Differential Equations Quiz: Superposition And Particular Solutions

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Superposition And Particular SolutionsQuestion 1 of 9

Consider the differential equation $y'' + 4y = 2\cos(2x) + 3e^x$ . Given that $y_{p1} = \frac{1}{4}x\sin(2x)$ is a particular solution to $y'' + 4y = 2\cos(2x)$ and $y_{p2} = \frac{3}{5}e^x$ is a particular solution to $y'' + 4y = 3e^x$ , which statement about the superposition principle is correct?

The function

y_p = \frac{1}{4}x\sin(2x) + \frac{3}{5}e^x

satisfies the original equation due to linearity of the differential operator

The function

y_p = \frac{1}{4}x\sin(2x) \cdot \frac{3}{5}e^x

satisfies the original equation because particular solutions multiply under superposition

The superposition principle applies only to homogeneous solutions, so we cannot combine these particular solutions

The function

y_p = \max\{\frac{1}{4}x\sin(2x), \frac{3}{5}e^x\}

satisfies the original equation by the maximum principle for linear operators

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Differential Equations Quiz

Differential Equations Quiz: Superposition And Particular Solutions

Practice Superposition And Particular Solutions 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 Superposition And Particular Solutions, 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

The function $y_p = \frac{1}{4}x\sin(2x) + \frac{3}{5}e^x$ satisfies the original equation due to linearity of the differential operator (correct answer)
The function $y_p = \frac{1}{4}x\sin(2x) \cdot \frac{3}{5}e^x$ satisfies the original equation because particular solutions multiply under superposition
The superposition principle applies only to homogeneous solutions, so we cannot combine these particular solutions
The function $y_p = \max\{\frac{1}{4}x\sin(2x), \frac{3}{5}e^x\}$ satisfies the original equation by the maximum principle for linear operators

Explanation: The superposition principle for linear differential equations states that if

L[y_1] = f_1

and

L[y_2] = f_2

, then

L[y_1 + y_2] = f_1 + f_2

. Since our equation has the form

L[y] = 2\cos(2x) + 3e^x

, the particular solution is the sum

y_{p1} + y_{p2}

. Choice B incorrectly suggests multiplication of solutions. Choice C incorrectly limits superposition to homogeneous solutions. Choice D invents a non-existent maximum principle.

Question 2

For the differential equation $y'' - 2y' - 3y = 4e^{3x} + 2e^{-x}$ , the homogeneous solution is $y_h = c_1e^{3x} + c_2e^{-x}$ . A student finds particular solutions $y_{p1} = 2xe^{3x}$ for the equation $y'' - 2y' - 3y = 4e^{3x}$ and $y_{p2} = -xe^{-x}$ for $y'' - 2y' - 3y = 2e^{-x}$ . What can be concluded about the student's work?

Both particular solutions are correct, and $y_p = 2xe^{3x} - xe^{-x}$ solves the original equation (correct answer)
The first particular solution is correct, but the second should be $y_{p2} = \frac{1}{2}e^{-x}$ since $e^{-x}$ doesn't create resonance
Both particular solutions show incorrect resonance handling; they should be $y_{p1} = \frac{1}{2}e^{3x}$ and $y_{p2} = -\frac{1}{4}e^{-x}$
The second particular solution is correct, but the first should be $y_{p1} = \frac{2}{3}e^{3x}$ without the $x$ factor

Explanation: Since both

e^{3x}

and

e^{-x}

appear in the homogeneous solution, resonance occurs for both terms. The student correctly used

xe^{3x}

and

xe^{-x}

forms. We can verify: for

y_{p1} = 2xe^{3x}

, we get

y_{p1}'' - 2y_{p1}' - 3y_{p1} = 4e^{3x}

(correct). For

y_{p2} = -xe^{-x}

, we get

y_{p2}'' - 2y_{p2}' - 3y_{p2} = 2e^{-x}

(correct). By superposition,

y_p = y_{p1} + y_{p2} = 2xe^{3x} - xe^{-x}

solves the original equation. Choices B, C, and D incorrectly analyze the resonance conditions.

