Differential Equations Quiz: Variation Of Parameters

What this quiz covers

This quiz focuses on Variation Of Parameters, 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. $u_1'(x) = -e^{2x}$ (correct answer)
2. $u_1'(x) = -x e^{2x}$
3. $u_1'(x) = e^{2x}$
4. $u_1'(x) = -(1+x)e^{2x}$

Explanation: First, write the equation in standard form $y'' + P(x)y' + Q(x)y = g(x)$ by dividing by $x$: $y'' - \frac{1+x}{x} y' + \frac{1}{x} y = x e^{2x}$. Thus, $g(x) = x e^{2x}$. The solutions to the homogeneous equation are $y_1 = 1+x$ and $y_2 = e^x$. The Wronskian is $W(y_1, y_2) = y_1 y_2' - y_1' y_2 = (1+x)e^x - (1)e^x = xe^x$. The formula for $u_1'$ is $u_1' = -\frac{y_2 g(x)}{W}$. Substituting the known functions, we get $u_1' = -\frac{e^x (x e^{2x})}{xe^x} = -e^{2x}$.

Question 2

1. $\frac{1}{x}$ (correct answer)
2. $-1$
3. $1$
4. $\frac{e^{2x}}{x}$

Explanation: The homogeneous solutions are $y_1 = e^x$ and $y_2 = xe^x$. The equation is in standard form with $g(x) = \frac{e^x}{x}$. The Wronskian is $W(y_1, y_2) = y_1 y_2' - y_1' y_2 = e^x(e^x + xe^x) - e^x(xe^x) = e^{2x}$. The integrand for $u_2(x)$ is $u_2'(x) = \frac{y_1 g(x)}{W}$. Substituting the functions, we get $u_2' = \frac{e^x (e^x/x)}{e^{2x}} = \frac{e^{2x}/x}{e^{2x}} = \frac{1}{x}$.

Question 3

1. $2xe^{3x} - \frac{1}{3}e^{3x}$
2. $-2xe^{3x}$
3. $2xe^{3x}$ (correct answer)
4. $2e^{3x}$

Explanation: When you encounter a non-homogeneous differential equation where the forcing function appears in your fundamental set of solutions, you need the method of undetermined coefficients with modification. Since the forcing function $12e^{3x}$ contains $e^{3x}$, which matches $y_1 = e^{3x}$ from your homogeneous solutions, you must multiply your usual guess by $x$ to avoid duplication. For $y'' - 9y = 12e^{3x}$, your particular solution guess should be $y_p = Axe^{3x}$. Taking derivatives: $y_p' = Ae^{3x} + 3Axe^{3x}$ and $y_p'' = 6Ae^{3x} + 9Axe^{3x}$. Substituting into the differential equation: $6Ae^{3x} + 9Axe^{3x} - 9(Axe^{3x}) = 12e^{3x}$ The $9Axe^{3x}$ terms cancel, leaving $6Ae^{3x} = 12e^{3x}$, so $A = 2$. Therefore, $y_p = 2xe^{3x}$, which is answer C. Answer A $2xe^{3x} - \frac{1}{3}e^{3x}$ incorrectly adds an $e^{3x}$ term, which would be part of the homogeneous solution, not the particular solution. Answer B $-2xe^{3x}$ has the wrong sign—likely from a calculation error when solving for the coefficient. Answer D $2e^{3x}$ forgets the crucial $x$ modification needed when the forcing function matches a homogeneous solution. Key strategy: Always check if your forcing function appears in the homogeneous solutions. If it does, multiply your guess by $x$ (or $x^2$ if needed for repeated roots).

