Differential Equations Quiz: General Vs Particular Solutions

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General Vs Particular SolutionsQuestion 1 of 19

The general solution to $y'' - 4y' + 4y = 0$ is $y(x) = c_1 e^{2x} + c_2 x e^{2x}$ . Find the value of the particular solution satisfying $y(0) = 3$ and $y'(0) = 8$ at the point $x=1$ .

3e^2

8e^2

5e^2

11e^2

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Differential Equations Quiz

Differential Equations Quiz: General Vs Particular Solutions

Practice General Vs 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 General Vs Particular Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

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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.

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Question 1

The general solution to $y'' - 4y' + 4y = 0$ is $y(x) = c_1 e^{2x} + c_2 x e^{2x}$ . Find the value of the particular solution satisfying $y(0) = 3$ and $y'(0) = 8$ at the point $x=1$ .

$3e^2$
$8e^2$
$5e^2$ (correct answer)
$11e^2$

Explanation: When you encounter a differential equation problem with given initial conditions, you need to use those conditions to find the specific values of the arbitrary constants in the general solution. Given the general solution

y(x) = c_1 e^{2x} + c_2 x e^{2x}

, start by applying the first initial condition. At

x = 0

y(0) = c_1 e^0 + c_2 \cdot 0 \cdot e^0 = c_1 = 3

. So

c_1 = 3

. For the second condition, you need the derivative. Using the product rule on the second term:

y'(x) = 2c_1 e^{2x} + c_2(e^{2x} + 2xe^{2x})

. At

x = 0

y'(0) = 2c_1 + c_2 = 2(3) + c_2 = 8

. Therefore

c_2 = 2

. The particular solution is

y(x) = 3e^{2x} + 2xe^{2x}

. At

x = 1

y(1) = 3e^2 + 2(1)e^2 = 3e^2 + 2e^2 = 5e^2

. Answer A (

3e^2

) represents what you'd get if you forgot the second term entirely—just evaluating

3e^{2x}

x = 1

. Answer B (

8e^2

) might come from incorrectly using the initial condition

y'(0) = 8

as if it were the coefficient. Answer D (

11e^2

) could result from arithmetic errors when combining terms or incorrectly calculating the derivative. Always remember: with initial value problems, methodically apply each condition one at a time to find your constants, then substitute your target

x

-value into the complete particular solution.

Question 2

Consider the family of curves defined by $y = c_1 e^{3x} + c_2 x e^{3x} + c_3 e^{3x}$ . What is the order of the linear homogeneous differential equation for which this family represents the general solution?

1, because all terms contain the function $e^{3x}$ .
2, because there are only two functionally independent solution forms, $e^{3x}$ and $x e^{3x}$ . (correct answer)
3, because there are three constants, $c_1$ , $c_2$ , and $c_3$ , shown in the expression.
Cannot be determined without knowing the specific differential equation.

Explanation: The order of a linear differential equation is equal to the number of essential (linearly independent) arbitrary constants in its general solution. The given family of curves can be rewritten by combining the terms with

c_1

and

c_3

y = (c_1 + c_3) e^{3x} + c_2 x e^{3x}

. If we define a new constant

C_1 = c_1 + c_3

, the solution becomes

y = C_1 e^{3x} + c_2 x e^{3x}

. This form reveals that there are only two essential arbitrary constants,

C_1

and

c_2

. Therefore, the order of the corresponding differential equation is 2.

Question 3

Consider the differential equation $y' = 2\sqrt{y}$ . The family of functions $y(x) = (x+C)^2$ for $x \ge -C$ represents a collection of solutions. However, this one-parameter family does not constitute the general solution on the interval $(-\infty, \infty)$ . Which statement best explains why?

The family of functions does not actually satisfy the differential equation for all $x$ and $C$ .
The constant $C$ cannot be uniquely determined by a single initial condition like $y(x_0)=y_0$ .
The family fails to include the singular solution $y(x) = 0$ , which also satisfies the differential equation. (correct answer)
A general solution must be a single function, not an infinite family of functions.

