Calculus 1 Quiz: Verifying De Solutions

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Verifying De SolutionsQuestion 1 of 20

For what non-zero value of the constant $k$ is the function $y = e^{kx^2}$ a solution to the differential equation $y' = 4xy$ ?

k = 1/2

k = -2

k = 4

k = 2

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Calculus 1 Quiz

Calculus 1 Quiz: Verifying De Solutions

Practice Verifying De Solutions in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Verifying De Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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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

For what non-zero value of the constant $k$ is the function $y = e^{kx^2}$ a solution to the differential equation $y' = 4xy$ ?

$k = 1/2$
$k = -2$
$k = 4$
$k = 2$ (correct answer)

Explanation: When you're asked to verify whether a function satisfies a differential equation, you need to find the derivative of the given function and check if it matches the equation's requirements. Starting with

y = e^{kx^2}

, you need to find

y'

using the chain rule. The derivative of

e^{kx^2}

e^{kx^2}

times the derivative of the exponent

kx^2

. Since the derivative of

kx^2

2kx

, we get:

y' = e^{kx^2} \cdot 2kx = 2kx \cdot e^{kx^2}

For this to satisfy the differential equation

y' = 4xy

, you need:

2kx \cdot e^{kx^2} = 4x \cdot e^{kx^2}

Since

y = e^{kx^2}

, this becomes:

2kxy = 4xy

Dividing both sides by

xy

(since we want non-zero

k

), we get:

2k = 4

k = 2

This confirms answer choice D is correct. Let's check why the other options fail: If you substitute A)

k = 1/2

, you'd get

y' = xy

, not

4xy

. For B)

k = -2

, you'd get

y' = -4xy

, which has the wrong sign. For C)

k = 4

, you'd get

y' = 8xy

, which is too large by a factor of 2. Study tip: When verifying solutions to differential equations, always substitute your proposed function directly into the equation. The chain rule is crucial for exponential functions with composite exponents—practice recognizing when you need it.

Question 2

The function $y = \frac{1}{x}$ is a solution to the differential equation $y' + y^2 = 0$ . Which of the following functions is also a solution?

$y = \frac{1}{x+3}$ (correct answer)
$y = \frac{3}{x}$
$y = \frac{1}{x} + 3$
$y = -\frac{1}{x}$

Explanation: The question asks which other function is a solution. We must test each option. Let's test

y = \frac{1}{x+3} = (x+3)^{-1}

. The derivative is

y' = -1(x+3)^{-2} = -\frac{1}{(x+3)^2}

. The term

y^2

\left(\frac{1}{x+3}\right)^2 = \frac{1}{(x+3)^2}

. Substituting into the DE gives

y' + y^2 = -\frac{1}{(x+3)^2} + \frac{1}{(x+3)^2} = 0

. So,

y = \frac{1}{x+3}

is a solution. The other options are not solutions. For B,

y'=-3/x^2

y'+y^2 = -3/x^2+9/x^2 = 6/x^2 \neq 0

. For C,

y'=-1/x^2

y'+y^2 = -1/x^2+(1/x+3)^2 \neq 0

. For D,

y'=1/x^2

y'+y^2 = 1/x^2+(-1/x)^2 = 2/x^2 \neq 0

. Note that the general solution is

y = 1/(x+C)

Question 3

Let $S$ be the set of all integer values of $n$ for which $y=x^n$ is a solution to the differential equation $x^2y'' - 2xy' + 2y = 0$ for $x > 0$ . Which of the following is $S$ ?

$\{1, 2\}$ (correct answer)
$\{-1, -2\}$
$\{1\}$
$\{2\}$

Explanation: Given

y = x^n

, we find the derivatives

y' = nx^{n-1}

and

y'' = n(n-1)x^{n-2}

. Substitute these into the differential equation:

x^2(n(n-1)x^{n-2}) - 2x(nx^{n-1}) + 2(x^n) = 0

. Simplify the expression:

n(n-1)x^n - 2nx^n + 2x^n = 0

. Factor out

x^n

(which is not zero for

x>0

x^n[n(n-1) - 2n + 2] = 0

. This requires the term in the brackets to be zero:

n^2 - n - 2n + 2 = 0

, which simplifies to the quadratic equation

n^2 - 3n + 2 = 0

. Factoring this equation gives

(n-1)(n-2) = 0

. The integer solutions are

n=1

and

n=2

. So, the set

S

\{1, 2\}

Question 4

A function $y(x)$ is a solution to the initial value problem $y'' - 4y' + 4y = 0$ , with $y(0) = 1$ and $y'(0) = 3$ . Which of the proposed functions is the correct solution?

