Differential Equations Quiz: Solving Separable Equations

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Solving Separable EquationsQuestion 1 of 20

A population $y(t)$ is modeled by the logistic-type differential equation $\frac{dy}{dt} = y(4-y)$ . If the initial population is $y(0)=2$ , what is the limiting population as $t \to \infty$ ?

0

2

4

\infty

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Differential Equations Quiz

Differential Equations Quiz: Solving Separable Equations

Practice Solving Separable Equations 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 Solving Separable Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A population $y(t)$ is modeled by the logistic-type differential equation $\frac{dy}{dt} = y(4-y)$ . If the initial population is $y(0)=2$ , what is the limiting population as $t \to \infty$ ?

$0$
$2$
$4$ (correct answer)
$\infty$

Explanation: We can solve this by separating variables. Note that

y=0

and

y=4

are equilibrium solutions. Since the initial value is

y(0)=2

, the solution will not be an equilibrium solution.

\frac{dy}{y(4-y)} = dt

We use partial fraction decomposition for the left side:

\frac{1}{y(4-y)} = \frac{A}{y} + \frac{B}{4-y} \implies 1 = A(4-y) + By

Setting

y=0

gives

1=4A \implies A=1/4

. Setting

y=4

gives

1=4B \implies B=1/4

. So the equation becomes:

\frac{1}{4}\left(\frac{1}{y} + \frac{1}{4-y}\right) dy = dt

Integrating both sides:

\int \left(\frac{1}{y} + \frac{1}{4-y}\right) dy = \int 4 \, dt

\ln|y| - \ln|4-y| = 4t + C_1

\ln\left|\frac{y}{4-y}\right| = 4t + C_1

Exponentiating both sides:

\frac{y}{4-y} = C e^{4t}

Using the initial condition

y(0)=2

\frac{2}{4-2} = C e^0 \implies 1 = C

The solution satisfies

\frac{y}{4-y} = e^{4t}

. To find the limit, we solve for

y

y = (4-y)e^{4t} = 4e^{4t} - y e^{4t}

y(1+e^{4t}) = 4e^{4t} \implies y(t) = \frac{4e^{4t}}{1+e^{4t}} = \frac{4}{e^{-4t}+1}

Now we take the limit as

t \to \infty

\lim_{t\to\infty} y(t) = \lim_{t\to\infty} \frac{4}{e^{-4t}+1} = \frac{4}{0+1} = 4

(A) would be the limit if the initial population were between 0 and the other equilibrium solution, but with a different equation structure. For this equation, an initial population of 2 grows towards 4. This answer can arise from a sign error in the setup, leading to $y(t) = 4/(1+e^{4t})$ .
(B) is the initial value, not the limiting value.
(D) is what a student might guess if they see the term $e^{4t}$ and don't correctly simplify the rational expression.

Question 2

Find the explicit solution $y(x)$ for the initial value problem $e^y y' - x - x e^y = 0, \quad y(1)=0$

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

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

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

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

Now integrate both sides. For the left side, use the substitution

u = 1+e^y

, so

du = e^y dy

\int \frac{1}{u} du = \int x \, dx

\ln|u| = \frac{x^2}{2} + C

Substituting back

u=1+e^y

(which is always positive), we get:

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

Use the initial condition

y(1)=0

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

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

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

The implicit solution is

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

. To find the explicit solution, we exponentiate both sides:

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

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

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

(A) results from an incorrect integration of the left side, e.g., if $\int \frac{e^y}{1+e^y} dy$ was incorrectly calculated as $y$ .
(B) results from an incorrect integration of the right side, using $\int x dx = x^2$ instead of $x^2/2$ . This leads to $C=\ln(2)-1$ and $y(x) = \ln(e^{x^2+\ln(2)-1}-1)=\ln(2e^{x^2-1}-1)$ .
(D) results from multiple errors, such as forgetting to exponentiate to solve for $y$ and making an integration error like $\int x dx = x^2$ .

