Differential Equations Quiz: Initial Conditions Separable Equations

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Initial Conditions Separable EquationsQuestion 1 of 17

For the initial value problem $x \frac{dy}{dx} = 2y$ with $y(-1) = 3$ , what is the value of $y(2)$ ?

-12

12

6

-6

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Differential Equations Quiz

Differential Equations Quiz: Initial Conditions Separable Equations

Practice Initial Conditions 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 Initial Conditions 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.

All questions

Question 1

For the initial value problem $x \frac{dy}{dx} = 2y$ with $y(-1) = 3$ , what is the value of $y(2)$ ?

$-12$
$12$ (correct answer)
$6$
$-6$

Explanation: When you encounter a differential equation like

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

, you're dealing with a separable equation that can be solved by isolating variables on opposite sides. First, separate the variables by dividing both sides by

y

and multiplying by

dx

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

Now integrate both sides:

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

This gives you:

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

Using logarithm properties,

2\ln|x| = \ln|x^2|

, so:

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

Exponentiating both sides:

|y| = e^C \cdot |x^2| = K|x|^2

Since we can absorb the absolute value signs into the constant, the general solution is:

y = Kx^2

Apply the initial condition

y(-1) = 3

3 = K(-1)^2 = K(1) = K

K = 3

, making our particular solution

y = 3x^2

. Therefore:

y(2) = 3(2)^2 = 3(4) = 12

Choice A (

-12

) likely comes from incorrectly handling the negative initial condition point. Choice C (

6

) results from forgetting to square the

x

-value when substituting

x = 2

. Choice D (

-6

) combines both errors—mishandling the sign and forgetting to square. Study tip: For separable differential equations, always check your general solution against the initial condition to find the particular constant, and be extra careful with signs when dealing with negative initial values.

Question 2

A separable differential equation $\frac{dy}{dx} = g(x)h(y)$ has the particular solution $F(y) = G(x) + 3$ after applying the initial condition $y(1) = 2$ . If the general solution before applying initial conditions was $F(y) = G(x) + C$ , which of the following must be true?

$F(2) = G(1)$ , so the constant $C = 0$ in this case
$F(2) = G(1) + C$ and $C = 3$ , so $F(2) = G(1) + 3$
$F(2) - G(1) = 3$ , indicating the integration constant equals $3$ (correct answer)
$F(2) + G(1) = 3$ , which determines the relationship between functions

Explanation: For a separable equation

\frac{dy}{dx} = g(x)h(y)

, we separate to get

\frac{dy}{h(y)} = g(x)dx

. Integrating:

\int \frac{dy}{h(y)} = \int g(x)dx

gives

F(y) = G(x) + C

. Applying initial condition

y(1) = 2

F(2) = G(1) + C

. Since the particular solution is

F(y) = G(x) + 3

, we have

C = 3

. Therefore

F(2) = G(1) + 3

, which can be rewritten as

F(2) - G(1) = 3

. Choice A incorrectly sets C = 0. Choice B correctly identifies the relationship but doesn't simplify to the form requested. Choice D incorrectly uses addition instead of subtraction.

Question 3

Water is draining from a tank. The rate of change of the volume $V$ of water in the tank with respect to time $t$ is proportional to the square root of the volume, so $\frac{dV}{dt} = -k\sqrt{V}$ for some constant $k>0$ . At time $t=0$ , the volume is $100$ liters. At time $t=1$ , the volume is $64$ liters. At what time $t$ is the tank empty?

$4.5$
$25/9$
$5$ (correct answer)
$4$

Explanation: This is a separable differential equation problem where the rate of volume change depends on the current volume. When you see

\frac{dV}{dt} = -k\sqrt{V}

, you need to separate variables and solve for the constant using given conditions. Start by separating variables:

\frac{dV}{\sqrt{V}} = -k \, dt

. Rewriting the left side as

V^{-1/2} dV

and integrating both sides gives

\int V^{-1/2} dV = \int -k \, dt

, which yields

2\sqrt{V} = -kt + C

. Using the initial condition

V(0) = 100

2\sqrt{100} = C

, so

C = 20

. This gives us

2\sqrt{V} = -kt + 20

, or

\sqrt{V} = 10 - \frac{kt}{2}

. Now use the second condition

V(1) = 64

\sqrt{64} = 10 - \frac{k}{2}

, so

8 = 10 - \frac{k}{2}

, giving us

k = 4

. The complete solution is

\sqrt{V} = 10 - 2t

. The tank empties when

V = 0

, so

0 = 10 - 2t

, which means

t = 5

. Choice A (4.5) likely comes from arithmetic errors in solving for

k

or the final equation. Choice B (25/9) might result from incorrectly setting up the separation of variables. Choice D (4) probably comes from confusing the value of

k

with the final answer. When solving separable differential equations, always check your constant using both given conditions, and remember that "empty" means setting the quantity equal to zero.

Question 4

The solution to the initial value problem $\sec^2(x) dy + \csc(y) dx = 0$ , with $y(\pi/4) = \pi/6$ , can be written implicitly as $\cos(y) = \frac{x}{2} + \frac{1}{4}\sin(2x) + K$ . What is the value of the constant $K$ ?

$\frac{4\sqrt{3} - \pi - 2}{8}$ (correct answer)
$\frac{\pi + 2 - 4\sqrt{3}}{8}$
$\frac{2 - \pi}{8}$
$\frac{4\sqrt{3} - \pi}{8}$

Explanation: First, separate the variables:

\frac{dy}{\csc(y)} = -\frac{dx}{\sec^2(x)}

, which is

\sin(y) dy = -\cos^2(x) dx

. Integrate both sides:

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

. This gives

-\cos(y) = -\int \frac{1+\cos(2x)}{2} dx = -(\frac{x}{2} + \frac{1}{4}\sin(2x)) + C_1

. So,

-\cos(y) = -\frac{x}{2} - \frac{1}{4}\sin(2x) + C_1

. Apply the initial condition

y(\pi/4) = \pi/6

-\cos(\pi/6) = -\frac{\pi/4}{2} - \frac{1}{4}\sin(\pi/2) + C_1

. This becomes

-\frac{\sqrt{3}}{2} = -\frac{\pi}{8} - \frac{1}{4} + C_1

, so

C_1 = \frac{\pi}{8} + \frac{1}{4} - \frac{\sqrt{3}}{2} = \frac{\pi + 2 - 4\sqrt{3}}{8}

. The implicit solution is

-\cos(y) = -\frac{x}{2} - \frac{1}{4}\sin(2x) + \frac{\pi + 2 - 4\sqrt{3}}{8}

. Multiplying by -1 gives

\cos(y) = \frac{x}{2} + \frac{1}{4}\sin(2x) - \frac{\pi + 2 - 4\sqrt{3}}{8} = \frac{x}{2} + \frac{1}{4}\sin(2x) + \frac{4\sqrt{3} - \pi - 2}{8}

. Comparing this to the given form, we see that

K = \frac{4\sqrt{3} - \pi - 2}{8}

Question 5

Consider the separable differential equation $\frac{dy}{dx} = \frac{2xy}{x^2 + 1}$ with initial condition $y(2) = 4$ . If the solution is written in the form $y = f(x)$ , what is the value of $y(0)$ ?

$\frac{4}{5}$ (correct answer)
$\frac{16}{5}$
$4$
$\frac{20}{4} = 5$

Explanation: Separating variables:

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

. Integrating:

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

, so

y = K(x^2 + 1)

where

K = e^C

. Using

y(2) = 4

4 = K(4 + 1) = 5K

, so

K = \frac{4}{5}

. Therefore

y = \frac{4}{5}(x^2 + 1)

. At

x = 0

y(0) = \frac{4}{5}(0 + 1) = \frac{4}{5}

. Choice B incorrectly uses

K = \frac{4}{5}

but calculates

y(0) = \frac{4}{5} \cdot 4 = \frac{16}{5}

. Choice C assumes the solution is constant. Choice D makes an arithmetic error in finding K.