$y = e^{2x} + xe^{2x}$ (correct answer)
$y = e^{2x} + 2xe^{2x}$
$y = e^{2x}$
$y = (1+x)e^{-2x}$

Explanation: The general solution to

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

(with characteristic equation

r^2-4r+4=0

(r-2)^2=0

) is

y = C_1 e^{2x} + C_2 x e^{2x}

. We must check which option satisfies both initial conditions. Let's test option A:

y = e^{2x} + xe^{2x}

. Check ICs:

y(0) = e^0 + 0 \cdot e^0 = 1

. This condition is met. Now find

y'

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

. Check the second IC:

y'(0) = 3e^0 + 2(0)e^0 = 3

. This condition is also met. Now we must verify it solves the DE.

y'' = 6e^{2x} + (2e^{2x} + 4xe^{2x}) = 8e^{2x} + 4xe^{2x}

. Substitute into the DE:

(8e^{2x} + 4xe^{2x}) - 4(3e^{2x} + 2xe^{2x}) + 4(e^{2x} + xe^{2x}) = 8e^{2x} + 4xe^{2x} - 12e^{2x} - 8xe^{2x} + 4e^{2x} + 4xe^{2x} = (8-12+4)e^{2x} + (4-8+4)xe^{2x} = 0

. So, A is the correct solution.

Question 5

A function $y(x)$ has a derivative $y' = \arctan(x)$ . Which of the following differential equations does this function satisfy?

$(1+x^2)y' = 1$
$y'' = \frac{1}{\sqrt{1-x^2}}$
$y'' - y' = 0$
$y'' = \frac{1}{1+x^2}$ (correct answer)

Explanation: When you encounter a problem giving you a derivative and asking for a differential equation, you need to find relationships between the function and its derivatives. Since you know

y' = \arctan(x)

, the key insight is to find

y''

by differentiating the given derivative. To find

y''

, differentiate both sides of

y' = \arctan(x)

y'' = \frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}

This directly gives us the differential equation

y'' = \frac{1}{1+x^2}

, which matches choice D. Now let's examine why the other options fail: Choice A states

(1+x^2)y' = 1

, which would mean

y' = \frac{1}{1+x^2}

. But we're told

y' = \arctan(x)

, not

\frac{1}{1+x^2}

. This confuses the function with its derivative. Choice B gives

y'' = \frac{1}{\sqrt{1-x^2}}

, which is actually the derivative of

\arcsin(x)

, not

\arctan(x)

. This is a common mix-up between inverse trigonometric derivatives. Choice C suggests

y'' - y' = 0

, or

y'' = y'

. This would mean the second derivative equals

\arctan(x)

, but we just calculated that

y'' = \frac{1}{1+x^2}

. Since

\arctan(x) \neq \frac{1}{1+x^2}

, this is incorrect. Study tip: Memorize the derivatives of inverse trigonometric functions, especially that

\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}

and

\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}

. When given derivative information, always consider differentiating again to find higher-order derivatives.

Question 6

On which interval is the function $y(x) = \frac{1}{4}(x-1)^2$ a solution to the differential equation $y' = \sqrt{y}$ ?

$[1, \infty)$ (correct answer)
$(-\infty, 1]$
$(-\infty, \infty)$
Only at the point $x=1$ .

Explanation: First, find the derivative of the given function:

y' = \frac{d}{dx} \left[ \frac{1}{4}(x-1)^2 \right] = \frac{1}{4} \cdot 2(x-1) = \frac{x-1}{2}

. Next, find the right-hand side of the differential equation using the function:

\sqrt{y} = \sqrt{\frac{1}{4}(x-1)^2} = \frac{1}{2}|x-1|

. For the function to be a solution, the derivative must equal the square root of the function:

\frac{x-1}{2} = \frac{1}{2}|x-1|

, which simplifies to

x-1 = |x-1|

. This equality holds true only when the quantity

x-1

is non-negative, i.e.,

x-1 \ge 0

, which means

x \ge 1

. Therefore, the function is a solution on the interval

[1, \infty)

Question 7

Which of the following functions is a solution to the initial value problem $x \ln(x) \frac{dy}{dx} = y$ , with the initial condition $y(e) = 2$ ?