Question 4

1. $\frac{1}{6}(\ln x)^3$
2. $\frac{1}{2}x^2 (\ln x)^2$
3. $\frac{5}{6}x^2 (\ln x)^3$
4. $\frac{1}{6}x^2 (\ln x)^3$ (correct answer)

Explanation: When you encounter a Cauchy-Euler equation with a non-homogeneous term, you need to find a particular solution using the method of undetermined coefficients or variation of parameters. Since one of the homogeneous solutions is $y_2 = x^2 \ln(x)$, which matches the form of our forcing function $x^2 \ln(x)$, we have a resonance case that requires multiplying our trial solution by an additional $\ln(x)$ factor. For the forcing function $x^2 \ln(x)$, we would normally try $y_p = Ax^2 \ln(x)$, but since this matches $y_2$, we must try $y_p = Ax^2 (\ln x)^2$. However, this still creates linear dependence issues with our homogeneous solutions, so we need $y_p = Ax^2 (\ln x)^3$. Substituting this form into the differential equation and solving for the coefficient $A$, we find $A = \frac{1}{6}$, giving us $y_p = \frac{1}{6}x^2 (\ln x)^3$. This makes (D) correct. (A) $\frac{1}{6}(\ln x)^3$ lacks the $x^2$ factor needed to match the structure of the forcing function. (B) $\frac{1}{2}x^2 (\ln x)^2$ has the wrong coefficient and insufficient logarithmic power. (C) $\frac{5}{6}x^2 (\ln x)^3$ has the correct form but wrong coefficient—when substituted back into the equation, it won't satisfy the differential equation. Study tip: In resonance cases with Cauchy-Euler equations, always multiply your trial solution by enough logarithmic factors to ensure linear independence from all homogeneous solutions. The coefficient must then be determined by substitution.

Question 5

1. It simplifies the system of equations by making it diagonal, allowing $u_1'$ and $u_2'$ to be solved for independently.
2. It is a necessary condition to ensure that the Wronskian of $y_1$ and $y_2$ is non-zero.
3. It forces the particular solution $y_p$ to be linearly independent from the homogeneous solutions $y_1$ and $y_2$.
4. It simplifies the derivative of $y_p$, which prevents second derivatives of $u_1$ and $u_2$ from appearing in the calculations. (correct answer)

Explanation: The method of variation of parameters requires careful handling of derivatives to avoid unnecessarily complex calculations. When you assume a particular solution $y_p = u_1(x)y_1(x) + u_2(x)y_2(x)$, taking derivatives becomes the key challenge. Without any constraints, the first derivative would be $y_p' = u_1'y_1 + u_1y_1' + u_2'y_2 + u_2y_2'$. Taking the second derivative would then introduce terms like $u_1''y_1$ and $u_2''y_2$, making the system much more complicated since you'd have second derivatives of the unknown functions. The condition $u_1'y_1 + u_2'y_2 = 0$ eliminates the problematic terms. With this constraint, $y_p' = u_1y_1' + u_2y_2'$, and the second derivative becomes $y_p'' = u_1'y_1' + u_1y_1'' + u_2'y_2' + u_2y_2''$. This keeps the system manageable with only first derivatives of $u_1$ and $u_2$, which is exactly what answer D describes. Answer A is incorrect because the system doesn't become diagonal—you still have a coupled 2×2 system to solve. Answer B confuses cause and effect; the Wronskian being non-zero is a requirement for the homogeneous solutions, not something this condition ensures. Answer C is wrong because the particular solution is supposed to be different from the homogeneous solutions—linear independence isn't the goal here. Study tip: Remember that variation of parameters is all about managing complexity. The seemingly arbitrary condition $u_1'y_1 + u_2'y_2 = 0$ is actually a clever mathematical trick to keep derivatives tractable.

Question 6

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

Explanation: The method of undetermined coefficients is applicable only when the non-homogeneous term $g(x)$ is a polynomial, exponential, sine, cosine, or a finite sum/product of these. In choice C, $g(x) = \frac{1}{1+e^x}$ is not of this form. Therefore, variation of parameters is required. For choices A, B, and D, the non-homogeneous terms ($x^2 e^{-x}$, $\sin(x) + e^{2x}$, and $3\cos(2x)$) are all of the correct form for the method of undetermined coefficients to be applied, although A and D would require using the modification rule.

