ , which is not a constant.
There is a calculation error, as the method of undetermined coefficients cannot fail this way. (correct answer)
The equation has no particular solution of the assumed form.

Explanation: The method of undetermined coefficients guarantees a solution of the assumed form exists, provided the form is chosen correctly. Let's verify the calculation.

Y_p = (At+B)e^t

Y_p' = Ae^t + (At+B)e^t = (At+A+B)e^t

Y_p'' = Ae^t + (At+A+B)e^t = (At+2A+B)e^t

. Substitute into the ODE:

(At+2A+B)e^t - 2(At+A+B)e^t + 2(At+B)e^t = 5te^t

. Divide by

e^t

(At+2A+B) - 2(At+A+B) + 2(At+B) = 5t

. Combine terms:

(A-2A+2A)t + (2A+B-2A-2B+2B) = 5t

. This simplifies to

At + B = 5t

. Equating coefficients gives

A=5

and

B=0

. The calculation in the stem, which leads to

Ae^t=5te^t

, is incorrect. A calculation error must have occurred. The guess does not need modification as

y_c=e^t(c_1\cos t + c_2\sin t)

has no overlap.

Question 3

A student attempts to find the form of a particular solution for $y''-y=te^{-t}$ . They note the complementary solution is $y_c = c_1 e^t + c_2 e^{-t}$ . Their initial guess is $Y_p=(At+B)e^{-t}$ . They claim this guess must be multiplied by $t$ because the $Be^{-t}$ term duplicates a term in $y_c$ . Which statement best evaluates this reasoning?

The reasoning is correct; the entire guess must be multiplied by $t$ to give $Y_p=t(At+B)e^{-t}$ . (correct answer)
The reasoning is partially correct; only the conflicting term should be multiplied by $t$ , giving $Y_p=Ate^{-t} + Bte^{-t}$ .
The reasoning is incorrect; the term $Ate^{-t}$ does not duplicate a term in $y_c$ , so no modification is needed.
The reasoning is correct, but the proposed solution is wrong; the modification rule does not apply to polynomial factors.

Explanation: The student's reasoning is correct. The initial guess for a non-homogeneous term of the form

P_n(t)e^{\alpha t}

(A_n t^n + \dots + A_0)e^{\alpha t}

. In this case,

g(t)=te^{-t}

, so the initial guess is

Y_p = (At+B)e^{-t}

. The complementary solution is

y_c = c_1 e^t + c_2 e^{-t}

. The term

Be^{-t}

in the expansion of the guess,

Ate^{-t}+Be^{-t}

, is a solution to the homogeneous equation. The modification rule states that if any term in the initial guess is a solution to the homogeneous equation, the entire initial guess must be multiplied by

t^k

until no term is a solution. Therefore, the entire expression

(At+B)e^{-t}

must be multiplied by

t

. The correct form is

Y_p = t(At+B)e^{-t} = (At^2+Bt)e^{-t}

Question 4

The complementary solution to $ay'' + by' + cy = g(t)$ is $y_c(t) = c_1e^{2t}\cos(t) + c_2e^{2t}\sin(t)$ . For which of the following functions $g(t)$ would the form of the particular solution be $Y_p(t) = t e^{2t} (A\cos(t) + B\sin(t))$ ?

$g(t) = 7e^{2t}\cos(t)$ (correct answer)
$g(t) = t\cos(t)$
$g(t) = 4e^{2t}$
$g(t) = t e^{2t} \sin(t)$

Explanation: The form of the particular solution

Y_p(t)

has been multiplied by

t

. This indicates that the initial, unmodified guess, which would be

e^{2t}(A\cos(t) + B\sin(t))

, is a solution to the homogeneous equation. The given complementary solution

y_c(t)

confirms this. The method of undetermined coefficients dictates that the initial guess is based on the form of the non-homogeneous term

g(t)

. A term like

g(t) = 7e^{2t}\cos(t)

(or any linear combination of

e^{2t}\cos(t)

and

e^{2t}\sin(t)

) would lead to this initial guess, thus requiring multiplication by

t

Question 5

Consider $y'' + 4y = \sin(2x) + 3\cos(2x) + 5e^{2x}$ . If the particular solution is written as $y_p = A\sin(2x) + B\cos(2x) + Ce^{2x}$ , what happens when this form is substituted into the differential equation?