$y = \frac{2x}{e}$
$y = \ln(x) + 1$
$y = 2\ln(x)$ (correct answer)
$y = (\ln x)^2 + 1$

Explanation: When you encounter an initial value problem with a differential equation, you need to verify that a proposed solution satisfies both the differential equation and the initial condition. Let's check option C:

y = 2\ln(x)

. First, we'll find its derivative:

\frac{dy}{dx} = \frac{2}{x}

. Now substitute into the original equation

x \ln(x) \frac{dy}{dx} = y

: Left side:

x \ln(x) \cdot \frac{2}{x} = 2\ln(x)

Right side:

y = 2\ln(x)

The equation is satisfied! Next, check the initial condition

y(e) = 2

y(e) = 2\ln(e) = 2 \cdot 1 = 2

✓ Now let's examine why the other options fail: Option A (

y = \frac{2x}{e}

): The derivative is

\frac{dy}{dx} = \frac{2}{e}

. Substituting gives

x \ln(x) \cdot \frac{2}{e} = \frac{2x\ln(x)}{e}

, which doesn't equal

\frac{2x}{e}

unless

\ln(x) = 1

for all

x

. Option B (

y = \ln(x) + 1

): With

\frac{dy}{dx} = \frac{1}{x}

, we get

x \ln(x) \cdot \frac{1}{x} = \ln(x)

, but we need

\ln(x) + 1

. This fails the differential equation. Option D (

y = (\ln x)^2 + 1

): The derivative is

\frac{dy}{dx} = \frac{2\ln(x)}{x}

. Substituting gives

2(\ln(x))^2

, which doesn't match

(\ln x)^2 + 1

. Study tip: Always verify both parts of an initial value problem—substitute into the differential equation first, then check the initial condition. Don't assume a function works just because it "looks right."

Question 8

Consider the differential equation $y' = 3y^{2/3}$ with initial condition $y(0)=0$ . Let $y_1(x) = x^3$ and $y_2(x) = 0$ for all $x$ . Which statement is correct?

Both $y_1(x)$ and $y_2(x)$ are solutions to the initial value problem. (correct answer)
Only $y_1(x)$ is a solution to the initial value problem.
Only $y_2(x)$ is a solution to the initial value problem.
Neither function is a solution to the initial value problem.

Explanation: First, check the trivial solution

y_2(x) = 0

. The derivative is

y_2' = 0

. The right side of the DE is

3(0)^{2/3} = 0

. So

0=0

, and the DE is satisfied. The initial condition is

y_2(0) = 0

, which is also satisfied. So

y_2(x)=0

is a solution. Next, check

y_1(x) = x^3

. The derivative is

y_1' = 3x^2

. The right side of the DE is

3(y_1)^{2/3} = 3(x^3)^{2/3} = 3x^2

. So

3x^2 = 3x^2

, and the DE is satisfied. The initial condition is

y_1(0) = 0^3 = 0

, which is also satisfied. So

y_1(x)=x^3

is also a solution. This is an example of an IVP with a non-unique solution.

Question 9

The functions $y_1 = e^{3x}$ and $y_2 = e^{-3x}$ are both solutions to the differential equation $y'' - 9y = 0$ . By the principle of superposition, $y = C_1 e^{3x} + C_2 e^{-3x}$ is the general solution. Which of the following hyperbolic functions is also a solution?