Question 7

1. $\frac{7}{6} - 4\ln(2)$
2. $4\ln(2) - \frac{7}{6}$ (correct answer)
3. $4\ln(2) - \frac{4}{3}$
4. $\ln(2) - \frac{1}{3}$

Explanation: When you encounter a second-order linear differential equation with constant coefficients and a non-homogeneous term, you need to find both the homogeneous solution and a particular solution, then apply initial conditions. First, solve the homogeneous equation $y'' - y' - 2y = 0$. The characteristic equation is $r^2 - r - 2 = 0$, which factors as $(r-2)(r+1) = 0$, giving roots $r = 2$ and $r = -1$. So the homogeneous solution is $y_h = c_1e^{2x} + c_2e^{-x}$. For the particular solution, since the right side is $3e^{2x}$ and $e^{2x}$ is already in the homogeneous solution, you must use the method of undetermined coefficients with modification. Try $y_p = Axe^{2x}$. Substituting into the differential equation yields $A = 1$, so $y_p = xe^{2x}$. The general solution is $y = c_1e^{2x} + c_2e^{-x} + xe^{2x}$. Applying $y(0) = 0$ gives $c_1 + c_2 = 0$. From $y'(0) = 0$, you get $2c_1 - c_2 + 1 = 0$. Solving these gives $c_1 = -\frac{1}{3}$ and $c_2 = \frac{1}{3}$. Therefore, $y(x) = -\frac{1}{3}e^{2x} + \frac{1}{3}e^{-x} + xe^{2x}$. At $x = \ln 2$: $y(\ln 2) = -\frac{1}{3} \cdot 4 + \frac{1}{3} \cdot \frac{1}{2} + \ln(2) \cdot 4 = -\frac{4}{3} + \frac{1}{6} + 4\ln(2) = 4\ln(2) - \frac{7}{6}$. Answer A has the terms with opposite signs. Answers C and D contain computational errors in combining fractions or applying initial conditions. Remember: when the forcing function matches a homogeneous solution term, multiply your particular solution guess by $x$ to avoid linear dependence.

Question 8

1. $\frac{x^5}{15}$
2. $\frac{x^4}{8}$ (correct answer)
3. $\frac{x^4}{4}$
4. $\frac{x^3}{3}$

Explanation: When you encounter variation of parameters problems, you're finding a particular solution to a non-homogeneous differential equation using the known homogeneous solutions. The method involves constructing $y_p = u_1y_1 + u_2y_2$ where $u_1$ and $u_2$ are functions to be determined. For $xy'' - y' = x^3$ with homogeneous solutions $y_1 = 1$ and $y_2 = x^2$, you need to solve the system: $u_1'y_1 + u_2'y_2 = 0$ $u_1'y_1' + u_2'y_2' = \frac{x^3}{x} = x^2$ Substituting $y_1 = 1$, $y_1' = 0$, $y_2 = x^2$, and $y_2' = 2x$: $u_1' + u_2'x^2 = 0$ $u_2'(2x) = x^2$ From the second equation: $u_2' = \frac{x}{2}$, so $u_2 = \frac{x^2}{4}$ From the first equation: $u_1' = -u_2'x^2 = -\frac{x^3}{2}$, so $u_1 = -\frac{x^4}{8}$ Therefore: $y_p = u_1y_1 + u_2y_2 = -\frac{x^4}{8}(1) + \frac{x^2}{4}(x^2) = -\frac{x^4}{8} + \frac{x^4}{4} = \frac{x^4}{8}$ Choice A ($\frac{x^5}{15}$) results from integration errors or incorrect setup. Choice C ($\frac{x^4}{4}$) comes from forgetting the negative term from $u_1$. Choice D ($\frac{x^3}{3}$) suggests attempting undetermined coefficients incorrectly or algebraic mistakes. The correct answer is B: $\frac{x^4}{8}$. Study tip: In variation of parameters, always double-check your Wronskian calculation and be meticulous with the integration steps—small algebraic errors compound quickly in this method.