The coefficients $A$ , $B$ , and $C$ can be uniquely determined since no resonance occurs
Both trigonometric and exponential terms create contradictions requiring complete reformulation of the particular solution
The trigonometric terms work correctly, but the exponential term creates a contradiction requiring modification
The exponential term works correctly, but the trigonometric terms create a contradiction requiring modification (correct answer)

Explanation: When solving non-homogeneous differential equations using the method of undetermined coefficients, you must check whether any terms in your proposed particular solution are already solutions to the homogeneous equation. This is the resonance condition that determines whether your initial guess will work. Let's examine what happens when we substitute

y_p = A\sin(2x) + B\cos(2x) + Ce^{2x}

into

y'' + 4y = \sin(2x) + 3\cos(2x) + 5e^{2x}

. First, check the homogeneous equation

y'' + 4y = 0

. Its characteristic equation is

r^2 + 4 = 0

, giving

r = \pm 2i

. The homogeneous solutions are

\sin(2x)

and

\cos(2x)

. For the exponential term

Ce^{2x}

: Since

e^{2x}

is not a solution to the homogeneous equation (

r = 2

is not a root), this term works perfectly and

C

can be determined uniquely. For the trigonometric terms

A\sin(2x) + B\cos(2x)

: These are solutions to the homogeneous equation, creating resonance. When you substitute these into the differential equation, they satisfy

y'' + 4y = 0

, not the non-zero right-hand side. This creates a contradiction that cannot be resolved. Answer D correctly identifies this situation. Answer A misses the resonance problem entirely. Answer B incorrectly claims both terms fail. Answer C reverses which terms work. Study tip: Always solve the homogeneous equation first and compare its solutions to your particular solution guess. If there's overlap, multiply the problematic terms by

x

(or higher powers) until resonance is eliminated.

Question 6

The equation $y'' + y = \sin(x) + x\cos(x)$ is being solved using undetermined coefficients. If a student attempts to use $y_p = A\sin(x) + B\cos(x) + C x\cos(x) + D x\sin(x)$ as the particular solution form, what issue will arise?

The form is correct and will yield a unique solution for all coefficients
The $x\cos(x)$ and $x\sin(x)$ terms are unnecessary since no resonance occurs for these terms
The $A\sin(x)$ and $B\cos(x)$ terms will lead to contradictory equations that cannot be satisfied (correct answer)
The $C x\cos(x)$ term is sufficient, but the $D x\sin(x)$ term creates an overdetermined system

Explanation: The characteristic equation is

r^2 + 1 = 0

with roots

r = \pm i

, so

y_h = c_1\cos(x) + c_2\sin(x)

. The forcing term

\sin(x)

resonates with

y_h

, requiring

x\sin(x)

and

x\cos(x)

terms. For

x\cos(x)

, we need

x^2\cos(x)

and

x^2\sin(x)

terms due to resonance. The proposed form mixes resonant and non-resonant terms incorrectly. When

A\sin(x) + B\cos(x)

is substituted, these terms cancel with parts of the left side, leading to

0 = \sin(x)

, which is impossible. Choice A ignores this contradiction. Choice B misidentifies which terms cause resonance. Choice D incorrectly describes the mathematical issue.

Question 7

Consider $y'' + 4y' + 4y = te^{-2t} + e^{-2t}\cos(3t)$ . When determining the particular solution using undetermined coefficients, what is the correct total number of undetermined coefficients needed?