Question 3

The solution $y(x)$ to the initial value problem $\frac{dy}{dx} = 2x(y-1)^2, \quad y(0)=2$ has vertical asymptotes at $x=a$ and $x=b$ . Which of the following statements about the solution is true for $x$ in the interval $(a, b)$ ?

$\lim_{x \to 1^-} y(x) = +\infty$ (correct answer)
$y(x)$ has a local maximum at $x=0$ .
The solution is defined for all real $x$ .
$\lim_{x \to -1^+} y(x) = -\infty$

Explanation: First, we solve the initial value problem. Separating variables gives:

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

Integrating both sides:

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

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

Using the initial condition

y(0)=2

-(2-1)^{-1} = 0^2 + C \implies -1 = C

So the particular solution is:

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

Solving for

y(x)

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

The solution has vertical asymptotes where the denominator is zero, i.e., at

x=\pm 1

. The interval of existence

(a, b)

containing the initial point

x=0

(-1, 1)

. Now we analyze the given statements.

(A) Let's check the limit as $x \to 1^-$ . As $x$ approaches 1 from the left, $x^2$ approaches 1 from the left, so $1-x^2$ approaches $0^+$ . The numerator $2-x^2$ approaches $1$ . Therefore, $\lim_{x \to 1^-} y(x) = \frac{1}{0^+} = +\infty$ . This statement is true.
(B) To check for a local maximum, we find the derivative, which is given as $y'(x) = 2x(y-1)^2$ . Substituting our solution for $y$ : $y'(x) = 2x\left(\frac{1}{1-x^2}\right)^2 = \frac{2x}{(1-x^2)^2}$ . At $x=0$ , $y'(0)=0$ . For $x$ near 0, the denominator is positive. If $x<0$ , $y'<0$ (decreasing). If $x>0$ , $y'>0$ (increasing). By the first derivative test, $y(x)$ has a local minimum at $x=0$ , so this statement is false.
(C) This is false, as there are vertical asymptotes at $x=\pm 1$ .
(D) Let's check the limit as $x \to -1^+$ . As $x$ approaches -1 from the right, $x^2$ approaches 1 from the left, so $1-x^2$ approaches $0^+$ . The numerator approaches 1. Therefore, $\lim_{x \to -1^+} y(x) = \frac{1}{0^+} = +\infty$ . This statement is false.

Question 4

A 100-liter tank initially contains brine with 10 kg of salt. Pure water flows into the tank at a rate of 5 L/min. The mixture is kept uniform by stirring and flows out at the same rate. Let $A(t)$ be the amount of salt in kg after $t$ minutes. How long does it take for the amount of salt in the tank to be reduced to 1 kg?

$10 \ln(10)$ minutes
$20 \ln(9)$ minutes
$20 \ln(10)$ minutes (correct answer)
$18$ minutes

Explanation: The rate of change of salt in the tank is given by

\frac{dA}{dt} = \text{Rate in} - \text{Rate out}

. The rate of salt flowing in is 0, since pure water is entering. Rate in = 0. The rate of salt flowing out depends on the concentration of salt in the tank at time

t

. The volume of the mixture is constant at 100 L. The concentration is

\frac{A(t)}{100}

kg/L. The mixture flows out at 5 L/min. So, Rate out =

5 \times \frac{A(t)}{100} = \frac{A}{20}

kg/min. The differential equation is

\frac{dA}{dt} = 0 - \frac{A}{20} = -\frac{A}{20}

. This is a separable equation. We solve it:

\frac{dA}{A} = -\frac{1}{20} dt

Integrating both sides:

\ln(A) = -\frac{t}{20} + C

(We can use

A

instead of

|A|

since amount of salt is non-negative). Exponentiating gives:

A(t) = e^{-t/20 + C} = K e^{-t/20}

The initial condition is

A(0)=10

kg.

10 = K e^0 \implies K=10

The amount of salt at time

t

A(t) = 10e^{-t/20}

. We want to find the time

t

when

A(t)=1

kg.