Question 9

1. $-\frac{4}{3x}$
2. $4x - 4x \ln(x)$
3. $\frac{4}{3}x \ln(x) - \frac{4}{3}x$
4. $\frac{4}{3}x - \frac{4}{3}x \ln(x)$ (correct answer)

Explanation: When you encounter a non-homogeneous differential equation like this, the method of variation of parameters is your tool for finding a particular solution. You'll use the fundamental solutions from the homogeneous equation and vary their constants. First, solve the homogeneous equation $y'' + y' - 2y = 0$. The characteristic equation $r^2 + r - 2 = 0$ factors as $(r-1)(r+2) = 0$, giving $r = 1, -2$. So your fundamental solutions are $y_1 = e^x$ and $y_2 = e^{-2x}$. For variation of parameters, you need the Wronskian: $W = y_1 y_2' - y_1' y_2 = e^x(-2e^{-2x}) - e^x(e^{-2x}) = -3e^{-x}$. The formula gives $u_2'(x) = \frac{y_1(x) \cdot g(x)}{W(x)} = \frac{e^x \cdot 4e^{-2x}\ln(x)}{-3e^{-x}} = \frac{4e^{-x}\ln(x)}{-3e^{-x}} = -\frac{4}{3}\ln(x)$. Integrating by parts: $u_2(x) = -\frac{4}{3}\int \ln(x)dx = -\frac{4}{3}[x\ln(x) - x] = \frac{4}{3}x - \frac{4}{3}x\ln(x)$. This matches answer choice D. Choice A gives just $-\frac{4}{3x}$, which would come from incorrectly integrating $\ln(x)$. Choice B has the wrong sign and missing coefficient. Choice C has the correct terms but wrong signs throughout. Study tip: Always double-check your integration by parts formula: $\int \ln(x)dx = x\ln(x) - x + C$. The negative sign in front often causes sign errors in variation of parameters problems.

Question 10

1. $\cos(x) \int \sin(x) dx + \sin(x) \int -\sin(x)\tan(x) dx$
2. $\cos(x) \int \sin(x)\tan(x) dx + \sin(x) \int \sin(x) dx$
3. $\cos(x) \int -\sin(x)\tan(x) dx + \sin(x) \int \sin(x) dx$ (correct answer)
4. $-\cos(x) \int \sin(x)\tan(x) dx + \sin(x) \int \sin(x) dx$

Explanation: When you encounter a second-order linear non-homogeneous differential equation, variation of parameters is a systematic method for finding particular solutions. The technique involves assuming your particular solution has the form $y_p = u_1(x)y_1(x) + u_2(x)y_2(x)$, where $y_1 = \cos(x)$ and $y_2 = \sin(x)$ are your homogeneous solutions. The variation of parameters formula gives you: $u_1'(x) = -\frac{y_2(x)f(x)}{W(x)}$ and $u_2'(x) = \frac{y_1(x)f(x)}{W(x)}$ where $f(x) = \tan(x)$ is your non-homogeneous term and $W(x)$ is the Wronskian. For $y_1 = \cos(x)$ and $y_2 = \sin(x)$, the Wronskian equals 1. Therefore: $u_1'(x) = -\sin(x)\tan(x)$ and $u_2'(x) = \cos(x)\tan(x) = \sin(x)$ This gives $y_p = \cos(x) \int -\sin(x)\tan(x) dx + \sin(x) \int \sin(x) dx$, which is answer C. Answer A incorrectly uses $\int \sin(x) dx$ for the first integral instead of the required $\int -\sin(x)\tan(x) dx$. Answer B has the signs wrong in both integrals - it uses positive $\sin(x)\tan(x)$ in the first term instead of negative. Answer D correctly identifies the integrands but incorrectly places a negative sign in front of the entire first term rather than just within the integral. Remember: variation of parameters requires careful attention to the formula's signs and the order of the functions - always double-check your Wronskian calculation and substitute methodically.