5 coefficients (correct answer)
4 coefficients
6 coefficients
7 coefficients

Explanation: When solving nonhomogeneous linear differential equations using undetermined coefficients, you need to carefully analyze each term on the right-hand side and account for any overlap with the homogeneous solution. First, find the homogeneous solution. The characteristic equation

r^2 + 4r + 4 = 0

factors as

(r + 2)^2 = 0

, giving a repeated root

r = -2

. So the homogeneous solution is

y_h = (c_1 + c_2t)e^{-2t}

. Now examine each term in

te^{-2t} + e^{-2t}\cos(3t)

: For

te^{-2t}

: Normally you'd try

Ate^{-2t}

, but since

e^{-2t}

and

te^{-2t}

are already in the homogeneous solution, you must multiply by

t^2

(the multiplicity of the repeated root). This gives

At^3e^{-2t} + Bt^2e^{-2t}

, requiring 2 coefficients. For

e^{-2t}\cos(3t)

: You'd normally try

(C\cos(3t) + D\sin(3t))e^{-2t}

, but since

e^{-2t}

appears in the homogeneous solution, multiply by

t^2

t^2(E\cos(3t) + F\sin(3t))e^{-2t}

, requiring 2 more coefficients. Wait—you also need the

t^2e^{-2t}\sin(3t)

term that appears when differentiating, giving you one additional coefficient:

G

. Total: 5 coefficients. Answer A (5 coefficients) is correct. Answer B (4 coefficients) misses the adjustment for repeated roots. Answer C (6 coefficients) likely double-counts somewhere. Answer D (7 coefficients) overcomplicates the overlap resolution. Remember: always check if your trial solution overlaps with the homogeneous solution—multiply by the appropriate power of

t

to eliminate that overlap.

Question 8

What is the form of a particular solution $Y_p(t)$ for the differential equation $y'' - 6y' + 9y = 5e^{3t}$ ?

$Y_p(t) = Ae^{3t}$
$Y_p(t) = Ate^{3t}$
$Y_p(t) = At^2e^{3t}$ (correct answer)
$Y_p(t) = (At+B)e^{3t}$

Explanation: First, find the complementary solution. The characteristic equation is

r^2 - 6r + 9 = 0

, which factors as

(r-3)^2 = 0

. This gives a repeated root

r=3

. The complementary solution is

y_c(t) = c_1e^{3t} + c_2te^{3t}

. The non-homogeneous term is

g(t) = 5e^{3t}

. The initial guess for the particular solution would be

Y_p(t) = Ae^{3t}

. However, this term is present in the complementary solution. According to the modification rule, we must multiply the guess by

t

, giving

Ate^{3t}

. This new guess is also present in the complementary solution. Therefore, we must multiply by

t

again, which gives the correct form

Y_p(t) = At^2e^{3t}

Question 9

What is the correct form of a particular solution $Y_p(t)$ for the differential equation $y'' + 3y' = 4 - 2e^{-3t}$ ?

$Y_p(t) = A + Be^{-3t}$
$Y_p(t) = At + Be^{-3t}$
$Y_p(t) = A + Bte^{-3t}$
$Y_p(t) = At + Bte^{-3t}$ (correct answer)

Explanation: The characteristic equation is

r^2 + 3r = 0

, or

r(r+3)=0

, giving roots

r=0

and

r=-3

. The complementary solution is

y_c(t) = c_1e^{0t} + c_2e^{-3t} = c_1 + c_2e^{-3t}

. The non-homogeneous term is

g(t)=4 - 2e^{-3t}

. We consider each part. For the constant term 4, the initial guess is

A

. This is a solution to the homogeneous equation (the

c_1

term), so we must modify it to

At

. For the term

-2e^{-3t}

, the initial guess is

Be^{-3t}

. This is also a solution to the homogeneous equation (the

c_2e^{-3t}

term), so we must modify it to

Bte^{-3t}

. Combining these gives the full form for the particular solution:

Y_p(t) = At + Bte^{-3t}

Question 10

The differential equation $y'' + 4y' + 4y = g(t)$ has a particular solution $Y_p(t) = 3t^2e^{-2t}$ . What is the non-homogeneous term $g(t)$ ?