Explanation: A general solution to a differential equation is a family of solutions that contains every possible solution, except possibly for some 'singular' solutions. In this case, the function

y(x) = 0

for all

x

is also a solution to

y' = 2\sqrt{y}

, since

0 = 2\sqrt{0}

. However, there is no value of the constant

C

for which the family

y(x) = (x+C)^2

yields the function

y(x) = 0

. Because this valid solution is omitted, the family is not the general solution. This missing solution is known as a singular solution.

Question 4

The differential equation $\frac{dy}{dx} = -\frac{x^2+y^2}{2xy}$ can be solved by recognizing it as a homogeneous equation. Which of the following equations implicitly defines the family of general solutions, where $C$ is an arbitrary constant?

$x^3 + 3xy^2 = C$ (correct answer)
$y^3 + 3yx^2 = C$
$x^2 + y^2 = Cxy$
$\ln(x^2+y^2) + 2y/x = C$

Explanation: This is a homogeneous equation. Let

y = vx

, so

\frac{dy}{dx} = v + x\frac{dv}{dx}

. The equation becomes

v + x\frac{dv}{dx} = -\frac{x^2+(vx)^2}{2x(vx)} = -\frac{1+v^2}{2v}

. Rearranging gives

x\frac{dv}{dx} = -\frac{1+v^2}{2v} - v = -\frac{1+3v^2}{2v}

. This is separable:

\frac{2v}{1+3v^2} dv = -\frac{1}{x} dx

. Integrating both sides gives

\frac{1}{3}\ln(1+3v^2) = -\ln|x| + C_1

. Multiplying by 3 and exponentiating gives

1+3v^2 = K|x|^{-3}

. Substituting back

v=y/x

gives

1+3(y/x)^2 = K/|x^3|

. Multiplying by

|x^3|

gives

|x^3| + 3|x|y^2 = K

. Assuming

x>0

for simplicity and renaming the constant, we have

x^3 + 3xy^2 = C

Question 5

A particle's position is given by the general solution $x(t) = c_1 \cos(3t) + c_2 \sin(3t)$ . The particle starts at position $x(0) = -2$ . Its velocity is zero at time $t = \pi/12$ . What is the position of the particle at time $t = \pi/3$ ?

$2$ (correct answer)
$-2$
$-2\sqrt{2}$
$0$

Explanation: First, use the initial position

x(0)=-2

x(0) = c_1\cos(0) + c_2\sin(0) = c_1

. So,

c_1=-2

. Next, find the velocity function:

v(t) = x'(t) = -3c_1\sin(3t) + 3c_2\cos(3t)

. Use the condition

v(\pi/12)=0

v(\pi/12) = -3c_1\sin(3\pi/12) + 3c_2\cos(3\pi/12) = -3c_1\sin(\pi/4) + 3c_2\cos(\pi/4) = 0

. This simplifies to

-3c_1(\sqrt{2}/2) + 3c_2(\sqrt{2}/2) = 0

, which implies

-c_1+c_2=0

, or

c_1=c_2

. Since

c_1=-2

, we have

c_2=-2

. The particular solution is

x(t) = -2\cos(3t) - 2\sin(3t)

. Finally, evaluate at

t=\pi/3

x(\pi/3) = -2\cos(3\pi/3) - 2\sin(3\pi/3) = -2\cos(\pi) - 2\sin(\pi) = -2(-1) - 2(0) = 2

Question 6

The general solution to a certain second-order linear homogeneous differential equation is $y(x) = c_1 e^{2x} + c_2 e^{-3x}$ . Which of the following functions cannot be a particular solution to this differential equation, regardless of the initial conditions?

$y(x) = 0$
$y(x) = 5e^{2x} - 2e^{-3x}$
$y(x) = 4e^{-3x}$
$y(x) = e^{2x} + e^{-3x} + 1$ (correct answer)

Explanation: A particular solution is obtained by assigning specific values to the constants

c_1

and

c_2

in the general solution

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

. A)

y(x)=0

corresponds to

c_1=0

and

c_2=0

. B)

y(x) = 5e^{2x} - 2e^{-3x}

corresponds to

c_1=5

and

c_2=-2

. C)

y(x) = 4e^{-3x}

corresponds to

c_1=0

and

c_2=4

. D)

y(x) = e^{2x} + e^{-3x} + 1

cannot be a solution because of the constant term '+1'. No choice of

c_1

and

c_2

in the general solution can produce this term. This function would be a solution to a non-homogeneous equation, not the given homogeneous one.