Question 11

1. $\begin{cases} u_1' \cos(2x) + u_2' \sin(2x) = 0 \\ -2u_1' \sin(2x) + 2u_2' \cos(2x) = \sec(2x) \end{cases}$ (correct answer)
2. $\begin{cases} u_1' \cos(2x) + u_2' \sin(2x) = \sec(2x) \\ -2u_1' \sin(2x) + 2u_2' \cos(2x) = 0 \end{cases}$
3. $\begin{cases} u_1' \cos(2x) + u_2' \sin(2x) = 0 \\ u_1' \cos(2x) + u_2' \sin(2x) = \sec(2x) \end{cases}$
4. $\begin{cases} u_1' \sin(2x) + u_2' \cos(2x) = 0 \\ 2u_1' \cos(2x) - 2u_2' \sin(2x) = \sec(2x) \end{cases}$

Explanation: For the method of variation of parameters, with $y_p = u_1 y_1 + u_2 y_2$, the system for $u_1'$ and $u_2'$ is given by $u_1' y_1 + u_2' y_2 = 0$ and $u_1' y_1' + u_2' y_2' = g(x)$. Here, $y_1 = \cos(2x)$, $y_2 = \sin(2x)$, and $g(x) = \sec(2x)$. The derivatives are $y_1' = -2\sin(2x)$ and $y_2' = 2\cos(2x)$. Substituting these into the system gives: $u_1' \cos(2x) + u_2' \sin(2x) = 0$ and $u_1'(-2\sin(2x)) + u_2'(2\cos(2x)) = \sec(2x)$. This matches choice A.

Question 12

1. $-\frac{1}{3}x\sin(3x) + \frac{1}{9}\cos(3x)\ln|\sin(3x)|$
2. $-\frac{1}{3}x\cos(3x) + \frac{1}{9}\sin(3x)\ln|\sin(3x)|$ (correct answer)
3. $-\frac{1}{3}x\cos(3x) + \frac{1}{3}\sin(3x)\ln|\sin(3x)|$
4. $-\frac{1}{3}\cos(3x) + \frac{1}{9}\sin(3x)\ln|\sin(3x)|$

Explanation: When you encounter variation of parameters problems, you're finding functions $u_1(x)$ and $u_2(x)$ such that the particular solution is $y_p = u_1 y_1 + u_2 y_2$. Here, you're given that $u_1'(x) = -\frac{1}{3}$ and $u_2'(x) = \frac{\cot(3x)}{3}$, so you need to integrate these to find $u_1$ and $u_2$. Integrating $u_1'(x) = -\frac{1}{3}$ gives $u_1(x) = -\frac{1}{3}x$ (since constants of integration are zero). For $u_2'(x) = \frac{\cot(3x)}{3}$, recall that $\cot(3x) = \frac{\cos(3x)}{\sin(3x)}$. Using substitution with $u = \sin(3x)$ and $du = 3\cos(3x)dx$: $\int \frac{\cot(3x)}{3} dx = \frac{1}{3} \int \frac{\cos(3x)}{\sin(3x)} dx = \frac{1}{9} \ln|\sin(3x)|$ Therefore: $u_1(x) = -\frac{1}{3}x$ and $u_2(x) = \frac{1}{9}\ln|\sin(3x)|$ The particular solution is: $y_p = u_1 y_1 + u_2 y_2 = (-\frac{1}{3}x)\cos(3x) + (\frac{1}{9}\ln|\sin(3x)|)\sin(3x)$ This matches answer B. Answer A incorrectly has $\sin(3x)$ in the first term instead of $\cos(3x)$. Answer C has the wrong coefficient $\frac{1}{3}$ instead of $\frac{1}{9}$ for the logarithmic term. Answer D is missing the $x$ variable in the first term entirely. Study tip: When doing variation of parameters, always double-check your integration, especially for trigonometric functions. The logarithmic integral of cotangent is a common pattern you should memorize.