$g(t) = 3t^2e^{-2t}$
$g(t) = 6e^{-2t}$ (correct answer)
$g(t) = (12t^2 - 24t + 6)e^{-2t}$
$g(t) = 6te^{-2t}$

Explanation: To find

g(t)

, we must compute

Y_p'' + 4Y_p' + 4Y_p

. Given

Y_p(t) = 3t^2e^{-2t}

. First derivative:

Y_p' = 6te^{-2t} - 6t^2e^{-2t} = (6t-6t^2)e^{-2t}

. Second derivative:

Y_p'' = (6-12t)e^{-2t} - 2(6t-6t^2)e^{-2t} = (6 - 12t - 12t + 12t^2)e^{-2t} = (12t^2 - 24t + 6)e^{-2t}

. Now substitute into the left side of the ODE:

g(t) = (12t^2 - 24t + 6)e^{-2t} + 4(6t-6t^2)e^{-2t} + 4(3t^2e^{-2t})

. Factor out

e^{-2t}

g(t) = [ (12t^2 - 24t + 6) + (24t - 24t^2) + (12t^2) ] e^{-2t}

. The terms in the brackets simplify:

(12t^2 - 24t^2 + 12t^2) + (-24t + 24t) + 6 = 6

. Therefore,

g(t) = 6e^{-2t}

Question 11

Consider the equation $y'' + 4y = 8 \sin^2(t)$ . Which of the following is the correct form for a particular solution $Y_p(t)$ ?

$Y_p(t) = A\sin^2(t)$
$Y_p(t) = A + B\cos(2t) + C\sin(2t)$
$Y_p(t) = A + t(B\cos(2t) + C\sin(2t))$ (correct answer)
$Y_p(t) = A\cos(2t) + B\sin(2t)$

Explanation: The method of undetermined coefficients cannot be directly applied to

g(t) = 8\sin^2(t)

. We must first use a trigonometric identity to rewrite

g(t)

in a suitable form. Using the power-reducing identity

\sin^2(t) = \frac{1 - \cos(2t)}{2}

, the non-homogeneous term becomes

g(t) = 8 \left( \frac{1 - \cos(2t)}{2} \right) = 4 - 4\cos(2t)

. Now we can determine the form of

Y_p(t)

. The characteristic equation is

r^2+4=0

, with roots

r=\pm 2i

, so

y_c(t) = c_1\cos(2t)+c_2\sin(2t)

. For the constant term 4, the guess is

A

. This does not conflict with

y_c

. For the term

-4\cos(2t)

, the guess is

B\cos(2t)+C\sin(2t)

. This form is present in

y_c

, so we must multiply this part of the guess by

t

. The total form for

Y_p(t)

is the sum of the forms for each part:

Y_p(t) = A + t(B\cos(2t) + C\sin(2t))

Question 12

Find the particular solution $Y_p(t)$ for the differential equation $y'' - y' = 3t^2$ .

$Y_p(t) = -t^3 - 3t^2 - 6t$ (correct answer)
$Y_p(t) = t^3 + 3t^2 + 6t$
$Y_p(t) = -t^3 - 3t^2$
$Y_p(t) = -3t^2 - 6t - 6$

Explanation: The characteristic equation is

r^2 - r = 0

, or

r(r-1)=0

, with roots

r=0, 1

. The complementary solution is

y_c(t) = c_1e^{0t} + c_2e^t = c_1 + c_2e^t

. The non-homogeneous term is

g(t) = 3t^2

. The initial guess for

Y_p

is a full quadratic,

At^2 + Bt + C

. However, the constant term

C

is a solution to the homogeneous equation (corresponding to the

c_1

term). Therefore, we must multiply the entire guess by

t

, giving

Y_p(t) = t(At^2 + Bt + C) = At^3 + Bt^2 + Ct

. We find the derivatives:

Y_p' = 3At^2 + 2Bt + C

and

Y_p'' = 6At + 2B

. Substituting into the ODE:

(6At + 2B) - (3At^2 + 2Bt + C) = 3t^2

. Grouping terms:

-3At^2 + (6A - 2B)t + (2B - C) = 3t^2

. Equating coefficients gives:

-3A = 3 \implies A = -1

6A - 2B = 0 \implies 6(-1) - 2B = 0 \implies B = -3

2B - C = 0 \implies 2(-3) - C = 0 \implies C = -6

. Thus,

Y_p(t) = -t^3 - 3t^2 - 6t

Question 13

Consider the differential equation $y'' + 4y = t + \cos(2t)$ . What is the correct form of the particular solution $Y_p(t)$ that should be used for the method of undetermined coefficients?

$Y_p(t) = At + B + C\cos(2t) + D\sin(2t)$
$Y_p(t) = At + B + t(C\cos(2t) + D\sin(2t))$ (correct answer)
$Y_p(t) = t(At + B + C\cos(2t) + D\sin(2t))$
$Y_p(t) = At + B + Ct\cos(2t) + D\sin(2t)$

Explanation: The characteristic equation for the homogeneous part is

r^2 + 4 = 0

, which has roots

r = \pm 2i

. The complementary solution is

y_c(t) = c_1\cos(2t) + c_2\sin(2t)

. The non-homogeneous term is

g(t) = t + \cos(2t)

. We find the guess for the particular solution by considering each part of

g(t)

separately. For the term

t

, the guess is

At+B

. For the term

\cos(2t)

, the guess is

C\cos(2t) + D\sin(2t)

. We must check if any part of the guess is a solution to the homogeneous equation. The guess

At+B

does not overlap with

y_c

. However, the guess

C\cos(2t) + D\sin(2t)

does overlap with

y_c

. Therefore, this part of the guess must be multiplied by

t

, becoming

t(C\cos(2t) + D\sin(2t))

. The final form for the particular solution is the sum of the individual forms:

Y_p(t) = At + B + t(C\cos(2t) + D\sin(2t))

Question 14

What is the correct form of a particular solution for the differential equation $y'' + 2y' + 5y = 3e^{-t}\cos(2t)$ ?

$Y_p(t) = e^{-t}(A\cos(2t) + B\sin(2t))$
$Y_p(t) = Ate^{-t}\cos(2t)$
$Y_p(t) = te^{-t}(A\cos(2t) + B\sin(2t))$ (correct answer)
$Y_p(t) = A e^{-t} \cos(2t)$

Explanation: The characteristic equation is

r^2 + 2r + 5 = 0

. Using the quadratic formula, the roots are

r = \frac{-2 \pm \sqrt{4-20}}{2} = -1 \pm 2i

. The complementary solution is

y_c(t) = e^{-t}(c_1\cos(2t) + c_2\sin(2t))

. The non-homogeneous term is

g(t) = 3e^{-t}\cos(2t)

. The initial guess for the particular solution, based on the form of

g(t)

, is

Y_p(t) = e^{-t}(A\cos(2t) + B\sin(2t))

. This form is identical to the form of the complementary solution

y_c(t)

. Therefore, the modification rule applies, and the entire initial guess must be multiplied by

t

. The correct form is

Y_p(t) = te^{-t}(A\cos(2t) + B\sin(2t))

Question 15

A particular solution to the differential equation $y'' + y' - 2y = 3e^t$ is $Y_p(t) = te^t$ . If the non-homogeneous term is changed to $g(t) = 6e^t - 2$ , what is the new particular solution?

