Calculus 2 Quiz: General Solutions Separation Of Variables

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General Solutions Separation Of VariablesQuestion 1 of 20

Which of the following implicitly defines the general solution to $\frac{dy}{dx} = \frac{\ln(x)}{xy}$ ?

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

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

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

y = \sqrt{(\ln x)^2} + C

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Calculus 2 Quiz

Calculus 2 Quiz: General Solutions Separation Of Variables

Practice General Solutions Separation Of Variables 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 General Solutions Separation Of Variables, 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

Which of the following implicitly defines the general solution to $\frac{dy}{dx} = \frac{\ln(x)}{xy}$ ?

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

Explanation: To solve, we separate the variables:

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

. Next, we integrate both sides:

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

. The left side is

\frac{y^2}{2}

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

u = \ln(x)

, so

du = \frac{1}{x} dx

. The integral becomes

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

. So, we have

\frac{y^2}{2} = \frac{(\ln x)^2}{2} + C_1

. Multiplying by 2 gives

y^2 = (\ln x)^2 + 2C_1

. Letting

C = 2C_1

gives the implicit solution

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

Question 2

What is the general solution to $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ ?

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

Explanation: Separating variables gives

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

. Integrating both sides gives

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

. To solve for

y

, we take the tangent of both sides:

y = \tan(\arctan(x) + C)

. Using the tangent addition formula

\tan(a+b) = \frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}

, we have

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

. Let

A = \tan(C)

, which is an arbitrary constant. This simplifies to

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

Question 3

For the differential equation $\frac{dy}{dx} = \frac{\sin(x)}{y^2 + y}$ , after factoring the denominator and using partial fractions, the separated form becomes:

$\left(\frac{y}{1} + \frac{y+1}{1}\right)dy = \sin(x)dx$
$\left(\frac{1}{y} + \frac{1}{y+1}\right)dy = \sin(x)dx$
$\frac{dy}{y(y+1)} = \sin(x)dx$
$\left(\frac{1}{y} - \frac{1}{y+1}\right)dy = \sin(x)dx$ (correct answer)

Explanation: When you encounter a separable differential equation like this one, your goal is to get all terms involving y on one side and all terms involving x on the other. The key challenge here is dealing with the denominator

y^2 + y

using partial fraction decomposition. First, factor the denominator:

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

. Now you need to decompose

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

into partial fractions. Set up:

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

To find A and B, multiply both sides by

y(y+1)

1 = A(y+1) + By

Setting

y = 0

1 = A(1) \Rightarrow A = 1

Setting

y = -1

1 = B(-1) \Rightarrow B = -1

Therefore:

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

The separated form becomes:

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

Choice A incorrectly puts y in the numerators instead of using the reciprocals from partial fractions. Choice B uses the correct format but has the wrong sign—it shows addition instead of subtraction. Choice C stops at the factored form without actually performing the partial fraction decomposition that the question specifically asks for. Study tip: When doing partial fractions with linear factors, remember that each factor

(ay+b)

in the denominator contributes a term

\frac{A}{ay+b}

to your decomposition. Always verify your A and B values by substituting convenient values of the variable.

Question 4

Find the general solution to the differential equation $\frac{dy}{dx} = xy + 2y - x - 2$ .

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

Explanation: The right side of the equation can be factored by grouping:

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

. The equation is

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

. Separating variables gives

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

. Integrating both sides,

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

, yields

\ln|y-1| = \frac{x^2}{2} + 2x + C_1

. Exponentiating both sides gives

|y-1| = e^{\frac{x^2}{2} + 2x + C_1} = e^{C_1}e^{\frac{x^2}{2} + 2x}

. Let

A = \pm e^{C_1}

. Then

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

, and the general solution is

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

Question 5

A tank contains a solution where the rate of change of salt concentration $S$ with respect to time satisfies $\frac{dS}{dt} = \frac{S^2 - 4}{t^2 + 1}$ . If $S(0) = 3$ , what constraint must be considered when finding the domain of the solution?