Question 7

The equation $y(x) = c_1 x + c_2 x^2$ is the general solution to a second-order linear differential equation. Which of the following statements about its constants of integration is correct?

The value of $c_1$ is always equal to the initial value $y(0)$ for any particular solution.
The value of $c_1$ is uniquely determined by the initial velocity $y'(0)$ at the origin. (correct answer)
The values of $c_1$ and $c_2$ can both be found using a single initial condition such as $y(1)=5$ .
The values of $c_1$ and $c_2$ require initial conditions that must be specified away from $x=0$ .

Explanation: Let's analyze the initial conditions at

x=0

. The general solution is

y(x) = c_1 x + c_2 x^2

. The derivative is

y'(x) = c_1 + 2c_2 x

. Evaluating at

x=0

y(0) = c_1(0) + c_2(0)^2 = 0

and

y'(0) = c_1 + 2c_2(0) = c_1

. This shows that any particular solution from this family must pass through the origin (

y(0)=0

). The constant

c_1

is uniquely determined by the initial velocity at the origin,

y'(0) = c_1

. The constant

c_2

would require another condition, such as the value of

y

y'

at a point other than

x=0

Question 8

A student solves the initial value problem $y' + 2y = 0$ , $y(0) = 3$ and finds a particular solution. They then solve the problem $y' + 2y = 0$ , $y(1) = 3$ and find another particular solution. Which statement accurately compares these two particular solutions?

The two particular solutions are identical for all $x$ .
The two particular solutions are different functions. (correct answer)
The solutions are the same, but the constants of integration are different.
The existence and uniqueness theorem does not guarantee a solution for the second problem.

Explanation: The general solution of

y' + 2y = 0

y(x) = Ce^{-2x}

. For the first IVP,

y(0) = 3

, we find

3 = Ce^0

, so

C=3

. The solution is

y_1(x) = 3e^{-2x}

. For the second IVP,

y(1) = 3

, we find

3 = Ce^{-2}

, so

C = 3e^2

. The solution is

y_2(x) = (3e^2)e^{-2x} = 3e^{2-2x}

. Since

3e^{-2x}

and

3e^{2-2x}

are different functions, the two particular solutions are different. Choice C is contradictory; if the constants of integration are different, the particular solutions (which are specific functions) must also be different.

Question 9

The general solution to a first-order differential equation is given implicitly by the relation $\sin(y) - e^{-x^2} = C$ . A particular solution curve passes through the point $(0, \pi/6)$ . What is the y-coordinate of the point on this particular solution curve where $x = \sqrt{\ln(2)}$ ?

$0$ (correct answer)
$\pi/6$
$\pi/3$
$\pi/2$

Explanation: First, use the initial condition

(0, \pi/6)

to find the constant

C

for the particular solution. Substitute

x=0

and

y=\pi/6

into the general solution:

\sin(\pi/6) - e^{-0^2} = C

, which simplifies to

1/2 - 1 = C

, so

C = -1/2

. The particular solution is defined by

\sin(y) - e^{-x^2} = -1/2

. Now, find the value of

y

when

x = \sqrt{\ln(2)}

. Substitute this value of

x

into the particular solution's equation:

\sin(y) - e^{-(\sqrt{\ln(2)})^2} = -1/2

. This simplifies to

\sin(y) - e^{-\ln(2)} = -1/2

. Since

e^{-\ln(2)} = e^{\ln(2^{-1})} = 1/2

, the equation becomes

\sin(y) - 1/2 = -1/2

, which gives

\sin(y) = 0

. The simplest solution for

y

y=0

Question 10

A physical system is modeled by the nonlinear differential equation $(\frac{d^3y}{dt^3})^2 - t^4 (\frac{dy}{dt})^5 + y^3 = \cos(t)$ . How many arbitrary constants must be present in the general solution for this equation?