1 = 10e^{-t/20}

0.1 = e^{-t/20}

\ln(0.1) = -\frac{t}{20}

\ln(1/10) = -\frac{t}{20}

-\ln(10) = -\frac{t}{20}

t = 20 \ln(10)

(A) results from an algebraic error, possibly using 10 instead of 20 when solving for $t$ .
(B) results from a conceptual error, perhaps by considering the change in salt, $10-1=9$ , and using $\ln(9)$ instead of $\ln(10)$ .
(D) would be the result of assuming a constant rate of decay. The initial rate of salt leaving is $10/20 = 0.5$ kg/min. If this rate were constant, the time to lose 9 kg would be $9 / 0.5 = 18$ minutes. This linear model is incorrect.

Question 5

Find the implicit solution to the differential equation $\frac{dy}{dx} = \frac{x e^{x^2}}{\cos(y)}$ with the initial condition $y(0) = \frac{\pi}{2}$ .

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

Explanation: This is a separable differential equation. We can write it as

\cos(y) \, dy = x e^{x^2} \, dx

. Integrating both sides, we have

\int \cos(y) \, dy = \int x e^{x^2} \, dx

. The integral of the left side is

\sin(y)

. For the right side, we use the substitution

u = x^2

, so

du = 2x \, dx

. The integral becomes

\frac{1}{2} \int e^u \, du = \frac{1}{2} e^u = \frac{1}{2} e^{x^2}

. So, the general solution is

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

. Using the initial condition

y(0) = \frac{\pi}{2}

, we substitute

x=0

and

y=\frac{\pi}{2}

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

, which gives

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

, so

C = \frac{1}{2}

. Substituting

C

back, we get

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

. Multiplying by 2 gives the implicit solution

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

Question 6

A tank initially contains 100 L of pure water. A salt solution with a concentration of 0.5 kg/L is pumped into the tank at a rate of 2 L/min. The well-mixed solution is pumped out at the same rate. Let $S(t)$ be the amount of salt (in kg) in the tank at time $t$ (in minutes). At what time $t$ will the amount of salt in the tank reach 40 kg?

$50 \ln(1.8)$ minutes
$100 \ln(5/3)$ minutes
$50 \ln(5)$ minutes (correct answer)
$50 \ln(40)$ minutes

Explanation: The rate of change of salt is

\frac{dS}{dt} = (\text{rate in}) - (\text{rate out})

. The rate in is

(0.5 \text{ kg/L}) \times (2 \text{ L/min}) = 1 \text{ kg/min}

. The concentration in the tank at time

t

S(t)/100

kg/L. The rate out is

(S(t)/100 \text{ kg/L}) \times (2 \text{ L/min}) = S/50 \text{ kg/min}

. The differential equation is

\frac{dS}{dt} = 1 - \frac{S}{50} = \frac{50-S}{50}

. This is a separable equation:

\frac{dS}{50-S} = \frac{1}{50} dt

. Integrating both sides gives

-\ln|50-S| = \frac{t}{50} + C

. The initial condition is

S(0)=0

(pure water), so

-\ln|50-0| = C

, which means

C = -\ln(50)

. The equation becomes

-\ln|50-S| = \frac{t}{50} - \ln(50)

, or

\ln(50) - \ln|50-S| = \frac{t}{50}

. Since

S

will not exceed 50 kg, we can drop the absolute value:

\ln\left(\frac{50}{50-S}\right) = \frac{t}{50}

. We want to find

t

when

S=40

\ln\left(\frac{50}{50-40}\right) = \frac{t}{50}

, which simplifies to

\ln(5) = \frac{t}{50}

. Therefore,

t = 50 \ln(5)

minutes.

Question 7

Find the implicit solution to the initial value problem $(y^2 + 1) \frac{dy}{dx} = y e^x$ with $y(0) = 1$ .

