Differential Equations Quiz: Solving Exact Des

What this quiz covers

This quiz focuses on Solving Exact Des, 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.

Question 1

1. $y(x\sin(x)) = C$
2. $y(x\sin(x)+1) = C$ (correct answer)
3. $y(x\cos(x)-\sin(x)) = C$
4. $y(x\sin(x)-x\cos(x)+1) = C$

Explanation: When you encounter a differential equation in the form $M(x,y)dx + N(x,y)dy = 0$, you should first check if it's exact. An exact equation has the property that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, and its solution comes from finding a function $F(x,y) = C$ whose total differential equals the given equation. Here, $M(x,y) = y\sin(x) + xy\cos(x)$ and $N(x,y) = x\sin(x) + 1$. Let's verify exactness: $\frac{\partial M}{\partial y} = \sin(x) + x\cos(x)$ and $\frac{\partial N}{\partial x} = \sin(x) + x\cos(x)$. Since these are equal, the equation is exact. To find the solution, integrate $M$ with respect to $x$: $F(x,y) = y\int(\sin(x) + x\cos(x))dx = y(x\sin(x) + C_1) = yx\sin(x) + C_1y$. Adding any function of $y$ alone, we get $F(x,y) = yx\sin(x) + g(y)$. Using $\frac{\partial F}{\partial y} = N$: $x\sin(x) + g'(y) = x\sin(x) + 1$, so $g'(y) = 1$ and $g(y) = y$. Therefore, $F(x,y) = yx\sin(x) + y = y(x\sin(x) + 1)$, giving us the solution $y(x\sin(x) + 1) = C$. Answer A omits the constant term. Answer C uses cosine instead of sine and has incorrect signs. Answer D includes extra terms that don't belong in this solution. Remember: for exact equations, always verify the exactness condition first, then integrate systematically to build the solution function.

Question 2

1. $2$
2. $\frac{7 - 3e^2}{2}$ (correct answer)
3. $\frac{3e^2 - 7}{2}$
4. $-\frac{3e^2 + 1}{2}$

Explanation: Let $M = y/x + 6x$ and $N = \ln(x) - 2$. The equation is defined for $x>0$. Check for exactness: $\frac{\partial M}{\partial y} = 1/x$ and $\frac{\partial N}{\partial x} = 1/x$. The equation is exact. Find the potential function $F(x,y)$ by integrating $N$ with respect to $y$: $F(x,y) = \int (\ln(x)-2)dy = y(\ln(x)-2) + h(x)$. Differentiate with respect to $x$: $\frac{\partial F}{\partial x} = y(1/x) + h'(x)$. Set this equal to $M$: $y/x + h'(x) = y/x + 6x$. So, $h'(x) = 6x$, which gives $h(x) = 3x^2$. The general solution is $y(\ln(x)-2) + 3x^2 = C$. Apply the initial condition $y(e)=4$: $4(\ln(e)-2) + 3e^2 = C \implies 4(1-2) + 3e^2 = C \implies C = 3e^2 - 4$. The particular solution is $y(\ln(x)-2) + 3x^2 = 3e^2 - 4$. To find $y(1)$, substitute $x=1$: $y(1)(\ln(1)-2) + 3(1)^2 = 3e^2 - 4 \implies y(1)(0-2) + 3 = 3e^2 - 4 \implies -2y(1) = 3e^2 - 7 \implies y(1) = \frac{7-3e^2}{2}$.

Question 3

1. The equation becomes exact after multiplying by the integrating factor $\mu(x)=e^x$.
2. The equation is not exact, because $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$.
3. The equation is exact, and its general solution is $x^2+y^2=C$.
4. The equation is exact, and its general solution is $x^2+xy-\frac{1}{2}y^2=C$. (correct answer)

