Calculus 2 Quiz: Particular Solutions Initial Conditions

19 questions · exam conditions

0:00

Particular Solutions Initial ConditionsQuestion 1 of 19

A function $y=f(x)$ is a solution to the differential equation $y' = xe^{-y}$ . If the graph of the function passes through the point $(0, \ln 2)$ , what is the value of $f(2)$ ?

\ln(2 + \ln 2)

\ln 4

\ln 2

2

← Back to quizzes

Calculus 2 Quiz

Calculus 2 Quiz: Particular Solutions Initial Conditions

Practice Particular Solutions Initial Conditions in Calculus 2 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 Particular Solutions Initial Conditions, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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

A function $y=f(x)$ is a solution to the differential equation $y' = xe^{-y}$ . If the graph of the function passes through the point $(0, \ln 2)$ , what is the value of $f(2)$ ?

$\ln(2 + \ln 2)$
$\ln 4$ (correct answer)
$\ln 2$
$2$

Explanation: The differential equation is separable. We write

\frac{dy}{dx} = xe^{-y}

, which separates to

e^y dy = x dx

. Integrating both sides gives

\int e^y dy = \int x dx

, which results in

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

. Using the initial condition

y(0) = \ln 2

, we substitute

x=0

and

y=\ln 2

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

, which simplifies to

2 = C

. The particular solution is

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

. To find

f(2)

, we substitute

x=2

e^y = \frac{1}{2}(2)^2 + 2 = 2 + 2 = 4

. Solving for

y

gives

y = \ln 4

Question 2

Let $y=f(x)$ be the particular solution to $\frac{dy}{dx} = \frac{2x+1}{2y}$ that passes through the point $(0, -2)$ . Find $f(2)$ .

$\sqrt{10}$
$-\sqrt{10}$ (correct answer)
$-\sqrt{2}$
$-2$

Explanation: The equation separates to

2y \, dy = (2x+1) \, dx

. Integrating both sides gives

\int 2y \, dy = \int (2x+1) \, dx

, which results in

y^2 = x^2 + x + C

. Using the initial condition

y(0)=-2

, we find

C

(-2)^2 = 0^2 + 0 + C

, so

4=C

. The implicit particular solution is

y^2 = x^2 + x + 4

. Since the initial condition specifies

y=-2

(a negative value), we must choose the negative branch of the solution:

y = -\sqrt{x^2+x+4}

. Evaluating at

x=2

, we find

f(2) = -\sqrt{2^2+2+4} = -\sqrt{4+2+4} = -\sqrt{10}

Question 3

The equation $\frac{dy}{dx} = \frac{y+1}{x}$ has a particular solution $y=f(x)$ with $f(1)=1$ . What is $f(e)$ ?

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

Explanation: Separating variables, we get

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

. Integrating both sides gives

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

. Exponentiating both sides gives

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

, which simplifies to

|y+1| = e^C |x|

. Let

A = \pm e^C

, so

y+1 = Ax

. Using the initial condition

y(1)=1

1+1 = A(1)

, so

A=2

. The particular solution is

y+1 = 2x

, or

y = 2x-1

. Evaluating at

x=e

, we find

f(e) = 2e-1

Question 4

Let $y=f(x)$ be the particular solution to $\frac{dy}{dx} = (y-1)^2 \cos(x)$ with $f(0)=0$ . What is the value of $f(\pi/2)$ ?

1/2 (correct answer)
0
2
-1

Explanation: Separating variables gives

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

. Integrating both sides,

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

, we get

-(y-1)^{-1} = \sin(x) + C

, or

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

. Using the initial condition

y(0)=0

\frac{-1}{0-1} = \sin(0) + C \implies 1 = 0 + C \implies C=1

. The particular solution is

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

. To find

f(\pi/2)

, we set

x=\pi/2

\frac{-1}{y-1} = \sin(\pi/2) + 1 = 1+1=2

. Solving for

y

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

Question 5

The rate of change of $y$ with respect to $x$ is inversely proportional to $y$ and directly proportional to $x$ . The relationship is given by $\frac{dy}{dx} = \frac{kx}{y}$ . If $y(0)=5$ and $y(4)=3$ , what is the value of $k$ ?