$Y_p(t) = 2te^t + 1$ (correct answer)
$Y_p(t) = 2te^t - 2$
$Y_p(t) = 6te^t + 1$
$Y_p(t) = te^t + 1$

Explanation: The problem can be solved using the principle of superposition. We are given that for

g_1(t)=3e^t

, the particular solution is

Y_{p1}(t)=te^t

. By linearity, for

g(t)=2g_1(t)=6e^t

, the particular solution will be

2Y_{p1}(t)=2te^t

. Next, we need to find the particular solution for the term

g_2(t)=-2

. The characteristic equation is

r^2+r-2=0

, with roots

r=1, -2

, so

y_c=c_1e^t+c_2e^{-2t}

. For

g_2(t)=-2

, our guess is

Y_{p2}(t)=A

. Since a constant is not part of

y_c

, no modification is needed. Substituting

Y_{p2}=A

into the ODE gives

0+0-2A = -2

, so

A=1

. By superposition, the total particular solution is

Y_p(t) = 2te^t + 1

Question 16

For the equation $y'' - 2y' + y = e^x(3x^2 + 4x)$ , a student proposes the particular solution $y_p = e^x(Ax^2 + Bx + C)$ . What is the primary error in this approach?

The polynomial degree should be increased by one to account for the forcing term structure
A cosine and sine term should be included since complex roots are present in the characteristic equation
The exponential factor should be $e^{2x}$ instead of $e^x$ to match the characteristic roots
The entire expression should be multiplied by $x$ due to resonance with the homogeneous solution (correct answer)

Explanation: When solving non-homogeneous linear differential equations using the method of undetermined coefficients, you must check whether your proposed particular solution overlaps with the homogeneous solution. This overlap creates a critical issue called resonance. First, let's find the homogeneous solution. The characteristic equation

r^2 - 2r + 1 = 0

factors as

(r-1)^2 = 0

, giving a repeated root

r = 1

. This means the homogeneous solution is

y_h = (C_1 + C_2x)e^x

, which contains terms of the form

e^x

and

xe^x

. The proposed particular solution

y_p = e^x(Ax^2 + Bx + C)

contains the term

Ce^x

, which directly overlaps with the homogeneous solution. When you substitute this into the differential equation, the overlapping terms will vanish, making it impossible to solve for the coefficients. To fix this resonance problem, you must multiply the entire proposed solution by

x

raised to the power equal to the multiplicity of the repeated root. Since we have a repeated root of multiplicity 2, we multiply by

x^2

, giving

y_p = x^2e^x(Ax^2 + Bx + C)

. Choice A is incorrect because the polynomial degree is already appropriate for the forcing term

3x^2 + 4x

. Choice B is wrong since the characteristic equation has real, not complex roots. Choice C is incorrect because the exponential factor

e^x

correctly matches the forcing term's exponential part. Key strategy: Always solve the homogeneous equation first and check for resonance before finalizing your particular solution form.

Question 17

Consider the differential equation $y'' - 4y' + 4y = 8e^{2x} + 12x$ . When applying the method of undetermined coefficients, what form should the particular solution take?

$y_p = Ae^{2x} + Bx + C$
$y_p = Ax^2e^{2x} + Bx + C$ (correct answer)
$y_p = Axe^{2x} + Bx^2 + Cx$
$y_p = Ae^{2x} + Bx^2 + Cx$

Explanation: The homogeneous equation has characteristic equation

r^2 - 4r + 4 = 0

, giving

(r-2)^2 = 0

, so

r = 2

is a repeated root. The homogeneous solution is

y_h = (c_1 + c_2x)e^{2x}

. For the forcing term

8e^{2x}

, since

e^{2x}

appears in the homogeneous solution with multiplicity 2, we multiply by

x^2

to get

Ax^2e^{2x}

. For

12x

, we use

Bx + C

. Choice A fails to account for the repeated root. Choice C incorrectly uses

Bx^2 + Cx

for the polynomial term. Choice D uses

Bx^2 + Cx

unnecessarily for the polynomial and doesn't multiply

e^{2x}

x^2