2, because of the power on the highest-order derivative.
8, because this is the sum of the orders of all derivative terms.
5, because of the power on the first-order derivative term.
3, because the order of the equation is determined by the highest derivative. (correct answer)

Explanation: When you encounter a differential equation, the key concept to focus on is the order of the equation, which determines how many arbitrary constants appear in the general solution. The order is always determined by the highest derivative present, regardless of any powers, coefficients, or nonlinear terms. In this equation

\left(\frac{d^3y}{dt^3}\right)^2 - t^4 \left(\frac{dy}{dt}\right)^5 + y^3 = \cos(t)

, the highest derivative is

\frac{d^3y}{dt^3}

, making this a third-order differential equation. The fundamental theorem states that an nth-order differential equation has exactly n arbitrary constants in its general solution. Therefore, this third-order equation must have exactly 3 arbitrary constants. Let's examine why the other choices are incorrect. Choice A) suggests the answer is 2 because of the squared power on the third derivative, but powers on derivatives don't affect the number of constants—only the order matters. Choice B) incorrectly adds the orders of derivative terms (3 + 1 = 4, plus other terms to get 8), but this isn't how differential equation theory works. Choice C) focuses on the fifth power of the first derivative, but again, powers are irrelevant to determining the number of arbitrary constants. Study tip: Always identify the highest-order derivative first when analyzing any differential equation. The order of that derivative equals the number of arbitrary constants in the general solution, no matter how complex the equation appears with powers, products, or nonlinear terms.

Question 11

The general solution to the differential equation $y' = -x/y$ is the family of circles $x^2 + y^2 = C$ . For the particular solution that passes through the point $(3, 4)$ , what is the direct geometric meaning of the constant $C$ ?

The radius of the circular solution curve.
The curvature of the circular solution curve.
The square of the radius of the circular solution curve. (correct answer)
The circumference of the circular solution curve.

Explanation: When you encounter a differential equation whose general solution represents a geometric family like circles, you need to understand what the arbitrary constant represents geometrically, not just algebraically. To find the particular solution through

(3, 4)

, substitute these coordinates into the general solution

x^2 + y^2 = C

3^2 + 4^2 = C

9 + 16 = 25 = C

So the particular solution is

x^2 + y^2 = 25

, which represents a circle centered at the origin. Now consider what

C = 25

means geometrically. The standard form of a circle centered at the origin is

x^2 + y^2 = r^2

, where

r

is the radius. Comparing this with our solution

x^2 + y^2 = 25

, we see that

r^2 = 25

, so the radius is

r = 5

. Therefore,

C

represents the square of the radius, making (C) correct. Let's examine why the other choices fail: (A) claims

C

is the radius itself, but

C = 25

while the radius is only

5

. (B) suggests

C

represents curvature, but curvature for a circle equals

1/r = 1/5

, not

25

. (D) proposes

C

is the circumference, but circumference equals

2\pi r = 10\pi

, which doesn't match

25

. Study tip: When working with geometric solution families, always compare the given equation with the standard form to identify what each parameter represents. The constant in

x^2 + y^2 = C

will always be

r^2

, not

r

itself.

Question 12

A student claims that $y = 2x^2 + 3x + 5$ and $y = 2x^2 + 3x - 2$ are both particular solutions to the same differential equation. Another student argues that if this is true, then $y = 2x^2 + 3x + C$ must be the general solution. Which analysis of these claims is most accurate?

Both students are correct; the particular solutions differ only by a constant, confirming the general solution structure (correct answer)
The first student is correct, but the second student is wrong because general solutions require homogeneous equations
The first student is wrong; two different particular solutions cannot satisfy the same differential equation simultaneously
Both students are wrong; the given functions are solutions to different differential equations entirely

Explanation: Both particular solutions have the same derivative structure:

y' = 4x + 3

and

y'' = 4

. They satisfy the same second-order differential equation

y'' = 4

. The general solution to

\frac{d^2y}{dx^2} = 4

y = 2x^2 + C_1x + C_2

, and with appropriate initial conditions, this reduces to the form

y = 2x^2 + 3x + C

. The two particular solutions differ by a constant (

5 - (-2) = 7

), which is exactly what we expect for different particular solutions of the same differential equation.