-1/2
-1 (correct answer)
1/2
2

Explanation: Separating variables gives

y \, dy = kx \, dx

. Integrating both sides gives

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

, or

y^2 = kx^2 + C_1

where

C_1=2C

. Using the first initial condition,

y(0)=5

5^2 = k(0)^2 + C_1 \implies 25 = C_1

. The equation becomes

y^2 = kx^2 + 25

. Using the second condition,

y(4)=3

3^2 = k(4)^2 + 25 \implies 9 = 16k + 25

. Solving for

k

16k = 9 - 25 = -16

, so

k = -1

Question 6

The temperature $T(t)$ of a cooling object follows Newton's Law of Cooling: $\frac{dT}{dt} = -k(T-70)$ where the ambient temperature is 70°F. If $T(0) = 200°F$ and $T(10) = 150°F$ , what is the temperature after 30 minutes?

91.5°F
95.2°F
98.8°F (correct answer)
102.4°F

Explanation: The general solution is

T(t) = 70 + Ce^{-kt}

. From

T(0) = 200

C = 130

. From

T(10) = 150

150 = 70 + 130e^{-10k}

, so

80 = 130e^{-10k}

, giving

e^{-10k} = \frac{8}{13}

. Therefore

e^{-k} = \left(\frac{8}{13}\right)^{1/10}

. At

t = 30

T(30) = 70 + 130\left(\frac{8}{13}\right)^3 = 70 + 130 \cdot \frac{512}{2197} \approx 98.8°F

. Choice A uses

k = 0.1

. Choice B results from incorrect exponentiation. Choice D comes from sign errors in the exponential.

Question 7

A chemical reaction follows the rate law $\frac{dc}{dt} = -k c^2$ where $c(t)$ is the concentration at time $t$ . If the initial concentration is $c(0) = 0.8$ M and the concentration after 10 minutes is $c(10) = 0.4$ M, what is the concentration after 25 minutes?

0.20 M
0.32 M
0.27 M
0.24 M (correct answer)

Explanation: This is a separable differential equation problem involving second-order kinetics. When you see a rate equation of the form

\frac{dc}{dt} = -kc^2

, you're dealing with a reaction where the rate depends on the square of the concentration. To solve this, separate variables:

\frac{dc}{c^2} = -k \, dt

. Integrating both sides gives

-\frac{1}{c} = -kt + C

. Using the initial condition

c(0) = 0.8

, we find

C = -\frac{1}{0.8} = -1.25

. So our solution is

\frac{1}{c(t)} = kt + 1.25

. Next, use the condition

c(10) = 0.4

to find

k

. Substituting:

\frac{1}{0.4} = 10k + 1.25

, which gives

2.5 = 10k + 1.25

, so

k = 0.125

min⁻¹. Now we can find

c(25)

\frac{1}{c(25)} = 0.125(25) + 1.25 = 4.375

. Therefore,

c(25) = \frac{1}{4.375} = 0.229

M, which rounds to 0.24 M. Answer D (0.24 M) is correct. Answer A (0.20 M) likely comes from using first-order kinetics instead of second-order. Answer B (0.32 M) might result from incorrectly applying the half-life concept without accounting for the concentration dependence. Answer C (0.27 M) could arise from arithmetic errors in the integration or constant calculation. Remember: for second-order kinetics, the integrated rate law involves

\frac{1}{c}

, not

\ln c

like first-order reactions. Always check which order kinetics you're dealing with first.

Question 8

The differential equation $\frac{dy}{dx} = \frac{xy}{x^2+1}$ has a particular solution passing through the point $(2, 3)$ . What is the value of $y$ when $x = 4$ ?