$y = \sinh(x)$
$y = \tanh(3x)$
$y = \cosh(9x)$
$y = \cosh(3x)$ (correct answer)

Explanation: When you encounter a differential equation with known solutions, you can often express other functions as linear combinations of those solutions if they satisfy the same equation. The key insight here is recognizing the relationship between exponential and hyperbolic functions. Since

y = C_1 e^{3x} + C_2 e^{-3x}

is the general solution to

y'' - 9y = 0

, any function that can be written in this form will also be a solution. Recall that hyperbolic cosine is defined as

\cosh(ax) = \frac{e^{ax} + e^{-ax}}{2}

, which means

\cosh(3x) = \frac{e^{3x} + e^{-3x}}{2}

. This is exactly the form

C_1 e^{3x} + C_2 e^{-3x}

with

C_1 = C_2 = \frac{1}{2}

. Therefore, choice D is correct. Looking at the incorrect options: Choice A,

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

, involves

e^x

and

e^{-x}

rather than

e^{3x}

and

e^{-3x}

, so it won't satisfy our differential equation. Choice B,

\tanh(3x) = \frac{\sinh(3x)}{\cosh(3x)}

, is a quotient of exponential functions, not a linear combination, so it cannot be expressed in the required form. Choice C,

\cosh(9x) = \frac{e^{9x} + e^{-9x}}{2}

, has the wrong exponential terms (

9x

instead of

3x

). Remember this pattern: when you have a second-order linear differential equation with exponential solutions

e^{ax}

and

e^{-ax}

, the hyperbolic functions

\cosh(ax)

and

\sinh(ax)

are automatically solutions too, since they're just specific linear combinations of those exponentials.

Question 10

For the differential equation $y'' - 5y' + 6y = 0$ , the function $y = e^{rx}$ is a solution if $r$ is a root of the characteristic equation $r^2 - 5r + 6 = 0$ , which are $r=2$ and $r=3$ . Thus, $y_1 = e^{2x}$ and $y_2 = e^{3x}$ are solutions. Which of the following is NOT a solution?

$y = 2e^{2x} - 5e^{3x}$
$y = e^{2x} + e^{3x}$
$y = e^{5x}$ (correct answer)
$y = 10e^{2x}$

Explanation: By the principle of superposition for linear homogeneous differential equations, any linear combination of solutions is also a solution. The general solution is

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

. Options A, B, and D are all of this form. Option A corresponds to

C_1=2, C_2=-5

. Option B corresponds to

C_1=1, C_2=1

. Option D corresponds to

C_1=10, C_2=0

. Therefore, they are all solutions. Option C,

y = e^{5x}

, is not a linear combination of

e^{2x}

and

e^{3x}

. We can verify it is not a solution by substituting it into the DE:

y' = 5e^{5x}

y'' = 25e^{5x}

. So,

25e^{5x} - 5(5e^{5x}) + 6(e^{5x}) = (25-25+6)e^{5x} = 6e^{5x} \neq 0

Question 11

A student claims that if $y_1$ is a solution to $y' + p(x)y = q_1(x)$ and $y_2$ is a solution to $y' + p(x)y = q_2(x)$ , then their sum $y_s = y_1 + y_2$ is a solution to $y' + p(x)y = q_1(x) + q_2(x)$ . To test this claim, consider $y_1 = x$ as a solution to $y' + y = x+1$ . Which of the following statements correctly applies the claim?

Since $y_2 = e^{-x}$ solves $y'+y=0$ , their sum $y_s = x+e^{-x}$ must solve $y'+y=x+1$ .
Since $y_2 = 1$ solves $y'+y=1$ , their sum $y_s = x+1$ must solve $y'+y=x+2$ . (correct answer)
Since $y_2 = x^2$ solves $y'+y=2x+x^2$ , their sum $y_s = x+x^2$ must solve $y'+y=x^2+3x+1$ .
The student's claim is false and cannot be applied.

Explanation: The student's claim is a valid superposition principle for linear non-homogeneous equations. We are given that

y_1=x

solves

y' + y = x+1

. Let's check:

y_1'=1

, so

1+x = x+1

. This is true. So we have

p(x)=1

and

q_1(x)=x+1

. Now we must test the second function given in each option. For B: Let

y_2 = 1

. Then

y_2' = 0

. The DE is

y'+y=1

. Substituting

y_2

0+1=1

. This is true. So

q_2(x)=1

. According to the claim, the sum

y_s = y_1+y_2 = x+1

should be a solution to

y'+y = q_1(x)+q_2(x) = (x+1)+1 = x+2

. Let's verify: for

y_s=x+1

y_s'=1

. So

y_s'+y_s = 1+(x+1)=x+2

. This matches. So option B is a correct application of the claim. The other options contain incorrect statements about

y_2

or the resulting equation for the sum.

