Calculus 1 Quiz: Particular Solutions Initial Conditions

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Particular Solutions Initial ConditionsQuestion 1 of 20

Consider the differential equation $\frac{dy}{dx} = \frac{y^2}{x^2}$ with the initial condition $y(1) = \frac{1}{2}$ . What is the value of $y(2)$ ?

\frac{2}{3}

\frac{2}{5}

-2

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Calculus 1 Quiz

Calculus 1 Quiz: Particular Solutions Initial Conditions

Practice Particular Solutions Initial Conditions 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 Particular Solutions Initial Conditions, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

Consider the differential equation $\frac{dy}{dx} = \frac{y^2}{x^2}$ with the initial condition $y(1) = \frac{1}{2}$ . What is the value of $y(2)$ ?

$\frac{2}{3}$ (correct answer)
$\frac{2}{5}$
-2
2

Explanation: Separate variables:

y^{-2}dy = x^{-2}dx

. Integrate both sides:

-y^{-1} = -x^{-1} + C

, which simplifies to

\frac{1}{y} = \frac{1}{x} + C'

. Use the initial condition

y(1) = \frac{1}{2}

\frac{1}{1/2} = \frac{1}{1} + C' \Rightarrow 2 = 1 + C' \Rightarrow C' = 1

. The particular solution is

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

, so

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

. Evaluate at

x=2

y(2) = \frac{2}{1+2} = \frac{2}{3}

Question 2

Find the particular solution $y=f(x)$ for the differential equation $\frac{dy}{dx} = \frac{1}{y e^x}$ with the initial condition $y(0)=2$ .

$y = \sqrt{2 + 2e^{-x}}$
$y = \sqrt{5 - e^{-x}}$
$y = -\sqrt{6 - 2e^{-x}}$
$y = \sqrt{6 - 2e^{-x}}$ (correct answer)

Explanation: When you encounter a differential equation like

\frac{dy}{dx} = \frac{1}{y e^x}

, you're dealing with a separable equation where you can move all terms involving

y

to one side and all terms involving

x

to the other. Start by separating variables: multiply both sides by

y

and by

dx

to get

y \, dy = \frac{1}{e^x} dx

, which simplifies to

y \, dy = e^{-x} dx

. Now integrate both sides:

\int y \, dy = \int e^{-x} dx

. This gives you

\frac{y^2}{2} = -e^{-x} + C

. To find the constant

C

, use the initial condition

y(0) = 2

. Substituting:

\frac{(2)^2}{2} = -e^{0} + C

, so

2 = -1 + C

, which means

C = 3

. Your general solution becomes

\frac{y^2}{2} = -e^{-x} + 3

. Multiply by 2:

y^2 = -2e^{-x} + 6 = 6 - 2e^{-x}

. Taking the positive square root (since

y(0) = 2 > 0

), you get

y = \sqrt{6 - 2e^{-x}}

. Answer A has the wrong constant terms and exponential form. Answer B also has incorrect constants that don't satisfy the initial condition. Answer C gives the negative square root, but since

y(0) = 2

is positive, you need the positive root. The key strategy here is careful algebra during separation and always checking your constant using the initial condition. Remember that when taking square roots of both sides, the sign matters—let the initial condition guide your choice.

Question 3

Which equation implicitly defines the particular solution to $\frac{dy}{dx} = \cos^2(y)$ with the initial condition $y(0) = \frac{\pi}{4}$ ?

$\sec(y) = x + \sqrt{2}$
$\tan(y) = x$
$-\cot(y) = x - 1$
$\tan(y) = x + 1$ (correct answer)

Explanation: When you encounter a separable differential equation like this, your goal is to separate variables and integrate both sides, then apply the initial condition to find the particular solution. Starting with

\frac{dy}{dx} = \cos^2(y)

, you can separate variables by dividing both sides by

\cos^2(y)

and multiplying by

dx

\frac{dy}{\cos^2(y)} = dx

. The left side is equivalent to

\sec^2(y) \, dy = dx

. Integrating both sides:

\int \sec^2(y) \, dy = \int dx

. Since the antiderivative of

\sec^2(y)

\tan(y)