Question 6

A differential equation $\frac{dy}{dx} = f(x)g(y)$ satisfies the initial condition $y(x_0) = y_0$ . If $G(y) = \int \frac{1}{g(y)} dy$ and $F(x) = \int f(x) dx$ , then the particular solution can be written as:

$G(y) = F(x) + G(y_0) - F(x_0)$
$G(y) - G(y_0) = F(x) - F(x_0)$ (correct answer)
$G(y) + F(x) = G(y_0) + F(x_0)$
$G(y) = F(x) - G(y_0) + F(x_0)$

Explanation: For the separable equation

\frac{dy}{dx} = f(x)g(y)

, we separate to get

\frac{dy}{g(y)} = f(x)dx

. Integrating both sides from the initial condition:

\int_{y_0}^y \frac{1}{g(t)} dt = \int_{x_0}^x f(s) ds

. This gives us

G(y) - G(y_0) = F(x) - F(x_0)

, where the definite integrals are expressed as differences of antiderivatives. Choice A incorrectly adds

G(y_0)

instead of subtracting. Choice C incorrectly suggests the sum is constant. Choice D has incorrect signs for the terms involving the initial conditions.

Question 7

Consider the initial value problem given by $y' = \frac{y}{x^2 + 1}$ with the initial condition $y(0) = -2$ . What is the value of $y(1)$ ?

$-2e^{\pi/4}$ (correct answer)
$2e^{\pi/4}$
$-e^{\pi/4}/2$
$-2 + \pi/4$

Explanation: The differential equation is separable. We can write

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

. Integrating both sides gives

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

, which results in

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

. Using the initial condition

y(0)=-2

, we have

\ln|-2| = \arctan(0) + C

, which gives

C = \ln(2)

. The general solution is

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

. Exponentiating both sides gives

|y| = e^{\arctan(x) + \ln(2)} = e^{\arctan(x)}e^{\ln(2)} = 2e^{\arctan(x)}

. Since the initial condition

y(0)=-2

is negative, we must choose the negative solution, so

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

. Evaluating at

x=1

, we get

y(1) = -2e^{\arctan(1)} = -2e^{\pi/4}

Question 8

Given the initial value problem $y' = \frac{x}{4y}$ with $y(4)=3$ , find the value of $y(0)$ .

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

Explanation: When you encounter a differential equation with an initial condition asking for a value at a different point, you need to solve the equation completely and then use both the initial condition and the target point. This is a separable differential equation. Start by separating variables:

y' = \frac{dy}{dx} = \frac{x}{4y}

Cross-multiplying gives:

4y \, dy = x \, dx

Integrating both sides:

\int 4y \, dy = \int x \, dx

This yields:

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

Now use the initial condition

y(4) = 3

to find

C

2(3)^2 = \frac{(4)^2}{2} + C

18 = 8 + C

C = 10

So the general solution is:

2y^2 = \frac{x^2}{2} + 10

To find

y(0)

, substitute

x = 0

2y^2 = \frac{0^2}{2} + 10 = 10

y^2 = 5

y = \pm\sqrt{5}

Since the differential equation involves

\frac{x}{4y}

, we need

y \neq 0

. Given that

y(4) = 3 > 0

and the solution is continuous,

y

remains positive, so

y(0) = \sqrt{5}

. Choice A (

3

) incorrectly assumes

y

is constant. Choice B (

-\sqrt{5}

) gives the wrong sign—the solution must be positive throughout. Choice C (

1

) likely comes from algebraic errors in integration or applying the initial condition. Strategy tip: Always check that your solution satisfies both the differential equation and initial condition, and consider the domain restrictions (here,

y \neq 0

) when choosing between multiple solutions.

Question 9

The solution to the differential equation $\frac{dy}{dx} = \frac{kx}{y}$ passes through the points $(1, 2)$ and $(2, 4)$ . If $(1, 2)$ is the initial condition, what is the value of the constant $k$ ?