$y^2 + 2\ln(y) = 2e^x + 1$
$y^2 + 2\ln(y) = 2e^x - 1$ (correct answer)
$\frac{y^2}{2} + y = e^x + \frac{1}{2}$
$y^2 - \frac{2}{y^2} = 2e^x - 3$

Explanation: The differential equation is separable. We separate the variables by dividing by

y

and multiplying by

dx

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

. The left side can be written as

(y + \frac{1}{y}) dy

. Now we integrate both sides:

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

. This yields

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

. We use the initial condition

y(0)=1

to find

C

. Substituting

x=0

and

y=1

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

, which simplifies to

\frac{1}{2} + 0 = 1 + C

. Solving for

C

, we get

C = -\frac{1}{2}

. Since

y(0)=1

is positive, we can drop the absolute value for the solution near

x=0

. The implicit solution is

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

. To match the answer choices, we multiply the entire equation by 2:

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

Question 8

Find the value of $y(1)$ for the solution to the initial value problem $\frac{dy}{dx} - x(y+1) = y+1$ , with $y(0)=1$ .

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

Explanation: First, we rearrange the equation to see if it is separable.

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

. We can factor out

(y+1)

on the right side:

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

. This is a separable equation. We separate the variables to get

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

. Now, we integrate both sides:

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

. This gives

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

. We use the initial condition

y(0)=1

to find

C

\ln|1+1| = 0 + \frac{0^2}{2} + C

, which means

\ln(2) = C

. So, the solution is

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

. Since

y(0)=1> -1

, we can drop the absolute value sign. To find

y(1)

, we first solve for

y(x)

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

, so

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

. Exponentiating both sides gives

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

. Then

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

. Finally, we evaluate at

x=1

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

Question 9

The differential equation $\frac{dy}{dx} = \frac{1}{2}(y^2 - 1)$ has two constant solutions, $y=1$ and $y=-1$ . A particular solution to this equation passes through the point $(0, 3)$ . Which of the following statements is true about this particular solution $y(x)$ ?

The solution $y(x)$ is defined for all real numbers $x$ .
The solution is always increasing on its interval of existence. (correct answer)
$\lim_{x \to \infty} y(x) = -1$ .
The solution $y(x)$ approaches the constant solution $y=1$ as $x \to \infty$ .

Explanation: First, we solve the initial value problem. Separating variables gives

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

. Using partial fraction decomposition,

\frac{1}{y^2 - 1} = \frac{1/2}{y-1} - \frac{1/2}{y+1}

. Integrating gives

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

, which yields

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

, or

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

. Given

y(0)=3

, we have

\ln\left|\frac{3-1}{3+1}\right| = 0+C

, so

C = \ln(1/2) = -\ln(2)

. The solution is

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

. Solving for

y(x)

, we find

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

. The solution has a vertical asymptote where

2-e^x = 0

, i.e., at

x=\ln(2)

. Since the initial condition is at

x=0

, the interval of existence is

(-\infty, \ln(2))

. Thus, statement (A) is false. Statement (C) is false because the limit is taken outside the domain of the solution. To check statement (B), we find the derivative of the solution:

y'(x) = \frac{e^x(2-e^x) - (2+e^x)(-e^x)}{(2-e^x)^2} = \frac{4e^x}{(2-e^x)^2}

. Since

e^x > 0

and

(2-e^x)^2 > 0

on the interval of existence,

y'(x) > 0

. Therefore, the solution is always increasing. Statement (D) is false; as

x \to -\infty

y(x) \to \frac{2+0}{2-0} = 1

Question 10

The differential equation $\frac{dy}{dx} = \frac{y \ln y}{x \ln x}$ is separable for $x, y > 1$ . After separation and integration, if $y(e) = e^2$ , what is the relationship between $\ln(\ln y)$ and $\ln(\ln x)$ ?

$\ln(\ln y) = 2\ln(\ln x) + \ln 2$
$\ln(\ln y) = \ln(\ln x) + \ln 2$ (correct answer)
$\ln(\ln y) = 2\ln(\ln x) - \ln 2$
$\ln(\ln y) = \ln(\ln x) + 2\ln 2$

Explanation: Separating variables:

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

. Let

u = \ln y

, so

du = \frac{dy}{y}

and

\frac{dy}{y \ln y} = \frac{du}{u}

. Similarly, let

v = \ln x

, so

\frac{dx}{x \ln x} = \frac{dv}{v}

. The equation becomes

\frac{du}{u} = \frac{dv}{v}

, which integrates to

\ln|u| = \ln|v| + C

, or

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

. Using the initial condition

y(e) = e^2

: when

x = e

\ln x = 1

and

\ln(\ln x) = \ln 1 = 0

. When

y = e^2

\ln y = 2

and

\ln(\ln y) = \ln 2

. Therefore

\ln 2 = 0 + C

, so

C = \ln 2

. Choice A incorrectly has coefficient 2 on

\ln(\ln x)

. Choice C has the wrong sign. Choice D has

2\ln 2

instead of

\ln 2

Question 11

A population model follows the differential equation $\frac{dP}{dt} = kP(M-P)$ where $P(t)$ is population at time $t$ , $k > 0$ is a constant, and $M$ is the carrying capacity. If $P(0) = P_0$ where $0 < P_0 < M$ , what is the correct form of the solution after applying partial fractions and integrating?

$P(t) = \frac{MP_0}{P_0 + (M-P_0)e^{-kMt}}$ (correct answer)
$P(t) = \frac{MP_0}{(M-P_0) + P_0 e^{-kMt}}$
$P(t) = \frac{MP_0 e^{kMt}}{M + P_0(e^{kMt} - 1)}$
$P(t) = \frac{M}{1 + \frac{M-P_0}{P_0}e^{-kMt}}$

Explanation: Separating variables:

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

. Using partial fractions:

\frac{1}{P(M-P)} = \frac{1/M}{P} + \frac{1/M}{M-P}

. Integrating:

\frac{1}{M}[\ln|P| - \ln|M-P|] = kt + C

, which gives

\ln\left|\frac{P}{M-P}\right| = kMt + C_1

. Thus

\frac{P}{M-P} = Ae^{kMt}

where

A = e^{MC_1}

. Using initial condition

P(0) = P_0

A = \frac{P_0}{M-P_0}

. Solving for

P

P = (M-P)\frac{P_0}{M-P_0}e^{kMt}

, leading to

P(M-P_0)e^{-kMt} + P_0P = MP_0

, so

P = \frac{MP_0}{P_0 + (M-P_0)e^{-kMt}}

. Choice B has incorrect denominator arrangement. Choice C has the exponential in the wrong position. Choice D is algebraically equivalent to A but written in different form.

Question 12

A tank initially contains 100 gallons of pure water. Brine containing 2 pounds of salt per gallon flows in at 3 gallons per minute, and the well-mixed solution flows out at 2 gallons per minute. If $S(t)$ represents pounds of salt at time $t$ minutes, which differential equation correctly models this situation?

$\frac{dS}{dt} = 6 - \frac{2S}{100 + t}$ (correct answer)
$\frac{dS}{dt} = 6 - \frac{2S}{100 - t}$
$\frac{dS}{dt} = 6 - \frac{S}{50 + \frac{t}{2}}$
$\frac{dS}{dt} = 6 - \frac{3S}{100 + t}$

Explanation: Salt enters at rate: (2 lb/gal)(3 gal/min) = 6 lb/min. Volume at time

t

V(t) = 100 + (3-2)t = 100 + t

gallons. Concentration at time

t

\frac{S(t)}{100+t}

lb/gal. Salt leaves at rate:

\frac{S(t)}{100+t} \times 2 = \frac{2S}{100+t}

lb/min. Therefore:

\frac{dS}{dt} = 6 - \frac{2S}{100+t}

. Choice B incorrectly uses

100-t

(would apply if outflow exceeded inflow). Choice C has wrong coefficient and volume formula. Choice D uses rate 3 instead of 2 for the outflow concentration factor.

Question 13

The differential equation $\sin(x)\frac{dy}{dx} = y\cos(x) + y^2\cos(x)$ can be solved by separation of variables on the interval $(0, \pi)$ . After separation, which of the following represents the correct integral setup?