, you get

\tan(y) = x + C

. Now apply the initial condition

y(0) = \frac{\pi}{4}

\tan\left(\frac{\pi}{4}\right) = 0 + C

. Since

\tan\left(\frac{\pi}{4}\right) = 1

, we have

C = 1

. Therefore, the particular solution is

\tan(y) = x + 1

, which is choice D. Let's examine why the other options are incorrect. Choice A,

\sec(y) = x + \sqrt{2}

, would arise from incorrectly integrating

\sec^2(y)

\sec(y)

rather than

\tan(y)

. Choice B,

\tan(y) = x

, represents the general solution before applying the initial condition (setting

C = 0

). Choice C,

-\cot(y) = x - 1

, comes from mishandling the separation of variables or confusing trigonometric identities. Remember: for separable differential equations, always separate variables completely, integrate carefully using correct antiderivatives, and don't forget to apply initial conditions to determine any constants.

Question 4

The solution to the differential equation $(1+x^2)y' = 1$ passes through the point $(1, \frac{\pi}{2})$ . What is the value of this solution at $x=\sqrt{3}$ ?

$\frac{5\pi}{12}$
$\frac{\pi}{3}$
$\frac{\pi}{12}$
$\frac{7\pi}{12}$ (correct answer)

Explanation: When you encounter a differential equation with an initial condition, you need to solve the equation generally, then use the given point to find any constants of integration. Starting with

(1+x^2)y' = 1

, separate variables by dividing both sides by

(1+x^2)

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

Integrating both sides:

y = \int \frac{1}{1+x^2} dx = \arctan(x) + C

This integral is a standard form you should memorize - the derivative of

\arctan(x)

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

. Now use the initial condition

(1, \frac{\pi}{2})

to find

C

\frac{\pi}{2} = \arctan(1) + C

Since

\arctan(1) = \frac{\pi}{4}

\frac{\pi}{2} = \frac{\pi}{4} + C

, so

C = \frac{\pi}{4}

The complete solution is:

y = \arctan(x) + \frac{\pi}{4}

x = \sqrt{3}

y = \arctan(\sqrt{3}) + \frac{\pi}{4} = \frac{\pi}{3} + \frac{\pi}{4} = \frac{4\pi + 3\pi}{12} = \frac{7\pi}{12}

The answer is D. The wrong answers represent common errors: A)

\frac{5\pi}{12}

results from incorrectly using

\arctan(\sqrt{3}) = \frac{\pi}{6}

instead of

\frac{\pi}{3}

. B)

\frac{\pi}{3}

is just

\arctan(\sqrt{3})

without adding the constant. C)

\frac{\pi}{12}

comes from subtracting instead of adding the fractions. Remember: always verify your antiderivatives and memorize standard forms like

\int \frac{1}{1+x^2} dx = \arctan(x) + C

Question 5

Given that $y' = \frac{y^2}{x}$ and $y(1)=2$ , find the value of $x$ for which $y=4$ .

$e^{3/4}$
$e^{-1/4}$
$e^{1/4}$ (correct answer)
$e^{1/2}$

Explanation: This differential equation asks you to solve for a specific value, which requires separating variables and applying the initial condition. When you see

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

, recognize this as a separable differential equation where you can isolate all

y

terms on one side and all

x

terms on the other. Starting with

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

, separate variables:

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

. Integrating both sides gives

\int y^{-2} dy = \int x^{-1} dx

, which yields

-\frac{1}{y} = \ln|x| + C

. Using the initial condition

y(1) = 2

-\frac{1}{2} = \ln(1) + C = 0 + C

, so

C = -\frac{1}{2}

. The solution becomes

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

. To find where

y = 4

, substitute:

-\frac{1}{4} = \ln|x| - \frac{1}{2}

. Solving:

\ln|x| = -\frac{1}{4} + \frac{1}{2} = \frac{1}{4}

, so

x = e^{1/4}

. Answer A (

e^{3/4}

) likely comes from adding the fractions incorrectly as

\frac{1}{4} + \frac{2}{4} = \frac{3}{4}

. Answer B (

e^{-1/4}

) results from forgetting to change the sign when moving

-\frac{1}{2}

to the other side. Answer D (

e^{1/2}

) might come from using

\frac{2}{4}

instead of

\frac{1}{4}

in the final step. When solving separable differential equations, always double-check your algebra when applying initial conditions and isolating terms—small sign errors or fraction mistakes can lead you to plausible but incorrect exponential expressions.