$3\sqrt{\frac{17}{5}}$ (correct answer)
$9\sqrt{\frac{17}{5}}$
$\frac{27}{5}$
$\frac{51}{5}$

Explanation: Separating variables:

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

. Integrating:

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

. This gives

y = K\sqrt{x^2+1}

. Using the initial condition

(2,3)

3 = K\sqrt{5}

, so

K = \frac{3}{\sqrt{5}}

. Therefore

y = \frac{3\sqrt{x^2+1}}{\sqrt{5}}

. When

x = 4

y = \frac{3\sqrt{17}}{\sqrt{5}} = 3\sqrt{\frac{17}{5}}

. Choice B squares the constant incorrectly. Choice C results from arithmetic errors. Choice D comes from incorrect integration of the right side.

Question 9

A radioactive substance decays according to $\frac{dy}{dt} = -ky$ where $y(t)$ is the amount remaining at time $t$ hours. If the half-life is 8 hours and $y(4) = 30$ grams, what was the initial amount $y(0)$ ?

35.36 grams
42.43 grams (correct answer)
48.99 grams
51.21 grams

Explanation: From the half-life,

k = \frac{\ln 2}{8}

. The general solution is

y(t) = Ce^{-kt}

. Given

y(4) = 30

, we have

30 = Ce^{-4k} = Ce^{-4\ln 2/8} = Ce^{-\ln 2/2} = C\cdot 2^{-1/2} = \frac{C}{\sqrt{2}}

. Therefore

C = 30\sqrt{2} = 42.43

. Choice A uses

k = 8

instead of

k = \ln 2/8

. Choice C incorrectly applies the half-life formula. Choice D results from using

e^{\ln 2/2}

instead of

e^{-\ln 2/2}

Question 10

Consider the differential equation $\frac{dy}{dx} = \frac{\cos(x)}{y^2}$ . If $y = 3$ when $x = \frac{\pi}{2}$ , what is the value of $y^3$ when $x=0$ ?

24 (correct answer)
26
27
30

Explanation: Separating variables gives

y^2 dy = \cos(x) dx

. Integrating both sides, we get

\int y^2 dy = \int \cos(x) dx

, which yields

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

. Using the initial condition

y(\pi/2) = 3

, we substitute to find C:

\frac{1}{3}(3)^3 = \sin(\frac{\pi}{2}) + C

, which simplifies to

9 = 1 + C

, so

C=8

. The particular solution is

\frac{1}{3}y^3 = \sin(x) + 8

, or

y^3 = 3\sin(x) + 24

. When

x=0

y^3 = 3\sin(0) + 24 = 3(0) + 24 = 24

Question 11

The differential equation $xy' - 2y = xy$ describes a family of curves. Which curve in this family passes through the point $(1, e)$ ? Find the value of $y$ for this curve when $x=2$ .

$e^2 + 3$
$e^2/4$
$2e^2$
$4e^2$ (correct answer)

Explanation: First, rearrange the differential equation to separate variables:

xy' = y(x+2) \implies y' = \frac{y(x+2)}{x}

. Now separate:

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

. Integrating both sides gives

\ln|y| = x + 2\ln|x| + C

. This can be rewritten as

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

. Using the initial condition

y(1)=e

\ln|e| = 1 + \ln(1^2) + C \implies 1 = 1 + 0 + C \implies C=0

. The particular solution is

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

. Since

y(1)=e>0

, we use

y

. To solve for

y

, we write

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

. Exponentiating both sides gives

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

, so

y = x^2e^x

. At

x=2

y = 2^2 e^2 = 4e^2

Question 12

Find the particular solution to the differential equation $\frac{dy}{dx} = e^{x-y}$ that satisfies the initial condition $y(0) = \ln 3$ . Then, evaluate this solution at $x=\ln 2$ .

$\ln(2+\ln 3)$
$\ln(1+\ln 3)$
$\ln 6$
$\ln 4$ (correct answer)

Explanation: The equation can be written as

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

. Separating the variables gives

e^y dy = e^x dx

. Integrating both sides yields

\int e^y dy = \int e^x dx

, which results in

e^y = e^x + C

. Using the initial condition

y(0) = \ln 3

, we have

e^{\ln 3} = e^0 + C

, which simplifies to

3 = 1 + C

, so

C=2

. The particular solution is

e^y = e^x + 2

. To find

y

explicitly, we take the natural logarithm of both sides:

y = \ln(e^x + 2)

. Evaluating at

x=\ln 2

, we get

y = \ln(e^{\ln 2} + 2) = \ln(2+2) = \ln 4

Question 13

Given the differential equation $\frac{dy}{dx} = \frac{y}{x\ln(x)}$ and the initial condition $y(e)=e$ , what is the value of $y(e^2)$ ?