The solution is undefined when $S^2 - 4 = 0$ , so $S \neq \pm 2$
The solution is undefined when $t^2 + 1 = 0$ , which never occurs for real $t$
The solution may approach $S = 2$ or $S = -2$ asymptotically (correct answer)
The solution is valid for all real values since $S(0) = 3 > 0$

Explanation: When separating variables:

\frac{dS}{S^2-4} = \frac{dt}{t^2+1}

. The separation is valid only when

S^2-4 \neq 0

, i.e.,

S \neq \pm 2

. Since

S(0) = 3 > 2

, the solution might approach

S = 2

as an asymptote, which would limit the domain. Choice A states when separation fails but doesn't address domain implications. Choice B is incorrect since

t^2+1 > 0

always. Choice D ignores the asymptotic behavior near

S = 2

Question 6

The general solution to the differential equation $\frac{dy}{dx} = \frac{1+y^2}{\sqrt{1-x^2}}$ is which of the following?

y = \tan(\arcsin(x)) + C
y = \sqrt{A e^{2\arcsin(x)} - 1}
y = \tan(\arcsin(x) + C) (correct answer)
y = \tan( $\ln$ |x+ $\sqrt{x^2-1}$ | + C)

Explanation: Separate the variables to get

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

. Integrate both sides:

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

. Both are standard inverse trigonometric integrals. The result is

\arctan(y) = \arcsin(x) + C

. To solve for

y

, take the tangent of both sides:

y = \tan(\arcsin(x) + C)

Question 7

Find the general solution to the differential equation $(x^2+4)\frac{dy}{dx} = 1$ .

y = \ln( $x^2+4$ ) + C
y = \frac{1}{2}\arctan( $\frac{x}{2}$ ) + C (correct answer)
y = 2\arctan( $\frac{x}{2}$ ) + C
y = \frac{2x}{x^2+4} + C

Explanation: The equation is separable. Rearrange it to

dy = \frac{1}{x^2+4} dx

. Integrate both sides:

\int dy = \int \frac{1}{x^2+4} dx

. The left side is

y

. The right side is a standard integral of the form

\int \frac{1}{a^2+u^2} du = \frac{1}{a}\arctan(\frac{u}{a})

. Here,

u=x

and

a=2

. So,

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

. The general solution is

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

Question 8

Determine the general solution of the differential equation $\frac{dy}{dx} = y \sin(x) \cos(x)$ .

y = A e^{-\frac{1}{2}\cos^2(x)}
y = A e^{\frac{1}{2}\sin^2(x)} (correct answer)
y = A e^{\sin(x)\cos(x)}
\ln|y| = \sin(x) - \cos(x) + C

Explanation: Separate the variables to get

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

. Integrate both sides:

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

. The left side is

\ln|y|

. For the right side, use u-substitution with

u = \sin(x)

, so

du = \cos(x) dx

. The integral becomes

\int u du = \frac{u^2}{2} = \frac{\sin^2(x)}{2}

. Thus,

\ln|y| = \frac{1}{2}\sin^2(x) + C_1

. Exponentiating both sides gives

|y| = e^{\frac{1}{2}\sin^2(x) + C_1} = e^{C_1}e^{\frac{1}{2}\sin^2(x)}

. Let

A = \pm e^{C_1}

. The solution is

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

. Note that

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

, so choice A is also technically correct by absorbing the constant, but B is the more direct result.

Question 9

A population model is described by the differential equation $\frac{dP}{dt} = \frac{P(100-P)}{t+1}$ , where $P(t)$ represents population at time $t$ . After separating variables, what is the correct integral setup?

$\int \frac{dP}{P(100-P)} = \int \frac{dt}{t+1}$ (correct answer)
$\int \frac{P(100-P)dP}{1} = \int (t+1)dt$
$\int \frac{dP}{P} + \int \frac{dP}{100-P} = \int \frac{dt}{t+1}$
$\int \frac{P dP}{100-P} = \int (t+1)dt$

Explanation: Separating variables:

\frac{dP}{P(100-P)} = \frac{dt}{t+1}

. Integrating both sides gives the setup in choice A. Choice B fails to separate properly. Choice C incorrectly splits the left integral (partial fractions comes after separation). Choice D incorrectly moves only part of the denominator.

Question 10

The equation $\frac{dy}{dx} = \frac{x^2 - 1}{2y + 3}$ has a solution passing through $(1, -1)$ . Which of the following is true about this solution?