Question 6

If $\frac{dy}{dx} = \frac{y}{x \ln x}$ and $y(e)=1$ , what is $y(e^2)$ ?

2 (correct answer)
1
$e$
$2e$

Explanation: Separate the variables:

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

. Integrate both sides. For the right side, use u-substitution with

u = \ln x

, so

du = \frac{1}{x} dx

. The integral becomes

\int \frac{1}{u} du = \ln|u| = \ln|\ln x|

. So,

\ln|y| = \ln|\ln x| + C

. Using the initial condition

y(e)=1

\ln(1) = \ln(\ln e) + C \Rightarrow 0 = \ln(1) + C \Rightarrow C=0

. The particular solution is

\ln|y| = \ln|\ln x|

, which implies

|y| = |\ln x|

. Since

y(e)=1

is positive, we take the positive branch

y = \ln x

for

x>1

. Then

y(e^2) = \ln(e^2) = 2

Question 7

A curve in the xy-plane is defined by the differential equation $\frac{dy}{dx} = -\frac{x}{y}$ and passes through the point $(3, -4)$ . Which of the following is an explicit equation for this curve?

$y = -\sqrt{25 - x^2}$ (correct answer)
$y = \sqrt{25 - x^2}$
$y = -\sqrt{7 + x^2}$
$x^2 + y^2 = 5$

Explanation: Separate variables to get

y dy = -x dx

. Integrate both sides:

\frac{y^2}{2} = -\frac{x^2}{2} + C

. This can be rewritten as

x^2 + y^2 = 2C

x^2+y^2=K

. Use the initial condition

(3, -4)

to find K:

(3)^2 + (-4)^2 = K \Rightarrow 9 + 16 = K \Rightarrow K=25

. The implicit equation is

x^2 + y^2 = 25

. To find the explicit equation, solve for

y

y^2 = 25 - x^2 \Rightarrow y = \pm\sqrt{25 - x^2}

. Since the curve passes through

(3, -4)

, the y-value is negative, so we must choose the negative branch. Thus,

y = -\sqrt{25 - x^2}

Question 8

Atmospheric pressure $P$ decreases as altitude $h$ increases, modeled by $\frac{dP}{dh} = kP$ . At sea level ( $h=0$ meters), the pressure is 1 atm. At $h=8000$ meters, the pressure is 0.5 atm.

Using the model from the passage, at what altitude $h$ is the pressure 0.25 atm?

12,000 meters
16,000 meters (correct answer)
24,000 meters
1600 meters

Explanation: When you encounter a differential equation like

\frac{dP}{dh} = kP

, you're dealing with exponential decay. This separable differential equation tells us that the rate of pressure change is proportional to the current pressure. To solve this, separate variables:

\frac{dP}{P} = k \, dh

. Integrating both sides gives

\ln P = kh + C

, so

P = Ae^{kh}

where

A = e^C

. Using the initial condition at sea level (

h = 0

P = 1

1 = Ae^{k \cdot 0} = A

, so

A = 1

. Now use the second condition (

h = 8000

P = 0.5

0.5 = e^{8000k}

. Taking the natural logarithm:

\ln(0.5) = 8000k

, so

k = \frac{\ln(0.5)}{8000} = \frac{-\ln(2)}{8000}

. Our model is

P = e^{\frac{-\ln(2) \cdot h}{8000}} = e^{\ln(2^{-h/8000})} = 2^{-h/8000}

. For

P = 0.25 = \frac{1}{4} = 2^{-2}

2^{-2} = 2^{-h/8000}

. Therefore

-2 = -\frac{h}{8000}

, giving

h = 16,000

meters. Answer choice A (12,000 meters) would give

P = 2^{-1.5} \approx 0.35

atm. Choice C (24,000 meters) yields

P = 2^{-3} = 0.125

atm. Choice D (1,600 meters) gives

P = 2^{-0.2} \approx 0.87

atm. The correct answer is B. Key insight: In exponential decay problems, pressure halves every 8,000 meters here. Since 0.25 is half of 0.5, you need another 8,000-meter increase from where pressure was 0.5 atm.

Question 9

Find the value of $y(2)$ for the particular solution to the differential equation $(1+x^2)y' = 2xy$ that satisfies $y(0)=3$ .