$2$
$3$
$4$ (correct answer)
$4/3$

Explanation: When you encounter a separable differential equation like

\frac{dy}{dx} = \frac{kx}{y}

, your approach should be to separate variables, integrate both sides, then use given conditions to find unknown constants. First, separate the variables by multiplying both sides by

y

and

dx

y \, dy = kx \, dx

Integrating both sides:

\int y \, dy = \int kx \, dx

\frac{y^2}{2} = \frac{kx^2}{2} + C

Multiplying by 2:

y^2 = kx^2 + 2C

Let

A = 2C

, so:

y^2 = kx^2 + A

Now use the initial condition

(1, 2)

to find

A

2^2 = k(1)^2 + A

4 = k + A

A = 4 - k

So our solution becomes:

y^2 = kx^2 + (4 - k)

Since the solution also passes through

(2, 4)

4^2 = k(2)^2 + (4 - k)

16 = 4k + 4 - k

16 = 3k + 4

12 = 3k

k = 4

This confirms answer C is correct. Answer A (

k = 2

) would give us

y^2 = 2x^2 + 2

, which doesn't satisfy

(2, 4)

since

16 ≠ 10

. Answer B (

k = 3

) yields

y^2 = 3x^2 + 1

, failing at

(2, 4)

since

16 ≠ 13

. Answer D (

k = 4/3

) produces

y^2 = \frac{4x^2}{3} + \frac{8}{3}

, which also fails the second condition. Study tip: Always verify your constant using all given points, not just the initial condition. This catches calculation errors and confirms your solution satisfies the entire problem.

Question 10

Let $y(x)$ be the solution to the initial value problem $\frac{dy}{dx} = (y-1)^2$ with $y(0)=0$ . Find the value of $y(2)$ .

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

Explanation: This is a separable differential equation, one of the most fundamental types you'll encounter. When you see an equation of the form

\frac{dy}{dx} = f(y)

, you should immediately think about separating variables. Starting with

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

, separate the variables:

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

. Integrate both sides:

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

. The left side gives

-\frac{1}{y-1}

and the right side gives

x + C

, so

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

. Using the initial condition

y(0) = 0

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

, which gives

C = 1

. Therefore:

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

. Solving for

y

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

, so

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

, and

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

. At

x = 2

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

, confirming answer B. Answer A (

2

) likely comes from mishandling the integration or algebra. Answer C (

1-e^2

) suggests confusion with exponential solutions, which don't apply here. Answer D (

3/2

) might result from incorrectly inverting the final fraction. When solving separable equations, always double-check your integration of rational functions and verify your solution satisfies the initial condition. These algebraic steps are where most errors occur.

Question 11

A solution $y(x)$ to the differential equation $\frac{dy}{dx} = \frac{x^2 e^{x^3}}{3y^2}$ satisfies the initial condition $y(0)=2$ . For which positive value of $x$ does $y(x)=3$ ?

$( \ln(58) )^{1/3}$ (correct answer)
$( \ln(20) )^{1/3}$
$( \ln(4) )^{1/3}$
$( \ln(57) )^{1/3}$

Explanation: Separating variables gives

3y^2 dy = x^2 e^{x^3} dx

. Integrating both sides, we get

\int 3y^2 dy = \int x^2 e^{x^3} dx

. The left side is

y^3

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

u=x^3

, so

du=3x^2 dx

. The integral becomes

\frac{1}{3} \int e^u du = \frac{1}{3}e^u = \frac{1}{3}e^{x^3}

. Thus,

y^3 = \frac{1}{3}e^{x^3} + C

. Using the initial condition

y(0)=2

, we have

2^3 = \frac{1}{3}e^0 + C

, so

8 = \frac{1}{3} + C

, which gives

C = \frac{23}{3}

. The particular solution is

y^3 = \frac{1}{3}e^{x^3} + \frac{23}{3}

. We want to find

x

when

y=3

. So,

3^3 = \frac{1}{3}e^{x^3} + \frac{23}{3}

, which simplifies to

27 = \frac{1}{3}e^{x^3} + \frac{23}{3}

. Multiplying by 3 gives

81 = e^{x^3} + 23

. Then

e^{x^3} = 58

, so

x^3 = \ln(58)

, and

x = (\ln(58))^{1/3}

Question 12

The solution $y(x)$ to the initial value problem $\frac{dy}{dx} = \frac{x-1}{2-y}$ with $y(1)=4$ represents a semicircle. What is the value of $y(2)$ ?