Question 12

A student is verifying if $y=Ce^x - x - 1$ is a solution to the differential equation $y' - y = x$ . The student's work is shown below.

Step 1: Find the derivative of the proposed solution. $y' = Ce^x - 1$

Step 2: Substitute $y$ and $y'$ into the left side of the equation. LHS = $y' - y = (Ce^x - 1) - (Ce^x - x - 1)$

Step 3: Simplify the expression. LHS = $Ce^x - 1 - Ce^x + x - 1$

Step 4: Final result. LHS = $x - 2$

The student concludes that since $x-2 \neq x$ , the function is not a solution.

What is the error, if any, in the student's work?

There is no error; the function is not a solution.
The error is in Step 1; the derivative calculation is incorrect.
The error is in Step 3; a sign error was made during simplification. (correct answer)
The error is in Step 2; the substitution was performed incorrectly.

Explanation: Let's review the student's work. Step 1:

y=Ce^x - x - 1

, so

y' = Ce^x - 1

. This is correct. Step 2: Substituting into

y' - y

gives

(Ce^x - 1) - (Ce^x - x - 1)

. This is correct. Step 3: Simplifying the expression from Step 2:

Ce^x - 1 - Ce^x + x + 1

. The student wrote

-1

instead of

+1

when distributing the negative sign to the

-1

term in

y

. The correct simplification is

Ce^x - Ce^x - 1 + x + 1 = x

. This equals the right-hand side of the differential equation. Therefore, the function is a solution, and the student's error was a sign error in Step 3.

Question 13

Which of the following functions is a solution to the initial value problem $\frac{dy}{dx} = xy^2$ , with the initial condition $y(0) = 1$ ?

$y = \frac{2}{x^2+2}$
$y = \frac{2}{2-x^2}$ (correct answer)
$y = \frac{-2}{x^2+1}$
$y = \frac{1}{1-x^2}$

Explanation: First, check the initial condition

y(0)=1

. For choice B,

y(0) = \frac{2}{2-0^2} = 1

. The condition is met. Next, verify the differential equation. For

y = \frac{2}{2-x^2} = 2(2-x^2)^{-1}

, the derivative is

\frac{dy}{dx} = -2(2-x^2)^{-2}(-2x) = \frac{4x}{(2-x^2)^2}

. The right side of the DE is

xy^2 = x\left(\frac{2}{2-x^2}\right)^2 = x\frac{4}{(2-x^2)^2} = \frac{4x}{(2-x^2)^2}

. Since

\frac{dy}{dx} = xy^2

, this is the correct solution.
(A) satisfies the IC but gives

y' = -xy^2

. (C) solves the DE but fails the IC (

y(0)=-2

). (D) satisfies the IC but fails the DE.

Question 14

Which of the following functions is a solution to the differential equation $y' - \frac{1}{x} y = x \cos(x)$ for $x > 0$ ?

$y = -x \cos(x)$
$y = \cos(x) - x \sin(x)$
$y = x \cos(x)$
$y = x \sin(x)$ (correct answer)

Explanation: When you encounter a differential equation like this, you need to verify which function satisfies the equation by substituting it and its derivative back into the original equation. Let's check option D:

y = x \sin(x)

. First, find the derivative using the product rule:

y' = \sin(x) + x \cos(x)

. Now substitute both

y

and

y'

into the left side of the differential equation:

y' - \frac{1}{x}y = (\sin(x) + x \cos(x)) - \frac{1}{x}(x \sin(x)) = \sin(x) + x \cos(x) - \sin(x) = x \cos(x)

This matches the right side exactly, confirming that

y = x \sin(x)

is indeed the solution. Let's see why the other options fail. For option A,

y = -x \cos(x)

, the derivative is

y' = -\cos(x) + x \sin(x)

. Substituting gives us

-\cos(x) + x \sin(x) + \cos(x) = x \sin(x)

, which doesn't equal

x \cos(x)

. Option B,

y = \cos(x) - x \sin(x)

, has derivative

y' = -\sin(x) - \sin(x) - x \cos(x) = -2\sin(x) - x \cos(x)

. The substitution yields

-2\sin(x) - x \cos(x) - \frac{1}{x}(\cos(x) - x \sin(x))

, which is far from our target. Option C,

y = x \cos(x)

, gives

y' = \cos(x) - x \sin(x)

, leading to

\cos(x) - x \sin(x) - \cos(x) = -x \sin(x)

, not

x \cos(x)

. Always verify differential equation solutions by direct substitution—it's the most reliable method and helps you avoid algebraic mistakes.