Question 3

The differential equation $y'' + y = \cos(x) + x\sin(x)$ has homogeneous solution $y_h = c_1\cos(x) + c_2\sin(x)$ . If $y_{p1} = \frac{1}{2}x\sin(x)$ is a particular solution to $y'' + y = \cos(x)$ and $y_{p2} = -\frac{1}{4}x^2\cos(x)$ is a particular solution to $y'' + y = x\sin(x)$ , which analysis of the complete particular solution is correct?

$y_p = \frac{1}{2}x\sin(x) - \frac{1}{4}x^2\cos(x)$ by direct superposition of the given particular solutions
The given particular solutions cannot be combined because they were derived for different forcing functions
$y_p = \frac{1}{2}x\sin(x) - \frac{1}{4}x^2\cos(x)$ but only after verifying each particular solution independently satisfies its respective equation (correct answer)
The particular solution requires recalculation using $y_p = Ax\cos(x) + Bx\sin(x) + Cx^2\cos(x) + Dx^2\sin(x)$ for the combined forcing function

Explanation: While superposition allows us to add particular solutions for different forcing terms, we should verify each given solution before applying the principle. Checking

y_{p1} = \frac{1}{2}x\sin(x)

y_{p1}'' + y_{p1} = -\frac{1}{2}x\sin(x) + \cos(x) + \frac{1}{2}x\sin(x) = \cos(x)

✓. Checking

y_{p2} = -\frac{1}{4}x^2\cos(x)

y_{p2}'' + y_{p2} = \frac{1}{2}\cos(x) + \frac{1}{2}x\sin(x) - \frac{1}{4}x^2\cos(x) = x\sin(x)

✓. Both solutions are correct, so superposition gives the answer. Choice A assumes correctness without verification. Choice B incorrectly rejects superposition. Choice D unnecessarily complicates the approach.

Question 4

For the equation $y'' - y = f(x)$ , suppose $\{y_1, y_2, y_3\}$ are three different particular solutions corresponding to forcing functions $\{f_1(x), f_2(x), f_3(x)\}$ respectively. If $f(x) = 2f_1(x) - f_2(x) + 3f_3(x)$ , which statement about constructing a particular solution for $y'' - y = f(x)$ is most accurate?

$y_p = 2y_1 - y_2 + 3y_3$ is a particular solution only if $y_1$ , $y_2$ , and $y_3$ are linearly independent functions
$y_p = 2y_1 - y_2 + 3y_3$ is the unique particular solution since the linear combination uniquely determines the result
The particular solution must be found independently using undetermined coefficients because superposition only applies to homogeneous solutions
$y_p = 2y_1 - y_2 + 3y_3$ is a particular solution, and any other particular solution differs from this by a solution to the homogeneous equation (correct answer)

Explanation: When dealing with non-homogeneous linear differential equations, understanding the superposition principle is crucial. This principle states that if you have particular solutions to equations with different forcing functions, you can combine them linearly to solve equations with combined forcing functions. Since

y_1

y_2

, and

y_3

are particular solutions to

y'' - y = f_1(x)

y'' - y = f_2(x)

, and

y'' - y = f_3(x)

respectively, the linearity of the differential operator means that

y_p = 2y_1 - y_2 + 3y_3

satisfies:

(2y_1 - y_2 + 3y_3)'' - (2y_1 - y_2 + 3y_3) = 2f_1(x) - f_2(x) + 3f_3(x) = f(x)

This confirms

y_p

is indeed a particular solution. However, particular solutions to non-homogeneous equations are never unique—any two particular solutions differ by a homogeneous solution. Answer D correctly captures both facts:

y_p = 2y_1 - y_2 + 3y_3

works as a particular solution, and the general solution is

y_p + y_h

where

y_h

solves the homogeneous equation. Answer A incorrectly requires linear independence—superposition works regardless of whether the particular solutions are linearly independent. Answer B falsely claims uniqueness when particular solutions are never unique for non-homogeneous equations. Answer C misunderstands superposition, claiming it only applies to homogeneous solutions when it actually applies to the entire linear operator. Study tip: Remember that for linear differential equations, superposition works for both forcing functions and their corresponding particular solutions. Particular solutions are never unique—they always differ by homogeneous solutions.