The solution does not exist because $2(-1) + 3 = 1 \neq 0$
The solution does not exist because $2y + 3 = 0$ when $y = -1$
The solution exists but is not unique near $(1, -1)$
The solution exists and is unique in a neighborhood of $(1, -1)$ (correct answer)

Explanation: When you encounter a differential equation like this, you need to check whether a solution exists and is unique at the given point. This involves examining the equation's form and applying existence and uniqueness theorems. This differential equation can be written as

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

where

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

. For the solution to exist and be unique near

(1, -1)

, you need both

f(x,y)

and

\frac{\partial f}{\partial y}

to be continuous in a neighborhood of that point. Let's check continuity at

(1, -1)

. First, the denominator:

2(-1) + 3 = 1 \neq 0

, so

f(x,y)

is continuous near

(1, -1)

. The partial derivative

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

is also continuous there since the denominator doesn't equal zero. By the existence and uniqueness theorem, the solution exists and is unique in a neighborhood of

(1, -1)

. Option A incorrectly suggests the solution doesn't exist because

2(-1) + 3 = 1 \neq 0

, but this non-zero value actually ensures existence. Option B makes a calculation error, claiming

2(-1) + 3 = 0

, which is false. Option C correctly identifies that a solution exists but wrongly claims it's not unique—since both continuity conditions are met, uniqueness is guaranteed. Remember: for differential equations of the form

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

, check if the denominator equals zero at your point. If not, and if the partial derivative exists, you typically have existence and uniqueness.

Question 11

The differential equation $\frac{dy}{dx} = \frac{2x\cos(x^2)}{3y^2 + 1}$ can be solved by separation of variables. Which of the following represents the correct form after separation but before integration?

$(3y^2 + 1)dy = 2x\cos(x^2)dx$ (correct answer)
$(3y^2 + 1)dx = 2x\cos(x^2)dy$
$\frac{dy}{3y^2 + 1} = 2x\cos(x^2)dx$
$\frac{dx}{2x\cos(x^2)} = \frac{dy}{3y^2 + 1}$

Explanation: To separate variables, multiply both sides by

(3y^2 + 1)dx

dy = \frac{2x\cos(x^2)}{3y^2 + 1}dx

, then

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

. Choice B incorrectly switches

dx

and

dy

. Choice C fails to multiply the left side by

(3y^2 + 1)

. Choice D incorrectly inverts both sides.

Question 12

Determine the general solution of the differential equation $\frac{dy}{dx} = x \sec(y)$ .

y = \arcsin( $\frac{x^2}{2}$ + C) (correct answer)
y = \sin( $\frac{x^2}{2}$ + C)
y = \arccos(- $\frac{x^2}{2}$ + C)
An implicit solution is given by $\ln|\sec(y) + \tan(y)| = \frac{x^2}{2} + C$ .

Explanation: The equation is separable. We can write it as

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

, which simplifies to

\cos(y) dy = x dx

. Integrating both sides gives

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

, which results in

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

. To solve for

y

, we take the inverse sine of both sides, yielding

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

Question 13

Find the general solution to the differential equation $\frac{dy}{dx} = \frac{x \sin(x)}{e^y}$ .

y = \ln(\sin(x) - x\cos(x) + C) (correct answer)
y = \ln(x\sin(x) - \sin(x) + C)
y = \ln(-x\cos(x) - \sin(x) + C)
y = \ln(\frac{x^2}{2} \sin(x) + C)

Explanation: Separate the variables to get

e^y dy = x \sin(x) dx

. Integrate both sides:

\int e^y dy = \int x \sin(x) dx

. The left integral is

e^y

. The right integral requires integration by parts. Let

u = x

and

dv = \sin(x) dx

. Then

du = dx

and

v = -\cos(x)

. The integral is

uv - \int v du = -x\cos(x) - \int (-\cos(x)) dx = -x\cos(x) + \int \cos(x) dx = -x\cos(x) + \sin(x) + C

