Differential Equations Quiz: Superposition And Particular Solutions

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.

Question 1

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)
2. 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
3. The superposition principle applies only to homogeneous solutions, so we cannot combine these particular solutions
4. 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

1. Both particular solutions are correct, and $y_p = 2xe^{3x} - xe^{-x}$ solves the original equation (correct answer)
2. 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
3. Both particular solutions show incorrect resonance handling; they should be $y_{p1} = \frac{1}{2}e^{3x}$ and $y_{p2} = -\frac{1}{4}e^{-x}$
4. 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

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

1. $y_p = 2y_1 - y_2 + 3y_3$ is a particular solution only if $y_1$, $y_2$, and $y_3$ are linearly independent functions
2. $y_p = 2y_1 - y_2 + 3y_3$ is the unique particular solution since the linear combination uniquely determines the result
3. The particular solution must be found independently using undetermined coefficients because superposition only applies to homogeneous solutions
4. $y_p = 2y_1 - y_2 + 3y_3$ 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

1. The polynomial terms in $y_{p2}$ will interfere with the exponential terms in $y_{p1}$ when combined via superposition
2. The form $y_{p2} = Bx^2 + Cx + D$ is unnecessarily complex since the forcing function $x^2$ suggests only $y_{p2} = Bx^2$
3. Both proposed forms are incorrect because superposition requires solving for the combined forcing function directly
4. 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

1. $y_p = -3y_1 + 2y_2$ by reversing the signs to account for the operator's effect on linear combinations
2. $y_p = y_1 + y_2$ since the particular solution must include both exponential structures regardless of coefficients
3. $y_p = 3y_1 - 2y_2$ by direct linear combination matching the forcing function coefficients (correct answer)
4. $y_p = 3y_1 - 2y_2$ 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

1. The proposed form will lead to a contradiction when substituted because $e^{2x}$ satisfies the homogeneous equation exactly (correct answer)
2. The coefficient $A$ cannot be determined uniquely since the system of equations will be inconsistent
3. The particular solution must include both exponential and polynomial terms to match the forcing function
4. 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

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

Explanation: Since $e^{2x}$ appears in the homogeneous solution, we must multiply the assumed form $Ae^{2x}$ by $x$ 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

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

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.