Differential Equations Quiz: Variation Of Parameters

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Variation Of ParametersQuestion 1 of 20

The homogeneous solutions to $x y'' - y' = 0$ for $x > 0$ are $y_1 = 1$ and $y_2 = x^2$ . Using variation of parameters for the equation $x y'' - y' = x^3$ , what is a valid particular solution $y_p$ ?

\frac{x^5}{15}

\frac{x^4}{8}

\frac{x^4}{4}

\frac{x^3}{3}

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Differential Equations Quiz

Differential Equations Quiz: Variation Of Parameters

Practice Variation Of Parameters 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 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.

All questions

Question 1

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

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 2

Consider the differential equation $x y'' - (1+x) y' + y = x^2 e^{2x}$ for $x>0$ . Given that $y_1(x) = 1+x$ and $y_2(x) = e^x$ are solutions to the corresponding homogeneous equation, a particular solution $y_p = u_1 y_1 + u_2 y_2$ is sought using variation of parameters. Which of the following is the correct expression for $u_1'$ ?

$u_1'(x) = -e^{2x}$ (correct answer)
$u_1'(x) = -x e^{2x}$
$u_1'(x) = e^{2x}$
$u_1'(x) = -(1+x)e^{2x}$

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'

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 3

Given that $y_1 = e^{3x}$ and $y_2 = e^{-3x}$ form a fundamental set of solutions for the homogeneous equation $y'' - 9y = 0$ , find a particular solution $y_p$ for the non-homogeneous equation $y'' - 9y = 12e^{3x}$ .

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

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

Consider the Cauchy-Euler equation $x^2 y'' - 3xy' + 4y = x^2 \ln(x)$ for $x > 0$ . Given that the homogeneous solutions are $y_1 = x^2$ and $y_2 = x^2 \ln(x)$ , which function below is a valid particular solution $y_p$ ?

$\frac{1}{6}(\ln x)^3$
$\frac{1}{2}x^2 (\ln x)^2$
$\frac{5}{6}x^2 (\ln x)^3$
$\frac{1}{6}x^2 (\ln x)^3$ (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

In applying the method of variation of parameters to solve $y'' + P(x)y' + Q(x)y = g(x)$ , we assume a particular solution of the form $y_p = u_1(x)y_1(x) + u_2(x)y_2(x)$ . The derivation requires imposing the condition $u_1'y_1 + u_2'y_2 = 0$ . What is the primary reason for this condition?

It simplifies the system of equations by making it diagonal, allowing $u_1'$ and $u_2'$ to be solved for independently.
It is a necessary condition to ensure that the Wronskian of $y_1$ and $y_2$ is non-zero.
It forces the particular solution $y_p$ to be linearly independent from the homogeneous solutions $y_1$ and $y_2$ .
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

For which of the following differential equations is the method of variation of parameters necessary, as opposed to the method of undetermined coefficients?

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

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

Let $y(x)$ be the solution to the initial value problem $y'' - y' - 2y = 3e^{2x}$ , with $y(0)=0$ and $y'(0)=0$ . What is the value of $y(\ln 2)$ ?

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

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

A particular solution to $y'' + y' - 2y = 4e^{-2x} \ln(x)$ for $x>0$ is given by $y_p = u_1(x) e^x + u_2(x) e^{-2x}$ . Using the method of variation of parameters and setting the constant of integration to zero, what is $u_2(x)$ ?

$-\frac{4}{3x}$
$4x - 4x \ln(x)$
$\frac{4}{3}x \ln(x) - \frac{4}{3}x$
$\frac{4}{3}x - \frac{4}{3}x \ln(x)$ (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 9

The differential equation $y'' + y = \tan(x)$ has a homogeneous solution $y_h = c_1 \cos(x) + c_2 \sin(x)$ . Using variation of parameters, the particular solution $y_p$ can be expressed in integral form. Which of the following correctly represents $y_p(x)$ ?