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

Explanation: Separating variables, we get

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

. Integrating both sides:

\int \frac{1}{y} dy = \ln|y|

. For the right side, we use u-substitution with

u = \ln(x)

and

du = \frac{1}{x} dx

. So

\int \frac{1}{x\ln(x)} dx = \int \frac{1}{u} du = \ln|u| + C = \ln|\ln(x)| + C

. The general solution is

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

. Using the initial condition

y(e)=e

\ln|e| = \ln|\ln(e)| + C

\implies 1 = \ln|1| + C \implies 1 = 0 + C \implies C=1

. The particular solution is

\ln|y| = \ln|\ln(x)| + 1

. Since

y(e)=e>0

, we can write

\ln(y) = \ln(\ln(x)) + 1

. To solve for

y

, we exponentiate:

y = e^{\ln(\ln x) + 1} = e^{\ln(\ln x)} \cdot e^1 = e \ln(x)

. Finally, we evaluate at

x=e^2

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

Question 14

Consider the initial value problem $y' = 1 + y^2$ with $y(0)=1$ . What is the value of $y(\pi/6)$ ?

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

Explanation: Separating variables gives

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

. Integrating both sides yields

\arctan(y) = x + C

. Using the initial condition

y(0)=1

, we get

\arctan(1) = 0 + C

, which means

C = \pi/4

. The particular solution is

\arctan(y) = x + \pi/4

, so

y = \tan(x+\pi/4)

. We need to evaluate

y(\pi/6)

y = \tan(\frac{\pi}{6} + \frac{\pi}{4}) = \tan(\frac{2\pi+3\pi}{12}) = \tan(\frac{5\pi}{12})

. Using the tangent addition formula:

\tan(75°) = \tan(45° + 30°) = \frac{\tan 45° + \tan 30°}{1 - \tan 45° \tan 30°} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)^2}{2} = 2+\sqrt{3}

Question 15

The rate of change of a quantity $P$ is proportional to the square of $P$ , i.e., $\frac{dP}{dt} = kP^2$ . Given that $P(0) = 10$ and $P(1) = 5$ , for what value of $t > 0$ is $P(t) = 2$ ?

4 (correct answer)
9/2
$\frac{\ln(5)}{\ln(2)}$
3

Explanation: Separating variables in

\frac{dP}{dt} = kP^2

gives

P^{-2} dP = k dt

. Integrating yields

-P^{-1} = kt + C

, or

-\frac{1}{P} = kt + C

. Using

P(0)=10

, we get

-\frac{1}{10} = k(0) + C

, so

C = -\frac{1}{10}

. The equation becomes

-\frac{1}{P} = kt - \frac{1}{10}

. Using

P(1)=5

, we get

-\frac{1}{5} = k(1) - \frac{1}{10}

, which gives

-\frac{2}{10} = k - \frac{1}{10}

, so

k = -\frac{1}{10}

. The particular solution is

-\frac{1}{P} = -\frac{t}{10} - \frac{1}{10}

, which simplifies to

\frac{1}{P} = \frac{t+1}{10}

, or

P(t) = \frac{10}{t+1}

. We need to find

t

when

P(t)=2

2 = \frac{10}{t+1} \implies 2(t+1) = 10 \implies t+1 = 5 \implies t=4

Question 16

Find the solution $y=f(x)$ to the differential equation $(x^2+1)y' = 2x(y-1)$ that passes through $(0, 2)$ . What is $f(\sqrt{3})$ ?