Question 13

A differential equation has the general solution $y = C_1 x^3 + C_2 x^2 + C_3 x + C_4$ . A student is asked to find a particular solution satisfying $y(1) = 2$ , $y'(1) = 6$ , $y''(1) = 12$ , and $y'''(1) = 18$ . After solving, the student concludes that this particular solution is unique. What is the most precise reason for this uniqueness?

The solution is unique because fourth-order differential equations always have unique solutions when boundary conditions are given
The solution is unique because the differential equation is linear and the initial conditions are specified at the same point
The solution is unique because the initial conditions provide enough information to solve for all four unknown constants simultaneously
The solution is unique because the number of initial conditions equals the number of arbitrary constants in the general solution (correct answer)

Explanation: When you encounter a differential equation with a general solution containing arbitrary constants, the key to uniqueness lies in having sufficient conditions to determine all those constants. The general solution

y = C_1 x^3 + C_2 x^2 + C_3 x + C_4

contains four arbitrary constants:

C_1

C_2

C_3

, and

C_4

. To find a unique particular solution, you need exactly four independent conditions to create a system of four equations in four unknowns. The given conditions

y(1) = 2

y'(1) = 6

y''(1) = 12

, and

y'''(1) = 18

provide precisely this—four independent conditions that allow you to solve for all four constants simultaneously, yielding a unique solution. Option A is incorrect because uniqueness isn't guaranteed simply by having a fourth-order equation with boundary conditions—you need the right number and type of conditions. Option B misses the fundamental point: while linearity helps ensure solutions exist and behave nicely, and having conditions at the same point matters for initial value problems, neither directly explains why this particular solution is unique. Option C is partially correct but imprecise—it's not just that you have "enough information," but specifically that you have exactly the right amount. Option D captures this precisely: the number of conditions (four) equals the number of arbitrary constants (four). Study tip: Always count the arbitrary constants in a general solution, then verify you have that many independent conditions for uniqueness. This one-to-one correspondence is the fundamental principle for unique solutions in linear differential equations.

Question 14

Two students are working with the differential equation $\frac{dy}{dx} = y + e^x$ . Student A finds the general solution $y = Ce^x - xe^x$ , while Student B finds the general solution $y = De^x - xe^x$ (using a different constant label). When comparing their particular solutions for the initial condition $y(0) = 3$ , what should they observe?

They obtain different particular solutions because they used different forms of the general solution with different constants
They obtain the same particular solution despite using different constant labels, demonstrating equivalent general solution forms (correct answer)
Student A's solution is correct while Student B's contains an error in the constant relationship
Both solutions are incorrect because the general solution should not contain the term $-xe^x$ for this differential equation

Explanation: Both students have equivalent forms of the general solution. For Student A:

3 = Ce^0 - 0 \cdot e^0 = C

, so the particular solution is

y = 3e^x - xe^x

. For Student B:

3 = De^0 - 0 \cdot e^0 = D

, so the particular solution is

y = 3e^x - xe^x

. The different constant labels (

C

D

) don't affect the final answer. This demonstrates that general solutions can be written with different parameter names but still represent the same family of solutions.

Question 15

Consider the general solution $y = C_1 \sin(3x) + C_2 \cos(3x) + \frac{1}{4}$ to a nonhomogeneous differential equation. A student argues that since any particular solution can be written as $y = A\sin(3x + \phi) + \frac{1}{4}$ for appropriate constants $A$ and $\phi$ , this alternative form is also a valid general solution. Which statement best evaluates this argument?