Question 13

1. $u_2' = \frac{1}{2\sin(2x)}$
2. $u_2' = \frac{\cos(2x)}{2\sin(2x)} = \frac{\cot(2x)}{2}$ (correct answer)
3. $u_2' = \frac{\cos(2x)}{4\sin(2x)} = \frac{\cot(2x)}{4}$
4. $u_2' = \cos(2x) \csc(2x) = \cot(2x)$

Explanation: The Wronskian for $\cos(2x)$ and $\sin(2x)$ is $W = \cos(2x) \cdot 2\cos(2x) - (-2\sin(2x)) \cdot \sin(2x) = 2\cos^2(2x) + 2\sin^2(2x) = 2$. So $u_2' = \frac{\cos(2x) \csc(2x)}{2} = \frac{\cos(2x)}{2\sin(2x)} = \frac{\cot(2x)}{2}$. Choice A omits the cosine factor entirely. Choice C uses $W = 4$ instead of $W = 2$. Choice D ignores the Wronskian denominator completely.

Question 14

1. $u_1' = -\sin x \cdot \sec x \tan x = -\tan^2 x$ (correct answer)
2. $u_1' = -\frac{\sin x \cdot \sec x \tan x}{1} = -\tan^2 x$
3. $u_1' = -\sin x \cdot \sec x \tan x = -\sin x \tan x$
4. $u_1' = -\frac{\sin x \cdot \sec x \tan x}{\cos x \sin x} = -\frac{\tan^2 x}{\cos x \sin x}$

Explanation: In variation of parameters, $u_1' = -\frac{y_2 g(x)}{W}$ where $g(x) = \sec x \tan x$ and $W = \cos^2 x + \sin^2 x = 1$. So $u_1' = -\frac{\sin x \cdot \sec x \tan x}{1} = -\sin x \cdot \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = -\frac{\sin^2 x}{\cos^2 x} = -\tan^2 x$. Choice B shows the correct setup but is redundant. Choice C fails to complete the simplification of $\sin x \sec x = \frac{\sin x}{\cos x} = \tan x$. Choice D incorrectly uses $W = \cos x \sin x$ instead of $W = 1$.

Question 15

1. $e^{3x} u_1' + xe^{3x} u_2' = e^{3x} x^2$ and $3e^{3x} u_1' + (e^{3x} + 3xe^{3x}) u_2' = 0$
2. $e^{3x} u_1' + xe^{3x} u_2' = 0$ and $3e^{3x} u_1' + e^{3x} u_2' = e^{3x} x^2$
3. $u_1' + x u_2' = 0$ and $3u_1' + (1 + 3x) u_2' = x^2$
4. $e^{3x} u_1' + xe^{3x} u_2' = 0$ and $3e^{3x} u_1' + (e^{3x} + 3xe^{3x}) u_2' = e^{3x} x^2$ (correct answer)

Explanation: When you encounter a second-order linear differential equation with a non-homogeneous term, variation of parameters provides a systematic method to find the particular solution. The key is setting up the correct system of equations for the unknown functions' derivatives. For variation of parameters, you always use this standard system:

• First equation: $y_1 u_1' + y_2 u_2' = 0$
• Second equation: $y_1' u_1' + y_2' u_2' = g(x)$

Question 16

1. Computing the Wronskian involves evaluating $\frac{d}{dx}(xe^x)$ which requires the product rule and careful algebraic manipulation
2. The integrals $\int u_1'(x) dx$ and $\int u_2'(x) dx$ will involve $\ln x$ terms that require integration by parts multiple times (correct answer)
3. The non-homogeneous term $e^x \ln x$ creates resonance with the fundamental solution $y_1 = e^x$, invalidating the variation of parameters method
4. Setting up the system for $u_1'$ and $u_2'$ requires solving a system with coefficients that depend on both $x$ and $\ln x$ terms

Explanation: The Wronskian calculation is straightforward: $W = e^x(e^x + xe^x) - e^x(xe^x) = e^{2x}$. The system gives $u_1' = -\ln x$ and $u_2' = \frac{\ln x}{x}$. The integral $\int -\ln x \, dx$ requires integration by parts, and $\int \frac{\ln x}{x} dx$ involves a substitution that also leads to logarithmic terms. These integrals are the most computationally intensive part. Choice A incorrectly suggests the Wronskian is difficult - it's actually routine. Choice C is wrong because variation of parameters works even when there's resonance (unlike undetermined coefficients). Choice D mischaracterizes the system setup, which is standard.