Explanation: When you encounter a differential equation in the form $M(x,y)dx + N(x,y)dy = 0$, your first step should be checking if it's exact by testing whether $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Here, $M(x,y) = 2x+y$ and $N(x,y) = x-y$. Computing the partial derivatives: $\frac{\partial M}{\partial y} = 1$ and $\frac{\partial N}{\partial x} = 1$. Since these are equal, the equation is exact. For an exact equation, there exists a function $F(x,y)$ such that $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$. To find $F$, integrate $M$ with respect to $x$: $F(x,y) = \int(2x+y)dx = x^2 + xy + g(y)$ To find $g(y)$, use the condition $\frac{\partial F}{\partial y} = N$: $\frac{\partial F}{\partial y} = x + g'(y) = x - y$ Therefore, $g'(y) = -y$, so $g(y) = -\frac{1}{2}y^2$ This gives us $F(x,y) = x^2 + xy - \frac{1}{2}y^2$, and the general solution is $x^2 + xy - \frac{1}{2}y^2 = C$. Option A is wrong because no integrating factor is needed—the equation is already exact. Option B incorrectly claims the equation isn't exact when our calculation shows it is. Option C has the wrong solution form entirely. Always verify exactness first before seeking integrating factors. This systematic approach prevents unnecessary work and ensures you don't miss when an equation is already in its simplest exact form.

Question 4

1. Only if $f'(x) = g'(y)$ for all $x, y$ in the domain.
2. Only if $f(x)$ and $g(y)$ are both constants.
3. The new equation is always exact for any choice of $f(x)$ and $g(y)$. (correct answer)
4. Only if $f(x)$ and $g(y)$ are both identically zero.

Explanation: When you encounter questions about exactness of differential equations, remember that the key criterion is whether the mixed partial derivatives are equal: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Since the original equation $M(x,y)dx + N(x,y)dy = 0$ is exact, we know that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. For the new equation $(M(x,y)+f(x))dx + (N(x,y)+g(y))dy = 0$, we need to check if $\frac{\partial}{\partial y}[M(x,y)+f(x)] = \frac{\partial}{\partial x}[N(x,y)+g(y)]$. Computing these partial derivatives: the left side gives $\frac{\partial M}{\partial y} + \frac{\partial f(x)}{\partial y} = \frac{\partial M}{\partial y} + 0 = \frac{\partial M}{\partial y}$, since $f(x)$ doesn't depend on $y$. The right side gives $\frac{\partial N}{\partial x} + \frac{\partial g(y)}{\partial x} = \frac{\partial N}{\partial x} + 0 = \frac{\partial N}{\partial x}$, since $g(y)$ doesn't depend on $x$. Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ from the original exact equation, the new equation is also exact regardless of what $f(x)$ and $g(y)$ are. This makes C correct. A is wrong because the derivatives of $f$ and $g$ don't appear in the exactness condition. B and D are incorrect because they impose unnecessary restrictions—the functions can be any continuous functions, not just constants or zero. Study tip: Remember that adding functions of only one variable to each term preserves exactness because mixed partials of single-variable functions are always zero.

Question 5

1. $a=9, b=1$ (correct answer)
2. $a=9, b=-1$
3. $a=3, b=1$
4. $a=12, b=-1$

Explanation: For the differential equation $M(x,y)dx + N(x,y)dy = 0$ to be exact, the condition $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ must hold. Here, $M(x,y) = 6xy^3 + b\cos(y)$ and $N(x,y) = ax^2y^2 - x\sin(y)$. We compute the partial derivatives: $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(6xy^3 + b\cos(y)) = 18xy^2 - b\sin(y)$. $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(ax^2y^2 - x\sin(y)) = 2axy^2 - \sin(y)$. For the equation to be exact, these two expressions must be identical for all $x$ and $y$. This means we can equate the coefficients of like terms: $18xy^2 = 2axy^2 \implies 18 = 2a \implies a=9$. $-b\sin(y) = -\sin(y) \implies b=1$. Thus, $a=9$ and $b=1$.

Question 6

1. $\frac{\pi}{2}$
2. $\frac{\pi}{2} + 5$
3. $\frac{\pi}{2} + 4$
4. $\frac{\pi}{2} + 6$ (correct answer)