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

Explanation: When you encounter a separable differential equation like this one, the key insight is recognizing that it can be rearranged so all terms involving one variable are on one side. Here, you can rewrite

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

(2-y)dy = (x-1)dx

. Integrating both sides:

\int(2-y)dy = \int(x-1)dx

, which gives

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

. Using the initial condition

y(1) = 4

2(4) - \frac{16}{2} = \frac{1}{2} - 1 + C

, so

8 - 8 = -\frac{1}{2} + C

, giving

C = \frac{1}{2}

. The solution becomes

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

. Rearranging:

y^2 - 4y + x^2 - 2x = 0

. Completing the square:

(y-2)^2 + (x-1)^2 = 5

. This is indeed a circle with center

(1,2)

and radius

\sqrt{5}

. To find

y(2)

(2-2)^2 + (2-1)^2 = 5

becomes

1 + (y-2)^2 = 5

, so

(y-2)^2 = 4

, giving

y = 2 \pm 2

. Since we start at

(1,4)

and follow the semicircle, we get

y = 4

y = 0

. However, checking the differential equation's behavior shows we move toward

y = 2 + \sqrt{3}

. Answer choice A gives

2-\sqrt{3}

, which would be the other intersection point. Choice C gives 4, our starting point. Choice D gives just

\sqrt{3}

, missing the shift. Study tip: For separable equations, always verify your solution satisfies both the differential equation and initial condition, and pay attention to which branch of multi-valued solutions applies.

Question 13

Let $y(x)$ be the solution to the initial value problem $\frac{dy}{dx} = y^2$ with $y(1) = -1/2$ . What is the largest open interval containing $x=1$ on which the solution $y(x)$ is defined?

$(-\infty, -1) \cup (-1, \infty)$
$(-\infty, 3)$
$(-\infty, \infty)$
$(-1, \infty)$ (correct answer)

Explanation: When you encounter a separable differential equation like

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

, you need to solve it and then determine where the solution might "blow up" or become undefined. First, separate variables:

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

. Integrating both sides gives

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

. Using the initial condition

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

, we find

-\frac{1}{-1/2} = 1 + C

, so

2 = 1 + C

and

C = 1

. Therefore,

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

, which means

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

. The solution becomes undefined when the denominator equals zero, so

x + 1 = 0

x = -1

. Since we need the largest open interval containing

x = 1

where the solution exists, and the solution is undefined at

x = -1

, the interval is

(-1, \infty)

. Looking at the wrong answers: A)

(-\infty, -1) \cup (-1, \infty)

correctly identifies where the solution exists globally, but the question asks for the interval containing

x = 1

, not the entire domain. B)

(-\infty, 3)

incorrectly places the singularity at

x = 3

. C)

(-\infty, \infty)

ignores the vertical asymptote entirely. Key strategy: For rational solutions to differential equations, always check where denominators become zero to find vertical asymptotes. These create boundaries for your intervals of existence. The solution can only be continuous within each separate interval.

Question 14

The rate of change of a population $P(t)$ is modeled by the logistic equation $\frac{dP}{dt} = P(4-P)$ . If the initial population is $P(0)=1$ , at what time $t$ will the population reach $P(t)=2$ ?

$-\frac{\ln(3)}{4}$
$\ln(3)$
$\frac{\ln(3)}{4}$ (correct answer)
$\frac{\ln(9)}{4}$

Explanation: When you encounter a logistic differential equation like

\frac{dP}{dt} = P(4-P)

, you're dealing with a separable equation that models population growth with a carrying capacity. The key is to separate variables and integrate. First, rewrite the equation:

\frac{dP}{P(4-P)} = dt

. To integrate the left side, use partial fractions. Decompose

\frac{1}{P(4-P)} = \frac{A}{P} + \frac{B}{4-P}

. Solving gives us

A = \frac{1}{4}

and

B = \frac{1}{4}

, so

\frac{1}{P(4-P)} = \frac{1}{4}\left(\frac{1}{P} + \frac{1}{4-P}\right)

. Integrating both sides:

\frac{1}{4}\int\left(\frac{1}{P} + \frac{1}{4-P}\right)dP = \int dt

This gives:

\frac{1}{4}[\ln|P| - \ln|4-P|] = t + C

Simplifying:

\frac{1}{4}\ln\left|\frac{P}{4-P}\right| = t + C

Using the initial condition

P(0) = 1

\frac{1}{4}\ln\left(\frac{1}{3}\right) = C

, so

C = -\frac{\ln(3)}{4}

. The solution becomes:

\frac{1}{4}\ln\left(\frac{P}{4-P}\right) = t - \frac{\ln(3)}{4}

When

P(t) = 2

\frac{1}{4}\ln\left(\frac{2}{2}\right) = \frac{1}{4}\ln(1) = 0 = t - \frac{\ln(3)}{4}

Therefore,

t = \frac{\ln(3)}{4}

, which is answer C. Answer A gives the negative value (incorrect sign). Answer B omits the factor of

\frac{1}{4}

. Answer D results from incorrectly calculating

\ln(3^2)

instead of the proper partial fraction integration. Remember: logistic equations always require partial fractions for integration, and careful attention to initial conditions determines the integration constant.

Question 15

Consider the equation $\frac{dy}{dx} = \frac{2x}{3y^2 + 1}$ with initial condition $y(0) = 1$ . After applying the initial condition, the coefficient of $x^2$ in the implicit solution is:

$\frac{1}{2}$
$1$ (correct answer)
$\frac{3}{2}$
$2$

Explanation: Separating variables:

(3y^2 + 1)dy = 2x dx

. Integrating both sides:

\int (3y^2 + 1)dy = \int 2x dx

gives

y^3 + y = x^2 + C

. Using initial condition

y(0) = 1

1^3 + 1 = 0^2 + C

, so

C = 2

. The solution is

y^3 + y = x^2 + 2

. The coefficient of

x^2

1

. Choice A would result from incorrectly integrating

2x

x^2

instead of

x^2

. Choice C might come from confusion with the coefficient 3 in the original equation. Choice D would result from incorrectly keeping the factor of 2 from the integration of

2x

Question 16

The separable equation $\frac{dy}{dx} = \frac{y \ln y}{x}$ has initial condition $y(1) = e$ . Which of the following best describes a potential issue when solving this initial value problem?

The equation is undefined when $x = 0$ , creating a vertical asymptote in the solution
The separation process fails because $\ln y$ is undefined when $y \leq 0$ , restricting the domain (correct answer)
The initial condition $y(1) = e$ leads to $\ln y = 1$ , causing integration difficulties
Multiple solutions exist due to the nonlinear nature of the logarithmic function in the equation

Explanation: When separating variables, we get

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

. The left side requires

y > 0

and

y \neq 1

(since

\ln 1 = 0

). The function

\ln y

is only defined for

y > 0

, and the denominator

y \ln y = 0

when

y = 1

. This creates a restriction on the domain of the solution that must be carefully considered. Choice A incorrectly focuses only on

x = 0

. Choice C misunderstands that

\ln e = 1

is actually convenient for calculations. Choice D incorrectly suggests multiple solutions when the initial value problem has a unique solution in appropriate domains.

Question 17

The differential equation $\frac{dy}{dx} = \frac{y\cos x}{\sin x + 2}$ satisfies $y(\pi) = e^2$ . What constraint must be placed on the domain of the solution to ensure continuity?

$x \neq 0$ and $x \neq \pi$ for all integer multiples
$\sin x + 2 > 0$ , which is satisfied for all real $x$
$x \neq \frac{\pi}{2} + n\pi$ for integer $n$ , where $\cos x = 0$
$\sin x \neq -2$ , but since $-1 \leq \sin x \leq 1$ , no restriction needed (correct answer)

Explanation: Separating variables:

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

. The denominator

\sin x + 2

in the original equation could potentially cause issues if it equals zero. However, since

-1 \leq \sin x \leq 1

for all real

x

, we have

1 \leq \sin x + 2 \leq 3

, so

\sin x + 2 > 0

always. Therefore, no domain restrictions are needed for continuity. Choice A incorrectly identifies zeros of sine and cosine. Choice B correctly notes

\sin x + 2 > 0

but incorrectly presents this as a constraint rather than noting it's automatically satisfied. Choice C incorrectly focuses on where cosine equals zero.