$\int \frac{dy}{y + y^2} = \int \frac{\cos(x)}{\sin(x)} dx$
$\int \frac{dy}{y^2(1 + y)} = \int \cot(x) dx$
$\int \frac{dy}{y(1 + y)} = \int \cot(x) dx$ (correct answer)
$\int \frac{dy}{y + y^2} = \int \tan(x) dx$

Explanation: When you encounter a differential equation that can be solved by separation of variables, your goal is to algebraically manipulate it so that all terms involving one variable (and its differential) appear on one side, while all terms involving the other variable appear on the opposite side. Starting with

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

, first factor the right side:

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

Now separate variables by dividing both sides by

y(1 + y)

and multiplying both sides by

dx

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

Since

\frac{\cos(x)}{\sin(x)} = \cot(x)

, the integral setup becomes:

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

This matches answer choice C exactly. Answer A incorrectly writes the left side as

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

instead of factoring out the

y

, and uses

\frac{\cos(x)}{\sin(x)}

instead of simplifying to

\cot(x)

. Answer B has the wrong denominator

y^2(1 + y)

instead of

y(1 + y)

. Answer D makes the same factoring error as A and incorrectly identifies the trigonometric function as

\tan(x)

instead of

\cot(x)

. Strategy tip: Always factor completely when separating variables, and remember that

\frac{\cos(x)}{\sin(x)} = \cot(x)

, not

\tan(x)

. Double-check your algebra by substituting back into the original equation.

Question 14

Consider the initial value problem $\frac{dy}{dx} = 2x y^2, \quad y(0) = 1$ The solution $y(x)$ is defined on a maximal open interval $(a, b)$ containing $x=0$ . What is this interval?

$(-\infty, \infty)$
$(-1, 1)$ (correct answer)
$(1, \infty)$
$(0, 1)$

Explanation: To solve the differential equation, we separate variables:

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

Integrating both sides gives:

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

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

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

Now, we apply the initial condition

y(0) = 1

\frac{-1}{1} = 0^2 + C \implies C = -1

Substituting

C=-1

back into the equation gives the particular solution:

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

Solving for

y(x)

, we get:

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

The solution is defined for all

x

such that the denominator is not zero, i.e.,

1 - x^2 \neq 0

, which means

x \neq \pm 1

. This breaks the domain into three intervals:

(-\infty, -1)

(-1, 1)

, and

(1, \infty)

. The maximal open interval of existence for the solution to an initial value problem must contain the initial point, which is

x=0

. The interval containing

x=0

(-1, 1)

(A) is incorrect. This would be the domain if the denominator were never zero, for example, if $C=1$ , leading to $y = -1/(x^2+1)$ .
(C) is incorrect. While the function is defined on this interval, it does not contain the initial point $x=0$ .
(D) is incorrect. This is a subinterval of the correct maximal interval of existence.

Question 15

Consider the initial value problem $\frac{dy}{dx} = e^{x+y}, \quad y(0) = -\ln(2)$ What is the value of $y(\ln(2))$ ?

$-\ln(3)$
$0$ (correct answer)
$1$
$\ln(3)$

Explanation: The differential equation can be written as

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

. We separate the variables:

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

Integrating both sides, we get:

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

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

We use the initial condition

y(0) = -\ln(2)

to find

C

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

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

-2 = 1 + C \implies C = -3

The implicit solution is

-e^{-y} = e^x - 3

, or

e^{-y} = 3 - e^x

. To find the explicit solution, we take the natural logarithm of both sides:

-y = \ln(3 - e^x)

y(x) = -\ln(3 - e^x)

Finally, we evaluate

y(\ln(2))

y(\ln(2)) = -\ln(3 - e^{\ln(2)}) = -\ln(3 - 2) = -\ln(1) = 0

(A) results from a sign error in the integration of $e^{-y}$ , i.e., $\int e^{-y} dy = e^{-y}$ . This leads to $e^{-y} = e^x + C$ . The IC gives $e^{\ln(2)} = 1+C \implies C=1$ . The solution becomes $y(x) = -\ln(e^x+1)$ , and $y(\ln(2)) = -\ln(2+1) = -\ln(3)$ .
(C) results from an algebraic error when solving for $y$ from $e^{-y} = 3-e^x$ , such as incorrectly concluding $y = 1/(3-e^x)$ . This would give $y(\ln(2)) = 1/(3-2) = 1$ .
(D) results from two sign errors, such as integrating $e^{-y}$ incorrectly and also making a sign error when solving for $y$ .