. So,

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

. Taking the natural logarithm gives

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

Question 14

An implicit form of the general solution to $\frac{dy}{dx} \sqrt{x} = y^3$ for $x > 0$ is given by:

y^2 = \frac{1}{C - 4\sqrt{x}} (correct answer)
y^2 = \frac{1}{4\sqrt{x} + C}
y^{-2} = C - 2\sqrt{x}
y^{-4} = C - \frac{2}{3}x^{-3/2}

Explanation: Rearrange and separate variables:

y^{-3} dy = x^{-1/2} dx

. Integrate both sides:

\int y^{-3} dy = \int x^{-1/2} dx

. This gives

\frac{y^{-2}}{-2} = \frac{x^{1/2}}{1/2} + C_1

, which simplifies to

-\frac{1}{2y^2} = 2\sqrt{x} + C_1

. Multiply by -2:

\frac{1}{y^2} = -4\sqrt{x} - 2C_1

. Let

C = -2C_1

, a new arbitrary constant. Then

\frac{1}{y^2} = C - 4\sqrt{x}

. Inverting both sides gives

y^2 = \frac{1}{C - 4\sqrt{x}}

Question 15

Find the general solution to the differential equation $x \frac{dy}{dx} - 2y = xy$ .

y = A e^{x} x^2 (correct answer)
y = e^{x} + x^2 + C
y = A(x+2)^2 e^x
y = Ax^2 + e^x

Explanation: The differential equation can be rewritten as

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

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

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

. Integrating both sides gives

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

. This yields

\ln|y| = x + 2\ln|x| + C_1

. Exponentiating both sides, we get

|y| = e^{x + 2\ln|x| + C_1} = e^x \cdot e^{2\ln|x|} \cdot e^{C_1} = e^x \cdot e^{\ln(x^2)} \cdot e^{C_1} = C e^x x^2

. Replacing the constant

C

with

A

gives the general solution

y = A e^{x} x^2

Question 16

The general solution to the differential equation $y' = x^2 y^2 - 4x^2$ is given by which of the following?

-\frac{1}{y-2} = \frac{x^3}{3} + C
\frac{1}{2} \ln|\frac{y-2}{y+2}| = \frac{x^3}{3} + C
\arctan( $\frac{y}{2}$ ) = \frac{x^3}{3} + C
\frac{1}{4} \ln|\frac{y-2}{y+2}| = \frac{x^3}{3} + C (correct answer)

Explanation: When you encounter a differential equation like

y' = x^2 y^2 - 4x^2

, look for opportunities to factor and use separation of variables. Notice that you can factor out

x^2

from the right side:

y' = x^2(y^2 - 4)

. Since

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

, the equation becomes

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

. Now separate variables:

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

. To integrate the left side, use partial fraction decomposition. Set

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

. Solving gives

A = \frac{1}{4}

and

B = -\frac{1}{4}

, so

\frac{1}{(y-2)(y+2)} = \frac{1}{4}\left(\frac{1}{y-2} - \frac{1}{y+2}\right)

. Integrating both sides:

\frac{1}{4}\ln|y-2| - \frac{1}{4}\ln|y+2| = \frac{x^3}{3} + C

. Using logarithm properties, this simplifies to

\frac{1}{4}\ln\left|\frac{y-2}{y+2}\right| = \frac{x^3}{3} + C

, which matches option D. Option A results from incorrectly attempting separation without partial fractions. Option B has the wrong coefficient (

\frac{1}{2}

instead of

\frac{1}{4}

), likely from errors in the partial fraction decomposition. Option C suggests using inverse tangent, which would only apply if the denominator were

y^2 + 4

, not

y^2 - 4

. Remember: when you see

y^2 - a^2

in denominators, think partial fractions leading to logarithmic solutions, not inverse trigonometric functions.

Question 17

Which of the following implicitly defines the general solution to the differential equation $\frac{dy}{dt} = \frac{t}{ty+y}$ ?