Question 15

Which of the following functions is a solution to the differential equation $y' \cos(x) + y \sin(x) = 1$ ?

$y = \sec(x)$
$y = \cos(x)$
$y = \sin(x)$ (correct answer)
$y = \tan(x)$

y = x \sin(x)

. First, find the derivative using the product rule:

y' = \sin(x) + x \cos(x)

. Now substitute both

y

and

y'

into the left side of the differential equation:

y' - \frac{1}{x}y = (\sin(x) + x \cos(x)) - \frac{1}{x}(x \sin(x)) = \sin(x) + x \cos(x) - \sin(x) = x \cos(x)

This matches the right side exactly, confirming that

y = x \sin(x)

is indeed the solution. Let's see why the other options fail. For option A,

y = -x \cos(x)

, the derivative is

y' = -\cos(x) + x \sin(x)

. Substituting gives us

-\cos(x) + x \sin(x) + \cos(x) = x \sin(x)

, which doesn't equal

x \cos(x)

. Option B,

y = \cos(x) - x \sin(x)

, has derivative

y' = -\sin(x) - \sin(x) - x \cos(x) = -2\sin(x) - x \cos(x)

. The substitution yields

-2\sin(x) - x \cos(x) - \frac{1}{x}(\cos(x) - x \sin(x))

, which is far from our target. Option C,

y = x \cos(x)

, gives

y' = \cos(x) - x \sin(x)

, leading to

\cos(x) - x \sin(x) - \cos(x) = -x \sin(x)

, not

x \cos(x)

. Always verify differential equation solutions by direct substitution—it's the most reliable method and helps you avoid algebraic mistakes.

Question 16

Which of the following implicitly defined relations is a solution to the differential equation $(y-x) \frac{dy}{dx} = x+y$ ?

$y^2 - 2xy - x^2 = C$ (correct answer)
$y^2 + x^2 = C$
$y^2 - x^2 = C$
$y/x + x = C$

Explanation: We use implicit differentiation on the relation

y^2 - 2xy - x^2 = C

with respect to

x

. Differentiating term by term:

\frac{d}{dx}(y^2) - \frac{d}{dx}(2xy) - \frac{d}{dx}(x^2) = \frac{d}{dx}(C)

. This gives

2y\frac{dy}{dx} - (2y + 2x\frac{dy}{dx}) - 2x = 0

. Now, we solve for

\frac{dy}{dx}

. Group terms with

\frac{dy}{dx}

\frac{dy}{dx}(2y - 2x) = 2x + 2y

. Factoring out 2 from both sides gives

\frac{dy}{dx}(y - x) = x+y

. This matches the given differential equation.

Question 17

Given the differential equation $(x^2+1)y'' - 2xy' + 2y = 0$ , and that $y_1(x) = x$ is a solution, determine if $y_2(x) = x^2 - 1$ is also a solution.

Yes, $y_2(x)$ is a solution. (correct answer)
No, because $(x^2+1)y_2'' - 2xy_2' + 2y_2 = 2$ .
No, because $(x^2+1)y_2'' - 2xy_2' + 2y_2 = -2$ .
No, because $(x^2+1)y_2'' - 2xy_2' + 2y_2 = 4$ .

Explanation: We need to check if

y_2(x) = x^2 - 1

satisfies the differential equation. First, find the derivatives:

y_2' = 2x

and

y_2'' = 2

. Now, substitute these into the left-hand side of the differential equation:

(x^2+1)(2) - 2x(2x) + 2(x^2 - 1)

= 2x^2 + 2 - 4x^2 + 2x^2 - 2

. Combining like terms, we get

(2x^2 - 4x^2 + 2x^2) + (2 - 2) = 0

. Since the expression simplifies to 0, the equation

0=0

is true, and

y_2(x) = x^2 - 1

is indeed a solution. The distractors represent possible outcomes of common algebraic or differentiation errors.