$\cos(x) \int \sin(x) dx + \sin(x) \int -\sin(x)\tan(x) dx$
$\cos(x) \int \sin(x)\tan(x) dx + \sin(x) \int \sin(x) dx$
$\cos(x) \int -\sin(x)\tan(x) dx + \sin(x) \int \sin(x) dx$ (correct answer)
$-\cos(x) \int \sin(x)\tan(x) dx + \sin(x) \int \sin(x) dx$

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 10

The general solution to the homogeneous differential equation $y'' + 4y = 0$ is $y_h = c_1 \cos(2x) + c_2 \sin(2x)$ . To find a particular solution to $y'' + 4y = \sec(2x)$ using the method of variation of parameters, we assume $y_p = u_1(x) \cos(2x) + u_2(x) \sin(2x)$ . Which system of equations must be solved for $u_1'$ and $u_2'$ ?

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

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 11

When solving $y'' + 9y = \csc(3x)$ using variation of parameters with $y_1 = \cos(3x)$ and $y_2 = \sin(3x)$ , one finds that $u_1'(x) = -1/3$ and $u_2'(x) = \frac{\cot(3x)}{3}$ . Assuming constants of integration are zero, what is the resulting particular solution $y_p(x)$ ?

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

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 12

For the equation $y'' + 4y = \csc(2x)$ , after finding the fundamental solutions $y_1 = \cos(2x)$ and $y_2 = \sin(2x)$ , the variation of parameters method yields $u_2' = \frac{\cos(2x) \csc(2x)}{W}$ . What is the simplified form of this expression?

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

Explanation: The Wronskian for

\cos(2x)

and

\sin(2x)

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 13

Consider $y'' - 6y' + 9y = e^{3x} x^2$ . When setting up variation of parameters, if the fundamental solutions are $y_1 = e^{3x}$ and $y_2 = xe^{3x}$ , which system of equations must be solved for $u_1'$ and $u_2'$ ?

$e^{3x} u_1' + xe^{3x} u_2' = e^{3x} x^2$ and $3e^{3x} u_1' + (e^{3x} + 3xe^{3x}) u_2' = 0$
$e^{3x} u_1' + xe^{3x} u_2' = 0$ and $3e^{3x} u_1' + e^{3x} u_2' = e^{3x} x^2$
$u_1' + x u_2' = 0$ and $3u_1' + (1 + 3x) u_2' = x^2$
$e^{3x} u_1' + xe^{3x} u_2' = 0$ 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: $y_1' u_1' + y_2' u_2' = g(x)$

where

g(x)

is the right-hand side of your differential equation. Given

y_1 = e^{3x}

and

y_2 = xe^{3x}

, you need their derivatives:

y_1' = 3e^{3x}

and

y_2' = e^{3x} + 3xe^{3x}

(using the product rule). The right-hand side is

g(x) = e^{3x}x^2

. This gives you the system in answer D:

e^{3x} u_1' + xe^{3x} u_2' = 0

and

3e^{3x} u_1' + (e^{3x} + 3xe^{3x}) u_2' = e^{3x} x^2

. Answer A incorrectly puts the non-homogeneous term in the first equation instead of the second. Answer B uses the wrong derivative for

y_2'

in the second equation (missing the

3xe^{3x}

term). Answer C attempts to simplify by dividing out

e^{3x}

, but this changes the fundamental structure and incorrectly modifies the right-hand side. Remember: in variation of parameters, the first equation always equals zero, and the second equation always equals the non-homogeneous term. Never deviate from this standard form.

Question 14

For the differential equation $y'' - 2y' + y = e^x \ln x$ , the method of variation of parameters requires finding $u_1(x)$ and $u_2(x)$ such that $y_p = u_1 y_1 + u_2 y_2$ . Given that the fundamental solutions have the form $y_1 = e^x$ and $y_2 = xe^x$ , what is the primary computational challenge in this problem?

Computing the Wronskian involves evaluating $\frac{d}{dx}(xe^x)$ which requires the product rule and careful algebraic manipulation
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)
The non-homogeneous term $e^x \ln x$ creates resonance with the fundamental solution $y_1 = e^x$ , invalidating the variation of parameters method
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.

Question 15

Consider the equation $x^2 y'' - 2xy' + 2y = x^3$ . After transforming this to standard form and identifying fundamental solutions $y_1 = x$ and $y_2 = x^2$ , the variation of parameters method gives $u_1' = \frac{-x^2 \cdot x}{W}$ and $u_2' = \frac{x \cdot x}{W}$ . What is the correct value of the Wronskian $W$ ?