Explanation: When you encounter an exact differential equation, you're looking for a potential function $F(x,y)$ such that $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$, where the equation is written as $M dx + N dy = 0$. Here, $M = 1 + y^2\sin(2x)$ and $N = -2y\cos^2(x)$. To find $F(x,y)$, integrate $M$ with respect to $x$: $F(x,y) = \int (1 + y^2\sin(2x)) dx = x - \frac{y^2\cos(2x)}{2} + g(y)$ To find $g(y)$, use the condition $\frac{\partial F}{\partial y} = N$: $\frac{\partial F}{\partial y} = -y\cos(2x) + g'(y) = -2y\cos^2(x)$ Since $\cos(2x) = 2\cos^2(x) - 1$, we have $-y\cos(2x) = -y(2\cos^2(x) - 1) = -2y\cos^2(x) + y$. Therefore: $g'(y) = y$, so $g(y) = \frac{y^2}{2} + C$ Thus: $F(x,y) = x - \frac{y^2\cos(2x)}{2} + \frac{y^2}{2} + C$ Using $F(0,1) = 5$: $0 - \frac{1}{2} + \frac{1}{2} + C = 5$, so $C = 5$. Therefore: $F(x,y) = x + \frac{y^2(1-\cos(2x))}{2} + 5$ At $(\pi/2, 2)$: $F(\pi/2, 2) = \frac{\pi}{2} + \frac{4(1-(-1))}{2} + 5 = \frac{\pi}{2} + 4 + 5 = \frac{\pi}{2} + 6$ Choice A omits the constant and $y^2$ term. Choice B includes only the original constant. Choice C includes the $y^2$ contribution but misses part of the constant adjustment. Key strategy: Always verify your potential function by checking both partial derivative conditions, and don't forget to apply initial conditions to determine all constants.

Question 7

1. $x e^{xy^2} + x^4 - y^3 = 1$
2. $e^{xy^2} + x^4 - y^3 = 1$
3. $e^{xy^2} + x^4 - y^3 = 2$ (correct answer)
4. $e^{xy^2} + 4x^4 - 3y^3 = 1$

Explanation: When you encounter a differential equation of the form $M(x,y)dx + N(x,y)dy = 0$, check if it's exact by verifying whether $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Here, $M = y^2 e^{xy^2} + 4x^3$ and $N = 2xye^{xy^2} - 3y^2$. Taking partial derivatives: $\frac{\partial M}{\partial y} = 2ye^{xy^2} + y^2 \cdot e^{xy^2} \cdot 2xy = 2ye^{xy^2}(1 + xy^2)$ $\frac{\partial N}{\partial x} = 2ye^{xy^2} + 2xy \cdot e^{xy^2} \cdot y^2 = 2ye^{xy^2}(1 + xy^2)$ Since these are equal, the equation is exact. The solution has the form $F(x,y) = C$ where $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$. Integrating $\frac{\partial F}{\partial x} = y^2 e^{xy^2} + 4x^3$: $F = e^{xy^2} + x^4 + g(y)$ To find $g(y)$, use $\frac{\partial F}{\partial y} = N$: $2xye^{xy^2} + g'(y) = 2xye^{xy^2} - 3y^2$ This gives $g'(y) = -3y^2$, so $g(y) = -y^3$. Therefore: $F(x,y) = e^{xy^2} + x^4 - y^3$ Applying the initial condition $y(1) = 0$: $e^{1 \cdot 0^2} + 1^4 - 0^3 = e^0 + 1 = 2$ The solution is $e^{xy^2} + x^4 - y^3 = 2$, which is choice C. Choice A has an extra factor of $x$ in the exponential term. Choice B uses the wrong constant (1 instead of 2). Choice D has incorrect coefficients (4 and 3 instead of 1). Remember: always check if a differential equation is exact first, then integrate systematically and apply initial conditions carefully to determine the constant.

Question 8

1. $x^2y^3 + \tan(x) = C$ (correct answer)
2. $x^2y^3 - \tan(x) = C$
3. $y^3 + \frac{\tan(x)}{x^2} = C$
4. The equation is not exact and cannot be solved this way.

Explanation: First, rewrite the equation in the standard form $M(x,y)dx + N(x,y)dy = 0$. From $\frac{dy}{dx} = -\frac{2xy^3 + \sec^2(x)}{3x^2y^2}$, we get $(3x^2y^2)dy = -(2xy^3 + \sec^2(x))dx$, which is $(2xy^3 + \sec^2(x))dx + (3x^2y^2)dy = 0$. Let $M(x,y) = 2xy^3 + \sec^2(x)$ and $N(x,y) = 3x^2y^2$. Check for exactness: $\frac{\partial M}{\partial y} = 6xy^2$ and $\frac{\partial N}{\partial x} = 6xy^2$. The equation is exact. We find a potential function $F(x,y)$. Integrating $N$ with respect to $y$: $F(x,y) = \int 3x^2y^2 dy = x^2y^3 + h(x)$. Differentiating with respect to $x$: $\frac{\partial F}{\partial x} = 2xy^3 + h'(x)$. Setting this equal to $M$: $2xy^3 + h'(x) = 2xy^3 + \sec^2(x)$. This gives $h'(x) = \sec^2(x)$, so $h(x) = \tan(x)$. The potential function is $F(x,y) = x^2y^3 + \tan(x)$. The general solution is $F(x,y) = C$, which is $x^2y^3 + \tan(x) = C$.