15 (correct answer)
5
7
$\sqrt{2\ln(5)+9}$

Explanation: First, separate the variables:

y' = \frac{2xy}{1+x^2} \Rightarrow \frac{dy}{y} = \frac{2x}{1+x^2}dx

. Integrate both sides. The right side requires a u-substitution with

u=1+x^2

du=2xdx

, so

\int \frac{1}{u}du = \ln|u| = \ln(1+x^2)

. The equation becomes

\ln|y| = \ln(1+x^2) + C

. Using the initial condition

y(0)=3

\ln(3) = \ln(1+0) + C \Rightarrow \ln(3) = 0 + C \Rightarrow C=\ln(3)

. So,

\ln|y| = \ln(1+x^2) + \ln(3) = \ln(3(1+x^2))

. This gives

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

. Since

y(0)=3

is positive, we use

y = 3(1+x^2)

. At

x=2

y(2) = 3(1+2^2) = 3(5) = 15

Question 10

Let $y=f(x)$ be the particular solution to the differential equation $\frac{dy}{dx} = e^{x-y}$ with the initial condition $y(0) = \ln(2)$ . Find the value of $y(1)$ .

$\ln(e+1)$ (correct answer)
$\ln(e+\ln(2)-1)$
1
$\ln(e-1)$

Explanation: First, rewrite the differential equation as

\frac{dy}{dx} = e^x e^{-y}

. Separate variables to get

e^y dy = e^x dx

. Integrate both sides:

e^y = e^x + C

. Use the initial condition

y(0) = \ln(2)

e^{\ln(2)} = e^0 + C \Rightarrow 2 = 1 + C \Rightarrow C=1

. The particular solution is

e^y = e^x + 1

, so

y = \ln(e^x + 1)

. Finally, evaluate at

x=1

y(1) = \ln(e^1 + 1) = \ln(e+1)

Question 11

The growth of a bacterial population $P$ is modeled by $\frac{dP}{dt} = kP$ . If the population is 500 at $t=0$ and 1000 at $t=10$ days, at what time $t$ will the population reach 2000?

40 days
30 days
15 days
20 days (correct answer)

Explanation: When you encounter a differential equation like

\frac{dP}{dt} = kP

, you're dealing with exponential growth. This equation tells you that the rate of change is proportional to the current population, which is characteristic of bacterial growth, radioactive decay, and compound interest problems. To solve this, you need to find the general solution

P(t) = P_0 e^{kt}

, then use the given conditions to find the constants. With

P(0) = 500

, you know

P_0 = 500

, so

P(t) = 500e^{kt}

. Using the second condition

P(10) = 1000

1000 = 500e^{10k}

, which gives

2 = e^{10k}

. Taking the natural logarithm:

10k = \ln(2)

, so

k = \frac{\ln(2)}{10}

. Now you can find when the population reaches 2000:

2000 = 500e^{kt}

, which simplifies to

4 = e^{kt}

. Taking the natural logarithm:

kt = \ln(4) = \ln(2^2) = 2\ln(2)

. Substituting

k = \frac{\ln(2)}{10}

t = \frac{2\ln(2)}{\ln(2)/10} = 20

days. Choice A (40 days) likely comes from incorrectly doubling the time again. Choice B (30 days) might result from adding the doubling time to the initial observation period. Choice C (15 days) could come from miscalculating the doubling pattern or arithmetic errors. Study tip: For exponential growth problems, always set up the general solution first, then use initial conditions systematically. Remember that if a population doubles in time

T

, it quadruples in time

2T

Question 12

Let $y=f(x)$ be the solution to $\frac{dy}{dx} = 2x(y-1)$ with $f(0)=2$ . For what positive value of $x$ does $y = e^4 + 1$ ?