3
4
5 (correct answer)
6

Explanation: First, we separate the variables:

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

. Integrating both sides gives

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

. The left side is

\ln|y-1|

. The right side can be solved with a u-substitution

u=x^2+1

du=2x dx

, so the integral is

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

(since

x^2+1

is always positive). So,

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

. Exponentiating,

|y-1| = e^C(x^2+1)

. Let

A=\pm e^C

, so

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

. Using the initial condition

y(0)=2

2-1 = A(0^2+1) \implies 1 = A

. The particular solution is

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

, so

y = x^2+2

. Evaluating at

x=\sqrt{3}

, we get

f(\sqrt{3}) = (\sqrt{3})^2 + 2 = 3+2 = 5

Question 17

Let $y(x)$ be the solution to the initial value problem $y' = y^2 - 4$ with $y(0)=3$ . What is the value of $y(\frac{1}{4}\ln 2)$ ?

$\sqrt{14}$
$-\frac{22}{9}$
$\frac{14}{3}$ (correct answer)
4

Explanation: We separate variables:

\frac{dy}{y^2-4} = dx

. We use partial fractions for the left side:

\frac{1}{(y-2)(y+2)} = \frac{A}{y-2} + \frac{B}{y+2}

. This gives

A=1/4

and

B=-1/4

. The integral becomes

\frac{1}{4}\int (\frac{1}{y-2} - \frac{1}{y+2}) dy = \int dx

. This integrates to

\frac{1}{4}(\ln|y-2| - \ln|y+2|) = x+C

, or

\ln|\frac{y-2}{y+2}| = 4x+C_1

. Given

y(0)=3

\ln|\frac{3-2}{3+2}| = 0+C_1 \implies C_1=\ln(1/5)

. Since

y(0)=3>2

, the argument of the log is positive, so

\ln(\frac{y-2}{y+2}) = 4x+\ln(1/5)

. Exponentiating gives

\frac{y-2}{y+2} = e^{4x+\ln(1/5)} = \frac{1}{5}e^{4x}

. Let

x_0 = \frac{1}{4}\ln 2

. Then

e^{4x_0} = e^{\ln 2}=2

. So

\frac{y-2}{y+2} = \frac{2}{5}

. Solving for

y

5(y-2) = 2(y+2) \implies 5y-10 = 2y+4 \implies 3y=14 \implies y=14/3

Question 18

Let $y=f(x)$ be the particular solution to the differential equation $\frac{dy}{dx} = \frac{3x^2}{2y}$ with the initial condition $f(2)=3$ . What is the value of $f(3)$ ?

$3\sqrt{3}$
$\sqrt{26}$
$2\sqrt{7}$ (correct answer)
$-2\sqrt{7}$

Explanation: Separating variables gives

2y \, dy = 3x^2 \, dx

. Integrating both sides yields

\int 2y \, dy = \int 3x^2 \, dx

, which results in

y^2 = x^3 + C

. Using the initial condition

y(2)=3

, we have

3^2 = 2^3 + C

, so

9 = 8 + C

, which means

C=1

. The particular solution is

y^2 = x^3 + 1

. Since the initial value of

y

is positive (3), we take the positive square root:

y = \sqrt{x^3 + 1}

. Evaluating at

x=3

, we get

y = \sqrt{3^3 + 1} = \sqrt{27 + 1} = \sqrt{28} = 2\sqrt{7}

Question 19

The rate of change of a population $P(t)$ is modeled by $\frac{dP}{dt} = kP(1000-P)$ where $k > 0$ is a constant. If $P(0) = 200$ and $P(2) = 400$ , what is the population when $t = 4$ ?

600 individuals
640 individuals
667 individuals (correct answer)
680 individuals

Explanation: This is a logistic equation. Using separation of variables:

\frac{dP}{P(1000-P)} = k\,dt

. Partial fractions gives

\frac{1}{1000}\left(\frac{1}{P} + \frac{1}{1000-P}\right)dP = k\,dt

. Integrating and applying initial conditions

P(0) = 200

and

P(2) = 400

determines the constant. At

t = 4

P(4) = \frac{200000}{300} = 666.67 \approx 667

. Choice A assumes linear growth. Choice B results from incorrect partial fraction decomposition. Choice D comes from sign errors in the integration.