Question 16

What is the value of $y(\pi)$ for the solution to the initial value problem $\frac{dy}{dx} = (y-1)\cos(x), \quad y(0)=1$ ?

$0$
$1+e$
$2$
$1$ (correct answer)

Explanation: When you encounter a separable differential equation with an initial condition, your first step is to check if there's an equilibrium solution that might satisfy the initial condition directly. Looking at

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

with

y(0)=1

, notice that when

y = 1

, the right side equals

(1-1)\cos(x) = 0

. This means

\frac{dy}{dx} = 0

, so

y

remains constant. Since the initial condition gives us

y(0) = 1

, the constant solution

y(x) = 1

satisfies both the differential equation and initial condition. Therefore,

y(\pi) = 1

. To verify this is the unique solution, you could separate variables:

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

. Integrating both sides gives

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

. Applying the initial condition

y(0) = 1

would require

\ln|1-1| = \ln(0)

, which is undefined. This confirms that the equilibrium solution

y = 1

is indeed the appropriate solution. Choice A (

0

) might tempt you if you confused this with a different type of equation. Choice B (

1+e

) could arise from incorrectly handling the integration and exponential operations. Choice C (

2

) has no clear connection to this problem's structure. Study tip: Always check for equilibrium solutions first when the initial condition makes the derivative zero. These constant solutions often provide the quickest path to the answer and help you avoid unnecessary integration complications.

Question 17

Given the initial value problem $\frac{dy}{dx} = \sqrt{1-y^2}, \quad y(0) = \frac{\sqrt{3}}{2}$ find the first positive value of $x$ for which $y(x)=0$ .

$\frac{2\pi}{3}$ (correct answer)
$\frac{\pi}{3}$
$\frac{5\pi}{6}$
$\frac{5\pi}{3}$

Explanation: When you encounter a differential equation of the form

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

, recognize this as a separable equation that's closely related to inverse trigonometric functions. To solve this, separate variables:

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

. The left side integrates to

\arcsin(y)

, so integrating both sides gives

\arcsin(y) = x + C

. Using the initial condition

y(0) = \frac{\sqrt{3}}{2}

\arcsin\left(\frac{\sqrt{3}}{2}\right) = 0 + C

. Since

\arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}

, we have

C = \frac{\pi}{3}

. Therefore,

\arcsin(y) = x + \frac{\pi}{3}

, which gives us

y = \sin\left(x + \frac{\pi}{3}\right)

. To find where

y(x) = 0

, solve

\sin\left(x + \frac{\pi}{3}\right) = 0

. This occurs when

x + \frac{\pi}{3} = n\pi

for integer

n

. The first positive solution is when

n = 1

x + \frac{\pi}{3} = \pi

, so

x = \frac{2\pi}{3}

. Looking at the wrong answers: B)

\frac{\pi}{3}

is the phase shift but not where the function equals zero. C)

\frac{5\pi}{6}

and D)

\frac{5\pi}{3}

correspond to other points on the sine curve but not the first zero. The answer is A)

\frac{2\pi}{3}

. Study tip: For differential equations involving

\sqrt{1-y^2}

, immediately think inverse trig functions. The key is correctly applying initial conditions to find the phase shift.

Question 18

Given the initial value problem $x \frac{dy}{dx} = y(1+x)$ for $x > 0$ with $y(1) = 2e$ , what is the value of $y(2)$ ?