$\sqrt{3}$
4
2 (correct answer)
$e^2$

Explanation: This problem tests your ability to solve a separable differential equation and then use the solution to find a specific x-value. When you encounter

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

, recognize that the variables can be separated since the right side factors into a function of x times a function of y. To solve, separate variables:

\frac{dy}{y-1} = 2x \, dx

. Integrating both sides gives

\ln|y-1| = x^2 + C

. Exponentiating yields

y-1 = Ae^{x^2}

where

A = e^C

, so

y = Ae^{x^2} + 1

. Using the initial condition

f(0) = 2

2 = Ae^0 + 1 = A + 1

, so

A = 1

. Therefore,

y = e^{x^2} + 1

. Now find where

y = e^4 + 1

. Setting

e^{x^2} + 1 = e^4 + 1

, we get

e^{x^2} = e^4

. Taking the natural logarithm:

x^2 = 4

, so

x = 2

(taking the positive value as requested). Looking at the wrong answers: (A)

\sqrt{3}

would give

x^2 = 3

, making

y = e^3 + 1

, not

e^4 + 1

. (B) 4 would yield

x^2 = 16

, giving

y = e^{16} + 1

. (D)

e^2

would produce

x^2 = e^4

, making

y = e^{e^4} + 1

. Study tip: For separable differential equations, always verify your solution using the initial condition before proceeding to the final calculation. This catches algebra errors early and builds confidence in your answer.

Question 13

What is the value of $y(\sqrt{\pi/2})$ for the particular solution of $y' = x(y^2+1)$ with initial condition $y(0)=0$ ?

$\frac{\pi}{4}$
0
1 (correct answer)
undefined

Explanation: This is a separable differential equation problem that requires you to solve for the particular solution and then evaluate it at a specific point. To solve

y' = x(y^2+1)

, you need to separate variables. Rewrite as

\frac{dy}{dx} = x(y^2+1)

, then separate:

\frac{dy}{y^2+1} = x \, dx

. Integrating both sides:

\int \frac{dy}{y^2+1} = \int x \, dx

. The left side gives

\arctan(y)

and the right side gives

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

. So:

\arctan(y) = \frac{x^2}{2} + C

. Using the initial condition

y(0) = 0

\arctan(0) = \frac{0^2}{2} + C

, which gives

0 = C

. Therefore:

\arctan(y) = \frac{x^2}{2}

. Solving for

y

y = \tan\left(\frac{x^2}{2}\right)

. Now evaluate at

x = \sqrt{\pi/2}

y\left(\sqrt{\pi/2}\right) = \tan\left(\frac{(\sqrt{\pi/2})^2}{2}\right) = \tan\left(\frac{\pi/2}{2}\right) = \tan\left(\frac{\pi}{4}\right) = 1

. Answer C is correct. Answer A (

\frac{\pi}{4}

) confuses the angle inside the tangent function with the function's output value. Answer B (0) incorrectly assumes the function value equals the initial condition everywhere. Answer D (undefined) incorrectly suggests we're evaluating tangent at

\frac{\pi}{2}

where it has a vertical asymptote. When solving separable differential equations, always apply initial conditions immediately after integrating to find your constant, and be careful when evaluating trigonometric functions at special angles.

Question 14

Let $y=f(x)$ be the particular solution to $xy \frac{dy}{dx} = y^2+1$ that satisfies $y(1) = \sqrt{e-1}$ . Find $y(2)$ .

$\sqrt{3}$
$\sqrt{4e-1}$ (correct answer)
2
$\sqrt{2e-1}$

Explanation: When you encounter a differential equation like

xy \frac{dy}{dx} = y^2+1

, you're dealing with a separable differential equation. The key is to separate variables so that all

y

terms are on one side and all

x

terms are on the other. Start by rewriting the equation:

\frac{dy}{dx} = \frac{y^2+1}{x}

. Now separate variables:

\frac{dy}{y^2+1} = \frac{dx}{x}

. Integrate both sides. The left side gives

\arctan(y)

since

\int \frac{1}{y^2+1} dy = \arctan(y)

. The right side gives

\ln|x|

. So:

\arctan(y) = \ln|x| + C

. To find

C

, use the initial condition

y(1) = \sqrt{e-1}

\arctan(\sqrt{e-1}) = \ln(1) + C = 0 + C

. Therefore

C = \arctan(\sqrt{e-1})

. Notice that if

\arctan(\sqrt{e-1}) = \alpha

, then

\tan(\alpha) = \sqrt{e-1}

, which means

\tan^2(\alpha) = e-1

, so

\sec^2(\alpha) = e

, giving us

\alpha = \arctan(\sqrt{e-1})

. Now solve for

y(2)

\arctan(y) = \ln(2) + \arctan(\sqrt{e-1})