Question 5

A student solving $y'' + 2y' + y = e^{-x} + x^2$ finds the homogeneous solution $y_h = (c_1 + c_2 x)e^{-x}$ and attempts to find particular solutions separately. For $y'' + 2y' + y = e^{-x}$ , they propose $y_{p1} = Ae^{-x}$ , and for $y'' + 2y' + y = x^2$ , they propose $y_{p2} = Bx^2 + Cx + D$ . What is the most significant error in this approach?

The polynomial terms in $y_{p2}$ will interfere with the exponential terms in $y_{p1}$ when combined via superposition
The form $y_{p2} = Bx^2 + Cx + D$ is unnecessarily complex since the forcing function $x^2$ suggests only $y_{p2} = Bx^2$
Both proposed forms are incorrect because superposition requires solving for the combined forcing function directly
The form $y_{p1} = Ae^{-x}$ fails to account for resonance since $e^{-x}$ appears in the homogeneous solution (correct answer)

Explanation: When solving non-homogeneous linear differential equations, the method of undetermined coefficients requires careful attention to resonance - the situation where your proposed particular solution overlaps with the homogeneous solution. The critical issue here is with the proposed form

y_{p1} = Ae^{-x}

for the forcing function

e^{-x}

. Since the homogeneous solution is

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

, the term

e^{-x}

already appears in the homogeneous solution. When resonance occurs, you must multiply your initial guess by the lowest power of

x

that eliminates the overlap. Since both

e^{-x}

and

xe^{-x}

appear in

y_h

, the correct form should be

y_{p1} = Ax^2e^{-x}

. Choice A is wrong because superposition works perfectly fine when combining particular solutions - polynomial and exponential terms don't "interfere" with each other. Choice B is incorrect because when the forcing function is a polynomial, you need the complete polynomial of that degree and all lower degrees due to how derivatives work. Choice C misunderstands superposition entirely - you can absolutely solve for particular solutions separately and add them together. Choice D correctly identifies that

y_{p1} = Ae^{-x}

fails due to resonance with the homogeneous solution. Key strategy: Always check if your proposed particular solution has any terms that appear in the homogeneous solution. If so, multiply by the appropriate power of

x

to eliminate the overlap before proceeding with undetermined coefficients.

Question 6

Consider the system where $L[y] = y'' + 4y' + 4y$ and we know that $L[y_1] = e^x$ and $L[y_2] = x e^x$ for specific functions $y_1$ and $y_2$ . If we want to solve $L[y] = 3e^x - 2xe^x$ , which combination of $y_1$ and $y_2$ provides a particular solution?

$y_p = -3y_1 + 2y_2$ by reversing the signs to account for the operator's effect on linear combinations
$y_p = y_1 + y_2$ since the particular solution must include both exponential structures regardless of coefficients
$y_p = 3y_1 - 2y_2$ by direct linear combination matching the forcing function coefficients (correct answer)
$y_p = 3y_1 - 2y_2$ but only after adding the homogeneous solution $c_1e^{-2x} + c_2xe^{-2x}$ to form the complete solution

Explanation: When you encounter a linear differential operator problem like this, you're working with the principle of superposition. Since the operator

L

is linear, you can combine known solutions to match any linear combination of forcing functions. The key insight is that if

L[y_1] = e^x

and

L[y_2] = xe^x

, then by linearity,

L[ay_1 + by_2] = aL[y_1] + bL[y_2] = ae^x + bxe^x

. To solve

L[y] = 3e^x - 2xe^x

, you need to find coefficients

a

and

b

such that

ae^x + bxe^x = 3e^x - 2xe^x

. Matching coefficients directly: the coefficient of

e^x

gives

a = 3

, and the coefficient of

xe^x

gives

b = -2

. Therefore,

y_p = 3y_1 - 2y_2

is your particular solution. Choice A incorrectly reverses signs, misunderstanding how linear operators work with combinations. The operator doesn't require sign reversal—it preserves the linear relationship directly. Choice B ignores the specific coefficients needed, assuming any combination works regardless of the forcing function's structure. Choice D correctly identifies the particular solution but incorrectly suggests you must add the homogeneous solution; the question asks specifically for the particular solution, not the general solution. Study tip: For linear operator problems, always use direct coefficient matching. The linearity property

L[ay_1 + by_2] = aL[y_1] + bL[y_2]

means you can build solutions by matching coefficients term-by-term without any sign manipulation tricks.