The argument is incorrect; the phase-shifted form $A\sin(3x + \phi)$ represents fewer solutions than the original general form
The argument is incorrect; the trigonometric identity changes the fundamental structure of the differential equation's solution
The argument is correct for the homogeneous part but fails to account for the particular solution $\frac{1}{4}$ properly
The argument is correct; both forms span the same solution space and contain the same number of arbitrary constants (correct answer)

Explanation: When you encounter questions about general solutions to differential equations, focus on whether different forms represent the same solution space with the same degrees of freedom. The key insight is recognizing that

C_1 \sin(3x) + C_2 \cos(3x)

and

A\sin(3x + \phi)

are mathematically equivalent expressions for the homogeneous solution. Using the angle addition formula,

A\sin(3x + \phi) = A\sin(3x)\cos(\phi) + A\cos(3x)\sin(\phi)

. If we set

C_1 = A\cos(\phi)

and

C_2 = A\sin(\phi)

, we can convert between the forms. Both expressions contain exactly two arbitrary constants and span the same two-dimensional solution space. The complete general solution

y = A\sin(3x + \phi) + \frac{1}{4}

is therefore equivalent to the original form, making answer D correct. Answer A is wrong because both forms represent exactly the same number of solutions—the phase-shifted form doesn't reduce the solution space. Answer B incorrectly suggests that trigonometric identities somehow alter the differential equation's structure, when they simply provide alternative representations of the same mathematical object. Answer C misunderstands the relationship between homogeneous and particular solutions—the particular solution

\frac{1}{4}

is handled identically in both forms and doesn't affect the equivalence of the homogeneous parts. Remember: different algebraic forms of the same expression are equivalent as long as they contain the same number of independent arbitrary constants and span the same solution space.

Question 16

A third-order differential equation has the general solution $y = C_1 e^{2x} + C_2 xe^{2x} + C_3 e^{-x}$ . If initial conditions $y(0) = 1$ , $y'(0) = 0$ , and $y''(0) = -3$ are specified, which approach correctly determines whether a unique particular solution exists?

Verify that the three solution components $e^{2x}$ , $xe^{2x}$ , and $e^{-x}$ are linearly independent functions
Check that the number of initial conditions matches the order of the equation, ensuring automatic existence of a unique solution
Calculate $y'(x)$ and $y''(x)$ , then solve the resulting $3 \times 3$ linear system for $C_1$ , $C_2$ , and $C_3$ (correct answer)
Substitute the initial conditions directly into the general solution without computing derivatives, then solve for the constants

Explanation: When dealing with initial value problems for linear differential equations, the key question is whether you can uniquely determine all arbitrary constants using the given conditions. This requires setting up and solving a system of equations. The correct approach (C) involves finding the derivatives

y'(x)

and

y''(x)

of the general solution, then applying all three initial conditions to create a system of three equations in three unknowns. From

y = C_1 e^{2x} + C_2 xe^{2x} + C_3 e^{-x}

, you'd compute

y' = 2C_1 e^{2x} + C_2(e^{2x} + 2xe^{2x}) + C_3(-e^{-x})

and

y''

, then substitute

x = 0

into each expression along with the given values. This gives you three equations to solve for

C_1

C_2

, and

C_3

. If this system has a unique solution, then a unique particular solution exists. Option A is insufficient because linear independence of the solution components tells you the general solution is complete, but doesn't guarantee the initial value problem is solvable. Option B makes a false assumption—having the right number of conditions is necessary but not sufficient for uniqueness. The coefficient matrix of your system could still be singular. Option D fails because you need all three conditions (including those involving derivatives) to determine all three constants; using only

y(0) = 1

leaves you with one equation and three unknowns. Remember: for initial value problems, always set up the complete linear system using all given conditions before concluding anything about existence or uniqueness.

Question 17

The general solution to a differential equation is $y = C_1 x^2 + C_2 x + C_3$ . Two particular solutions are given: $y_1 = 2x^2 + 3x - 1$ and $y_2 = x^2 - x + 4$ . If $y_3 = ay_1 + by_2$ where $a$ and $b$ are constants with $a + b = 1$ , which statement about $y_3$ is most accurate?