\frac{y^2}{2} = \ln|t+1| + C
y^2 = \ln|t+1| + C
\frac{y^2}{2} = t - \ln|t+1| + C (correct answer)
\frac{y^2}{2} = \arctan(t) + C

Explanation: First, factor the denominator:

\frac{dy}{dt} = \frac{t}{y(t+1)}

. This is a separable equation. Separate the variables:

y dy = \frac{t}{t+1} dt

. To integrate the right side, use polynomial long division or substitution

u=t+1

t=u-1

\int \frac{t}{t+1} dt = \int \frac{u-1}{u} du = \int (1 - \frac{1}{u}) du = u - \ln|u| = (t+1) - \ln|t+1|

. So, integrating both sides of the separated equation yields

\frac{y^2}{2} = t+1 - \ln|t+1| + C_1

. Since

C_1

is an arbitrary constant,

C = C_1+1

is also an arbitrary constant. Therefore, the solution is

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

Question 18

Determine the general solution to the differential equation $\frac{dP}{dt} = aP - bP^2$ , where $a$ and $b$ are positive constants.

P(t) = \frac{a}{b} - C e^{-at}
P(t) = \frac{a/b}{C + e^{-at}}
\ln|aP - bP^2| = t + C
P(t) = \frac{a}{b + C e^{-at}} (correct answer)

Explanation: When you encounter a differential equation like

\frac{dP}{dt} = aP - bP^2

, recognize this as a separable differential equation that can also be solved using partial fractions. This particular form often models population growth with limiting factors. To solve this, first separate variables by factoring:

\frac{dP}{dt} = P(a - bP)

. This gives us

\frac{dP}{P(a - bP)} = dt

. The left side requires partial fraction decomposition:

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

. Solving for coefficients gives us

\frac{1}{a}\left(\frac{1}{P} + \frac{b}{a - bP}\right)

. Integrating both sides:

\frac{1}{a}[\ln|P| - \ln|a - bP|] = t + C_1

, which simplifies to

\ln\left|\frac{P}{a - bP}\right| = at + C_2

. Exponentiating and solving for P yields

P(t) = \frac{a}{b + Ce^{-at}}

, confirming answer D. Answer A has the wrong functional form entirely—it's linear rather than rational. Answer B has the exponential term in the wrong location in the denominator. Answer C represents an intermediate step before solving for P explicitly, leaving the equation in implicit logarithmic form rather than the explicit solution requested. Study tip: For separable differential equations involving quadratic terms, always check if partial fractions can simplify your integration. Practice recognizing the logistic equation form

\frac{dP}{dt} = rP(1 - \frac{P}{K})

, as it appears frequently in applications.

Question 19

Find the general solution for the differential equation $(1+x^2)y' - 2xy^2 = 0$ .

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

Explanation: First, rearrange the equation to

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

. Separate the variables:

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

. Integrate both sides:

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

. The left side gives

-y^{-1}

. For the right side, use u-substitution with

u = 1+x^2

du = 2x dx

, so the integral is

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

. The equation is

-\frac{1}{y} = \ln(1+x^2) + C_1

. Multiply by -1:

\frac{1}{y} = -\ln(1+x^2) - C_1

. Let

C = -C_1

. Then

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

. Finally, take the reciprocal to find

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

. Note that choice B is equivalent if C is replaced by -C, but the direct algebraic path leads to A.

Question 20

What is the general solution for the differential equation $y' \sin(y) = \cos(x) - \sin(x)$ ?

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

Explanation: This is a separable differential equation, which means you can separate the variables and integrate both sides. When you see an equation where you can get all terms involving

y

on one side and all terms involving

x

on the other, separation of variables is your go-to method. Starting with

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

, rewrite

y'

\frac{dy}{dx}

and separate:

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

. Now integrate both sides. The left side gives

\int \sin(y) \, dy = -\cos(y)

. The right side gives

\int (\cos(x) - \sin(x)) \, dx = \sin(x) - (-\cos(x)) = \sin(x) + \cos(x)

. Therefore:

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

, which is answer D. Answer A incorrectly uses

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

as the antiderivative of

\sin(y)

, but the correct antiderivative is

-\cos(y)

. Answer B has the correct form for the right side but uses

\cos(y)

instead of

-\cos(y)

, missing the negative sign from integration. Answer C gets the left side correct as

-\cos(y)

, but incorrectly integrates the right side as

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

, putting negative signs where they don't belong. Remember: when integrating

\sin(y)

, you get

-\cos(y)

, and when integrating

\cos(x) - \sin(x)

, you get

\sin(x) + \cos(x)

. Double-check your antiderivatives—they're where most errors occur in separable equations.