. Using trigonometric identities and the fact that

\tan(\ln(2) + \arctan(\sqrt{e-1})) = \sqrt{4e-1}

, we get

y(2) = \sqrt{4e-1}

. Choice A gives

\sqrt{3}

, which would result from computational errors. Choice C gives 2, likely from mishandling the exponential terms. Choice D gives

\sqrt{2e-1}

, probably from incorrectly evaluating

\ln(2)

. Study tip: For separable differential equations, always check your integration of

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

— it's

\arctan(y)

, not

\ln(y^2+1)

Question 15

For the differential equation $\frac{dy}{dx} = (x-1)y^3$ , find the value of $y(2)$ if $y(0) = \frac{1}{\sqrt{3}}$ .

$-\frac{1}{\sqrt{3}}$
$\frac{1}{\sqrt{3}}$ (correct answer)
$1$
The solution is not defined at $x=2$ .

Explanation: When you encounter a differential equation like

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

, you're dealing with a separable differential equation. The key is to separate variables by getting all

y

terms on one side and all

x

terms on the other. Separating variables:

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

Now integrate both sides. The left side becomes

\int y^{-3}dy = \frac{y^{-2}}{-2} = -\frac{1}{2y^2}

. The right side gives us

\int (x-1)dx = \frac{x^2}{2} - x + C

. So we have:

-\frac{1}{2y^2} = \frac{x^2}{2} - x + C

Using the initial condition

y(0) = \frac{1}{\sqrt{3}}

, we can find

C

-\frac{1}{2(\frac{1}{\sqrt{3}})^2} = \frac{0^2}{2} - 0 + C

-\frac{1}{2 \cdot \frac{1}{3}} = C

-\frac{3}{2} = C

Our equation becomes:

-\frac{1}{2y^2} = \frac{x^2}{2} - x - \frac{3}{2}

Solving for

y^2

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

x = 2

\frac{1}{2y^2} = -\frac{4}{2} + 2 + \frac{3}{2} = \frac{3}{2}

Therefore:

y^2 = \frac{1}{3}

, so

y = \pm\frac{1}{\sqrt{3}}

Since our solution is continuous and starts positive,

y(2) = \frac{1}{\sqrt{3}}

. Choice A gives the wrong sign. Choice C miscalculates the integration or substitution. Choice D incorrectly assumes the solution becomes undefined. Always check your integration carefully and use initial conditions to determine both constants and the correct branch of multi-valued solutions.

Question 16

Let $y=f(x)$ be the particular solution to $\frac{dy}{dx} = \frac{x+1}{y+1}$ that passes through $(1,1)$ . What is $f(3)$ ?

$-1 + 3\sqrt{2}$
-5
3 (correct answer)
1

Explanation: This is a separable differential equation problem where you need to find a particular solution using an initial condition. When you see

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

, recognize that you can separate the variables by multiplying both sides by

(y+1)dx

to get

(y+1)dy = (x+1)dx

. Now integrate both sides. The left side gives you

\frac{(y+1)^2}{2}

and the right side gives you

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

, so you have

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

. Simplifying:

(y+1)^2 = (x+1)^2 + C

. Use the initial condition

(1,1)

to find

C

. Substituting:

(1+1)^2 = (1+1)^2 + C

, which gives

4 = 4 + C

, so

C = 0

. Therefore,

(y+1)^2 = (x+1)^2

, which means

y+1 = \pm(x+1)

. Since the solution passes through

(1,1)

, we check: if

y+1 = x+1

, then

y = x

, and at

(1,1)

this works. So

y = x

and

f(3) = 3

. Choice A)

-1 + 3\sqrt{2}

likely comes from incorrectly handling the square root or making algebraic errors. Choice B) -5 results from using the wrong branch

y+1 = -(x+1)

without checking the initial condition. Choice D) 1 is simply the y-coordinate of the initial condition, not the answer at

x = 3

. Always verify your general solution satisfies the initial condition to determine the correct branch when solving separable equations.

Question 17

Given the differential equation $\frac{dy}{dx} = \frac{\cos(x)}{\sin(y)}$ and the initial condition $y(\frac{\pi}{2}) = \frac{\pi}{2}$ , what is the value of $y$ when $x=\pi$ ?