$y_3$ is a particular solution only for specific values of $a$ and $b$ that must be determined by substitution
$y_3$ is a particular solution only when $a$ and $b$ are both positive and $a + b = 1$
$y_3$ is always a particular solution to the same differential equation, regardless of the specific values of $a$ and $b$ (correct answer)
$y_3$ represents a linear combination that may not satisfy the original differential equation due to nonlinearity

Explanation: When you encounter questions about linear combinations of solutions to differential equations, you're dealing with the fundamental principle of superposition. This principle states that if you have a linear differential equation, any linear combination of its solutions is also a solution. Since both

y_1 = 2x^2 + 3x - 1

and

y_2 = x^2 - x + 4

are particular solutions to the same differential equation (they can both be written in the form

y = C_1x^2 + C_2x + C_3

), their linear combination

y_3 = ay_1 + by_2

will automatically satisfy the original equation. The constraint

a + b = 1

doesn't affect this property—it's simply given information that doesn't change the fundamental fact that linear combinations of solutions remain solutions. Let's examine why the other options miss the mark: Option A suggests you need specific values of

a

and

b

, but superposition works for any constants, not just particular ones. Option B incorrectly claims the constants must be positive—the sign of

a

and

b

is irrelevant to whether

y_3

satisfies the equation. Option D mentions "nonlinearity," but we're dealing with a linear differential equation here, as evidenced by the polynomial form of the general solution. Study tip: Remember that for linear differential equations, the superposition principle always holds. Any linear combination of solutions (particular or homogeneous) will be a solution to the same equation. This is a powerful tool for constructing new solutions from known ones.

Question 18

Consider the differential equation $\frac{d^2y}{dx^2} + 4y = 0$ . If $y_1 = A\cos(2x) + B\sin(2x)$ represents the general solution, and $y_2 = 3\cos(2x) - 2\sin(2x)$ represents a particular solution, what can be concluded about the relationship between the constants $A$ , $B$ and the specific values $3$ , $-2$ ?

$A = 3$ and $B = -2$ represent one possible assignment that makes the general solution equal the particular solution (correct answer)
$A$ and $B$ must always equal $3$ and $-2$ respectively for any solution to this differential equation
$A = 3$ and $B = -2$ are the only values that satisfy the differential equation's boundary conditions
The values $3$ and $-2$ are incorrect because they don't satisfy the characteristic equation of the differential equation

Explanation: The general solution

y_1 = A\cos(2x) + B\sin(2x)

represents all possible solutions to the differential equation, with

A

and

B

being arbitrary constants determined by initial conditions. The particular solution

y_2 = 3\cos(2x) - 2\sin(2x)

is obtained from the general solution by setting

A = 3

and

B = -2

. This represents just one specific member of the infinite family of solutions. Option B is wrong because

A

and

B

can take any real values. Option C is wrong because no boundary conditions are given. Option D is wrong because both

3\cos(2x)

and

-2\sin(2x)

satisfy the differential equation (the characteristic equation

r^2 + 4 = 0

gives

r = \pm 2i

, leading to the cosine and sine solutions).

Question 19

A student solving $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 12$ finds the homogeneous solution $y_h = C_1 e^{2x} + C_2 e^{3x}$ and a particular solution $y_p = 2$ . When writing the general solution as $y = C_1 e^{2x} + C_2 e^{3x} + 2$ , the student claims that the constants $C_1$ and $C_2$ can be chosen independently to satisfy any two initial conditions. Which analysis of this claim is most appropriate?

The claim is correct because the homogeneous and particular solutions address different aspects of the problem independently
The claim is correct because linear superposition allows the homogeneous solution constants to be determined separately from the particular solution (correct answer)
The claim is incorrect because the particular solution $y_p = 2$ constrains the possible values of $C_1$ and $C_2$
The claim is incorrect because the constant term $2$ should be absorbed into one of the arbitrary constants $C_1$ or $C_2$

Explanation: The student's claim is mathematically correct. For linear nonhomogeneous differential equations, the general solution is

y = y_h + y_p

where

y_h

contains all arbitrary constants and

y_p

is any particular solution to the nonhomogeneous equation. The particular solution

y_p = 2

is fixed and satisfies the nonhomogeneous equation, while the homogeneous solution

y_h = C_1 e^{2x} + C_2 e^{3x}

provides the freedom to satisfy initial conditions. The constants

C_1

and

C_2

can indeed be chosen independently because the exponential functions

e^{2x}

and

e^{3x}

are linearly independent. Option A is less precise about the mathematical principle involved. Option C incorrectly suggests constraint where none exists. Option D misunderstands the structure of the general solution.