Question 7

The general solution to the homogeneous equation $y'' - 4y' + 4y = 0$ is $y_h = (c_1 + c_2 x)e^{2x}$ . For the nonhomogeneous equation $y'' - 4y' + 4y = e^{2x}$ , a student proposes the particular solution form $y_p = Ae^{2x}$ . What is the fundamental issue with this approach?

The proposed form will lead to a contradiction when substituted because $e^{2x}$ satisfies the homogeneous equation exactly (correct answer)
The coefficient $A$ cannot be determined uniquely since the system of equations will be inconsistent
The particular solution must include both exponential and polynomial terms to match the forcing function
The method of undetermined coefficients requires the particular solution to be orthogonal to all homogeneous solutions

Explanation: Since

e^{2x}

is part of the homogeneous solution (when

c_1 = 1, c_2 = 0

), substituting

y_p = Ae^{2x}

into the nonhomogeneous equation yields

0 = e^{2x}

, which is impossible. The correct form should be

y_p = Ax^2e^{2x}

since both

e^{2x}

and

xe^{2x}

appear in

y_h

. Choice B is incorrect because the issue is contradiction, not indeterminacy. Choice C misunderstands the structure needed. Choice D invents a false orthogonality requirement.

Question 8

Consider the nonhomogeneous linear differential equation $y'' - 3y' + 2y = 6e^{2x} + 4\sin(x)$ . If $y_h = c_1 e^x + c_2 e^{2x}$ is the general solution to the homogeneous equation, which of the following represents the correct form for a particular solution using the method of undetermined coefficients?

$y_p = Ae^{2x} + B\sin(x) + C\cos(x)$
$y_p = Axe^{2x} + B\sin(x) + C\cos(x)$ (correct answer)
$y_p = Ae^{2x} + B\sin(x)$
$y_p = Axe^{2x} + Bx\sin(x) + Cx\cos(x)$

Explanation: Since

e^{2x}

appears in the homogeneous solution, we must multiply the assumed form

Ae^{2x}

x

to get

Axe^{2x}

. For the

\sin(x)

term, since neither

\sin(x)

nor

\cos(x)

appears in the homogeneous solution, we use

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

without modification. Choice A fails to account for the resonance with

e^{2x}

. Choice C omits the necessary

\cos(x)

term. Choice D incorrectly multiplies the trigonometric terms by

x

when no resonance exists.

Question 9

The functions $y_1(x) = e^{-x}\cos(2x)$ and $y_2(x) = e^{-x}\sin(2x)$ are solutions to a homogeneous linear differential equation. If $y_3(x) = x^2 e^{-x}$ is a particular solution to the nonhomogeneous equation $L[y] = f(x)$ , what is the general solution to $L[y] = 3f(x)$ ?

$y = c_1 e^{-x}\cos(2x) + c_2 e^{-x}\sin(2x) + x^2 e^{-x}$
$y = c_1 e^{-x}\cos(2x) + c_2 e^{-x}\sin(2x) + 3x^2 e^{-x}$ (correct answer)
$y = 3c_1 e^{-x}\cos(2x) + 3c_2 e^{-x}\sin(2x) + 3x^2 e^{-x}$
$y = c_1 e^{-x}\cos(2x) + c_2 e^{-x}\sin(2x) + 9x^2 e^{-x}$

Explanation: By the linearity of differential operators, if

y_3

is a particular solution to

L[y] = f(x)

, then

3y_3

is a particular solution to

L[y] = 3f(x)

. The general solution is the sum of the general homogeneous solution and any particular solution, so we get

y = c_1 e^{-x}\cos(2x) + c_2 e^{-x}\sin(2x) + 3x^2 e^{-x}

. Choice A uses the wrong particular solution. Choice C incorrectly scales the homogeneous solution. Choice D incorrectly squares the scaling factor.