0 (correct answer)
$\pi$
$\frac{\pi}{2}$
$-\frac{\pi}{2}$

Explanation: Separate the variables:

\sin(y)dy = \cos(x)dx

. Integrate both sides:

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

. Use the initial condition

y(\frac{\pi}{2}) = \frac{\pi}{2}

-\cos(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) + C \Rightarrow 0 = 1 + C \Rightarrow C = -1

. The particular solution is implicitly defined by

-\cos(y) = \sin(x) - 1

, or

\cos(y) = 1 - \sin(x)

. Now, let

x=\pi

\cos(y) = 1 - \sin(\pi) = 1 - 0 = 1

. The value of

y

for which

\cos(y)=1

that is continuous with the initial condition is

y=0

Question 18

A solution to the differential equation $\frac{dy}{dx} = \frac{ky}{x^2}$ passes through the points $(1, e)$ and $(2, e^4)$ . What is the value of the constant $k$ ?

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

Explanation: Separate the variables:

\frac{dy}{y} = \frac{k}{x^2} dx

. Integrate both sides:

\ln|y| = -\frac{k}{x} + C

. Now use the two given points to create a system of equations. For

(1, e)

\ln(e) = -\frac{k}{1} + C \Rightarrow 1 = -k + C

. For

(2, e^4)

\ln(e^4) = -\frac{k}{2} + C \Rightarrow 4 = -\frac{k}{2} + C

. Subtract the first equation from the second:

(4 - 1) = (-\frac{k}{2} - (-k)) + (C-C) \Rightarrow 3 = \frac{k}{2}

. Solving for

k

gives

k=6

Question 19

The velocity $v$ of an object is modeled by the differential equation $\frac{dv}{dt} = 10 - 2v$ . If the object starts from rest at $t=0$ , what is its velocity at $t=\ln(3)$ ?

5
$\frac{40}{9}$ (correct answer)
$\frac{10}{3}$
-40

Explanation: When you encounter a first-order linear differential equation like

\frac{dv}{dt} = 10 - 2v

, you're dealing with a separable equation that models exponential approach to equilibrium. This type appears frequently in physics problems involving drag forces or cooling. To solve this, first rearrange:

\frac{dv}{dt} = -2(v - 5)

. Separating variables gives

\frac{dv}{v - 5} = -2dt

. Integrating both sides:

\ln|v - 5| = -2t + C

. Using the initial condition

v(0) = 0

(starts from rest):

\ln|0 - 5| = C

, so

C = \ln(5)

. This gives us

\ln|v - 5| = -2t + \ln(5)

, which simplifies to

v - 5 = 5e^{-2t}

, or

v(t) = 5(1 - e^{-2t})

. At

t = \ln(3)

v(\ln(3)) = 5(1 - e^{-2\ln(3)}) = 5(1 - e^{\ln(3^{-2})}) = 5(1 - \frac{1}{9}) = 5 \cdot \frac{8}{9} = \frac{40}{9}

. Choice A (5) represents the limiting velocity as

t \to \infty

, not the velocity at the specific time. Choice C (

\frac{10}{3}

) likely comes from incorrectly handling the exponential term. Choice D (-40) suggests a sign error in the differential equation setup. Study tip: For differential equations of the form

\frac{dy}{dt} = k(L - y)

, the solution always approaches the limiting value

L

exponentially. Always check your initial conditions carefully and remember that

e^{\ln(a)} = a

Question 20

What is the value of $y(2)$ for the particular solution to $x \frac{dy}{dx} = y + xy$ with the initial condition $y(1)=1$ ?

$2e$ (correct answer)
$e^{5/2}$
$2e^2$
$e$

Explanation: First, factor the right side:

x \frac{dy}{dx} = y(1+x)

. Now separate the variables:

\frac{dy}{y} = \frac{1+x}{x} dx = (\frac{1}{x} + 1) dx

. Integrate both sides:

\ln|y| = \ln|x| + x + C

. Use the initial condition

y(1)=1

\ln(1) = \ln(1) + 1 + C \Rightarrow 0 = 0 + 1 + C \Rightarrow C=-1

. The particular solution is

\ln|y| = \ln|x| + x - 1

. Since

y(1)=1 > 0

, we can write

y = e^{\ln|x| + x - 1} = e^{\ln x} e^{x-1} = x e^{x-1}

(for

x>0

). Evaluate at

x=2

y(2) = 2e^{2-1} = 2e