$W = x \cdot x^2 - x^2 \cdot x = 0$ , indicating linear dependence
$W = x \cdot 2x - 1 \cdot x^2 = x^2$ after factoring
$W = x \cdot 2x - 1 \cdot x^2 = 2x^2 - x^2 = x^2$ (correct answer)
$W = x^2 \cdot 2x - x \cdot 1 = 2x^3 - x = x(2x^2 - 1)$

Explanation: When applying variation of parameters to solve differential equations, you need the Wronskian of the fundamental solutions to find the particular solution. The Wronskian is calculated as

W = y_1 y_2' - y_1' y_2

, where you must correctly identify each function and its derivative. Given the fundamental solutions

y_1 = x

and

y_2 = x^2

, you first find their derivatives:

y_1' = 1

and

y_2' = 2x

. Now substitute into the Wronskian formula:

W = y_1 y_2' - y_1' y_2 = x \cdot 2x - 1 \cdot x^2 = 2x^2 - x^2 = x^2

This confirms that answer C is correct. Let's examine why the other options fail. Option A incorrectly uses

W = y_1 y_2 - y_2 y_1

, which is just the product of the functions (always zero when rearranged), not the Wronskian formula. This shows a fundamental misunderstanding of the definition. Option B starts correctly but claims the answer is

x^2

"after factoring" without showing the crucial subtraction step

2x^2 - x^2

. Option D uses the wrong Wronskian formula

W = y_2 y_2' - y_1 y_1'

, mixing up which derivatives go with which functions. Study tip: Always write out the Wronskian formula explicitly as

W = y_1 y_2' - y_1' y_2

and carefully match each function with its correct derivative. The order matters—swapping terms will give you the negative of the correct answer.

Question 16

For $y'' + 9y = \tan(3x)$ , when applying variation of parameters with $y_1 = \cos(3x)$ and $y_2 = \sin(3x)$ , the expression for $u_2'$ simplifies to $u_2' = \frac{\cos(3x) \tan(3x)}{3}$ . To find $u_2(x)$ , which substitution strategy is most effective?

Use the identity $\cos(3x) \tan(3x) = \sin(3x)$ and integrate directly to get $u_2 = -\frac{\cos(3x)}{9}$
Use the substitution $w = 3x$ to transform the integral into a more standard form involving $\cos w \tan w$
Apply integration by parts with $u = \cos(3x)$ and $dv = \tan(3x) dx$
Rewrite as $\frac{\cos(3x) \sin(3x)}{3 \cos(3x)} = \frac{\sin(3x)}{3}$ and integrate to get $u_2 = -\frac{\cos(3x)}{9}$ (correct answer)

Explanation: When solving differential equations using variation of parameters, you'll often encounter integrals that require algebraic simplification before integration. The key insight here is recognizing when trigonometric expressions can be rewritten in simpler forms. Starting with

u_2' = \frac{\cos(3x) \tan(3x)}{3}

, you need to simplify this expression before integrating. Since

\tan(3x) = \frac{\sin(3x)}{\cos(3x)}

, you can rewrite the expression as:

u_2' = \frac{\cos(3x) \cdot \frac{\sin(3x)}{\cos(3x)}}{3} = \frac{\sin(3x)}{3}

Now integration is straightforward:

u_2 = \int \frac{\sin(3x)}{3} dx = -\frac{\cos(3x)}{9}

. This confirms answer D is correct. Answer A makes a fundamental error by claiming

\cos(3x) \tan(3x) = \sin(3x)

without the factor of 3 in the denominator. This algebraic mistake leads to an incorrect final answer. Answer B suggests substitution

w = 3x

, which is unnecessary here. While substitution can work, it adds complexity when direct algebraic simplification is simpler and more efficient. Answer C proposes integration by parts, but this approach is overly complicated for this problem. Integration by parts with

\tan(3x)

would create a more difficult integral than the original. Strategy tip: When encountering products involving tangent functions, always try rewriting tangent as

\frac{\sin}{\cos}

first. This often reveals immediate simplifications through cancellation, avoiding more complex integration techniques.