$2e^2 + e$
$2e^{7/2}$
$4e$
$4e^2$ (correct answer)

Explanation: First, we rearrange the equation to separate the variables. For

y \neq 0

and

x > 0

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

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

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

Next, we integrate both sides:

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

\ln|y| = \ln|x| + x + C_1

Since

x > 0

|x|=x

. Exponentiating both sides gives:

|y| = e^{\ln(x) + x + C_1} = e^{\ln(x)} e^x e^{C_1} = x e^x K

where

K = e^{C_1} > 0

. This implies

y = C x e^x

for some constant

C

. We use the initial condition

y(1) = 2e

2e = C(1)e^1 \implies C = 2

The particular solution is

y(x) = 2xe^x

. Finally, we evaluate this solution at

x=2

y(2) = 2(2)e^2 = 4e^2

(A) results from incorrectly handling the constant of integration, assuming a solution of the form $y=xe^x+C$ . Applying the IC gives $2e=1e^1+C \implies C=e$ , leading to $y(2)=2e^2+e$ .
(B) results from an incorrect separation of variables, such as $\frac{dy}{y} = (1+x) dx$ , which leads to an incorrect solution form.
(C) results from a calculation error when evaluating the final expression, such as $y(2)=2(2)e=4e$ .

Question 19

The solution to the initial value problem $\frac{dy}{dx} = -\frac{x}{y}, \quad y(3)=4$ describes a curve in the xy-plane. What is the value of $y$ on this curve when $x=0$ ?

$5$ (correct answer)
$-5$
$\pm 5$
The solution is not defined at $x=0$ .

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

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

, you can separate variables by multiplying both sides by

y

and

dx

y \, dy = -x \, dx

Integrating both sides:

\int y \, dy = \int -x \, dx

, which gives you

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

Multiplying by 2 and rearranging:

y^2 + x^2 = 2C

. Let

K = 2C

, so

x^2 + y^2 = K

This represents a circle centered at the origin! Now apply the initial condition

y(3) = 4

3^2 + 4^2 = K

, so

K = 9 + 16 = 25

The solution is therefore

x^2 + y^2 = 25

, a circle with radius 5. When

x = 0

0^2 + y^2 = 25

, so

y^2 = 25

and

y = \pm 5

However, since we have an initial value problem with a specific starting point

(3,4)

, we're looking at one continuous branch of the solution. The differential equation

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

is undefined when

y = 0

, which means the solution cannot cross the x-axis. Since we start with

y(3) = 4 > 0

, we must stay on the upper semicircle, giving us

y = 5

when

x = 0

. Choice A is correct. Choice B gives the lower semicircle, choice C ignores the initial condition's constraint, and choice D incorrectly assumes the solution breaks down at

x = 0

. Strategy tip: For initial value problems, always check whether the solution curve can cross regions where the differential equation is undefined—this often determines which branch to choose.

Question 20

The solution to the initial value problem $\frac{dy}{dx} = \frac{e^x}{2y+1}, \quad y(0) = 1$ satisfies the implicit equation $y^2 + y = e^x + 1$ . For what positive value of $x$ is $y=2$ ?

$\ln(4)$
$\ln(5)$ (correct answer)
$\ln(8)$
$5$

Explanation: The problem can be solved in two ways: by verifying the provided implicit solution or by solving from scratch. Let's assume the provided implicit solution is correct (a quick check confirms this: separating gives

(2y+1)dy = e^x dx

, integrating gives

y^2+y=e^x+C

. With

y(0)=1

1^2+1=e^0+C \implies 2=1+C \implies C=1

. So

y^2+y=e^x+1

is correct). We are asked to find the value of

x

for which

y=2

. We substitute

y=2

into the implicit solution:

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

4 + 2 = e^x + 1

6 = e^x + 1

e^x = 5

x = \ln(5)

(A) would result from making an error in evaluating the constant of integration, for example, by assuming $e^0=0$ . This would give $C=2$ and the equation $y^2+y=e^x+2$ . Setting $y=2$ gives $6=e^x+2$ , so $e^x=4$ and $x=\ln(4)$ .
(C) would result from an integration error, such as integrating $\int(2y+1)dy$ as $2y^2+y$ . This would lead to the implicit solution $2y^2+y=e^x+2$ . Setting $y=2$ gives $2(4)+2=e^x+2$ , so $10=e^x+2$ , $e^x=8$ , and $x=\ln(8)$ .
(D) results from the algebraic mistake of concluding $x=5$ from $e^x=5$ .