Question 18

The function $y(t) = 20 + 80e^{-kt}$ is a solution to the differential equation $\frac{dy}{dt} = -0.05(y-20)$ . What is the value of the constant $k$ ?

0.05 (correct answer)
-0.05
20
4

Explanation: We need to find the value of

k

for which the function solves the differential equation. First, calculate the left-hand side (LHS),

\frac{dy}{dt}

. For

y(t) = 20 + 80e^{-kt}

, the derivative is

\frac{dy}{dt} = -k \cdot 80e^{-kt}

. Next, calculate the right-hand side (RHS),

-0.05(y-20)

. Substitute the function for

y

-0.05((20 + 80e^{-kt}) - 20) = -0.05(80e^{-kt}) = -4e^{-kt}

. Now, set the LHS equal to the RHS:

-80ke^{-kt} = -4e^{-kt}

. Since

e^{-kt}

is never zero, we can divide both sides by it, leaving

-80k = -4

. Solving for

k

gives

k = \frac{-4}{-80} = \frac{1}{20} = 0.05

Question 19

Consider the differential equation $y' = y^2 - x$ . Which of the following statements is true about the function $y = \sqrt{x}$ for $x>0$ ?

It is a solution to the differential equation.
It is not a solution, and at $x=1$ , the left side is $1/2$ and the right side is $0$ . (correct answer)
It is not a solution, and at $x=4$ , the left side is $1/4$ and the right side is $0$ .
It is not a solution, because $y'$ and $y^2 - x$ differ by a constant value.

Explanation: Let's check if

y = \sqrt{x} = x^{1/2}

is a solution to

y' = y^2 - x

. First, find the derivative:

y' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}

. Now, evaluate the right side of the DE:

y^2 - x = (\sqrt{x})^2 - x = x - x = 0

. For the function to be a solution, we must have

\frac{1}{2\sqrt{x}} = 0

, which is not true for any

x>0

. So, it is not a solution. Now we check the specific claims in the distractors. At

x=1

, the left side is

y'(1) = \frac{1}{2\sqrt{1}} = 1/2

, and the right side is

y(1)^2 - 1 = (\sqrt{1})^2 - 1 = 1 - 1 = 0

. Statement B is correct. Statement C is incorrect because at

x=4

, the right side is

y(4)^2 - 4 = 2^2 - 4 = 0

, not -12. Statement D is incorrect because the difference

y' - (y^2-x) = \frac{1}{2\sqrt{x}} - 0

is not a constant.

Question 20

Let $y=f(x)$ be a solution to the differential equation $y' = y \cdot \cos(x)$ . Which of the following functions must also be a solution?

$y = (f(x))^3$
$y = f(x) + 3$
$y = f(3x)$
$y = 3f(x)$ (correct answer)

Explanation: When working with differential equations, you need to understand how transformations affect solutions. If a function satisfies a differential equation, certain transformations will preserve that property while others won't. Let's start with what we know:

f(x)

satisfies

y' = y \cdot \cos(x)

, meaning

f'(x) = f(x) \cdot \cos(x)

. For option D,

y = 3f(x)

, let's check if it satisfies the differential equation. Taking the derivative:

y' = 3f'(x) = 3f(x) \cdot \cos(x) = (3f(x)) \cdot \cos(x) = y \cdot \cos(x)

. This works! Multiplying a solution by a constant preserves the differential equation because derivatives distribute over constant multiples. Now let's see why the other options fail: Option A,

y = (f(x))^3

: Using the chain rule,

y' = 3(f(x))^2 \cdot f'(x) = 3(f(x))^2 \cdot f(x) \cdot \cos(x) = 3(f(x))^3 \cdot \cos(x) = 3y \cdot \cos(x)

. This gives us an extra factor of 3, so it doesn't satisfy our original equation. Option B,

y = f(x) + 3

: Here

y' = f'(x) = f(x) \cdot \cos(x)

, but we need

y' = (f(x) + 3) \cdot \cos(x)

. Adding a constant creates a mismatch. Option C,

y = f(3x)

: Using the chain rule,

y' = f'(3x) \cdot 3 = 3f(3x) \cdot \cos(3x)

. The argument of cosine changes, breaking the equation. Remember: linear differential equations preserve solutions under scalar multiplication, but be cautious with other transformations like addition, composition, or powers.