Question 17

Consider the differential equation $y'' - 4y' + 4y = e^{2x} \ln x$ . When using variation of parameters to find a particular solution, which expression correctly represents the Wronskian $W(y_1, y_2)$ of the fundamental solutions?

$W = e^{4x}$ (correct answer)
$W = xe^{4x}$
$W = e^{2x}$
$W = 2xe^{4x}$

Explanation: For the homogeneous equation

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

, the characteristic equation is

r^2 - 4r + 4 = 0

, which gives

(r-2)^2 = 0

, so

r = 2

is a repeated root. The fundamental solutions are

y_1 = e^{2x}

and

y_2 = xe^{2x}

. The Wronskian is

W = y_1 y_2' - y_1' y_2 = e^{2x}(e^{2x} + 2xe^{2x}) - 2e^{2x}(xe^{2x}) = e^{4x} + 2xe^{4x} - 2xe^{4x} = e^{4x}

. Choice B incorrectly includes the factor

x

that cancels out. Choice C uses only the exponential part without accounting for the derivative calculations. Choice D incorrectly retains the

2x

factor that should cancel.

Question 18

Consider $y'' - 4y' + 3y = e^{3x} \sinh x$ . The fundamental solutions are $y_1 = e^x$ and $y_2 = e^{3x}$ . When setting up the variation of parameters system, the function $g(x) = e^{3x} \sinh x$ can be rewritten in a form that simplifies the integration. Which rewritten form is most advantageous?

$g(x) = e^{3x} \sinh x$ should be left as is since hyperbolic functions integrate directly
$g(x) = e^{3x} \sinh x = e^{3x} \cdot \frac{e^x + e^{-x}}{2} = \frac{e^{4x} + e^{2x}}{2}$
$g(x) = e^{3x} \sinh x = e^{3x} \cdot \frac{e^x - e^{-x}}{2} = \frac{e^{4x} - e^{2x}}{2}$ (correct answer)
$g(x) = e^{3x} \sinh x = \frac{1}{2}(e^{4x} - e^{2x})$ after applying the identity $\sinh x = \frac{e^x - e^{-x}}{2}$

Explanation: When solving differential equations using variation of parameters, you need to integrate expressions involving the fundamental solutions and the forcing function. The key insight here is recognizing when rewriting the forcing function will make those integrations much easier. For this problem, you need to recall the correct definition of hyperbolic sine:

\sinh x = \frac{e^x - e^{-x}}{2}

, not

\frac{e^x + e^{-x}}{2}

(which is

\cosh x

). Using this definition,

g(x) = e^{3x} \sinh x = e^{3x} \cdot \frac{e^x - e^{-x}}{2} = \frac{e^{4x} - e^{2x}}{2}

. This exponential form will integrate cleanly when you apply variation of parameters, giving you simple rational coefficients rather than messy hyperbolic function integrals. Choice A is wrong because while hyperbolic functions do have known integrals, they're more complex than necessary here. Choice B contains a fundamental error—it uses the definition of

\cosh x

instead of

\sinh x

, leading to

\frac{e^{4x} + e^{2x}}{2}

rather than the correct difference form. Choice D, while mathematically correct about the identity, doesn't show the crucial algebraic steps that demonstrate why this rewriting helps—the expansion that reveals the clean exponential terms. The correct answer is C because it shows both the proper hyperbolic identity and the resulting simplified exponential form. Study tip: Always double-check hyperbolic function definitions (

\sinh

uses minus,

\cosh

uses plus), and in variation of parameters problems, look for ways to rewrite forcing functions as sums or differences of exponentials—they integrate much more cleanly than other forms.

Question 19

Using variation of parameters to solve $y'' - 2y' + y = \frac{e^x}{x}$ , a particular solution is sought in the form $y_p = u_1(x)e^x + u_2(x)xe^x$ . What is the integrand for calculating $u_2(x)$ ?

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

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)

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 20

For the equation $y'' + y = \sec x \tan x$ , when applying variation of parameters with fundamental solutions $y_1 = \cos x$ and $y_2 = \sin x$ , what is the correct expression for $u_1'(x)$ ?

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

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