Question 9

1. $k=5$
2. $k=8$
3. $k=10$ (correct answer)
4. $k=20$

Explanation: Let $M(x, y) = y^3 + kxy^4 - 2x$ and $N(x, y) = 3xy^2 + 20x^2y^3$. For the equation to be exact, the partial derivative of $M$ with respect to $y$ must equal the partial derivative of $N$ with respect to $x$. First, compute $\frac{\partial M}{\partial y}$: $\frac{\partial}{\partial y}(y^3 + kxy^4 - 2x) = 3y^2 + 4kxy^3$. Next, compute $\frac{\partial N}{\partial x}$: $\frac{\partial}{\partial x}(3xy^2 + 20x^2y^3) = 3y^2 + 40xy^3$. Set the two partial derivatives equal to each other: $3y^2 + 4kxy^3 = 3y^2 + 40xy^3$. Subtracting $3y^2$ from both sides gives: $4kxy^3 = 40xy^3$. Assuming $x$ and $y$ are not identically zero, we can divide by $4xy^3$ to find $k$: $k = \frac{40}{4} = 10$.

Question 10

1. $x^2y^2 + \sin x + \cos y = -1$ (correct answer)
2. $x^2y^2 + \sin x + \cos y = 1$
3. $x^2y^2 - \sin x + \cos y = -1$
4. $x^2y^2 + \sin x - \cos y = 1$

Explanation: The equation is exact because $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2xy^2 + \cos x) = 4xy$ and $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(2x^2y - \sin y) = 4xy$. To find the solution $F(x, y) = C$, we integrate $M(x, y)$ with respect to $x$: $F(x, y) = \int (2xy^2 + \cos x) \, dx = x^2y^2 + \sin x + g(y)$. To find $g(y)$, we differentiate $F(x, y)$ with respect to $y$ and set it equal to $N(x, y)$: $\frac{\partial F}{\partial y} = 2x^2y + g'(y) = 2x^2y - \sin y$. This implies $g'(y) = -\sin y$. Integrating with respect to $y$ gives $g(y) = \cos y$. The general solution is $F(x, y) = x^2y^2 + \sin x + \cos y = C$. Apply the initial condition $y(0) = \pi$: $0^2\pi^2 + \sin(0) + \cos(\pi) = C$ $0 + 0 + (-1) = C \implies C = -1$. Thus, the particular solution is $x^2y^2 + \sin x + \cos y = -1$. Distractor B arises from the common error $\cos(\pi)=1$.

Question 11

1. $y=2$
2. $y=1$
3. $y=0$
4. $y=-1$ (correct answer)

Explanation: When you encounter a differential equation like this, you need to recognize it's exact and solve by finding a potential function whose partial derivatives match the given coefficients. For an exact equation $M \, dx + N \, dy = 0$, there exists a function $F(x,y)$ where $\frac{\partial F}{\partial x} = M = y e^{xy} + \cos x$ and $\frac{\partial F}{\partial y} = N = x e^{xy} - 1$. To find $F$, integrate the first equation with respect to $x$: $F(x,y) = \int (y e^{xy} + \cos x) \, dx = e^{xy} + \sin x + g(y)$ To find $g(y)$, take the partial derivative with respect to $y$ and set it equal to $N$: $\frac{\partial F}{\partial y} = x e^{xy} + g'(y) = x e^{xy} - 1$ This gives us $g'(y) = -1$, so $g(y) = -y + C$. Therefore, $F(x,y) = e^{xy} + \sin x - y = C$. Using the initial condition $(\frac{\pi}{2}, 0)$: $e^{(\pi/2)(0)} + \sin(\frac{\pi}{2}) - 0 = 1 + 1 - 0 = 2$ So our solution is $e^{xy} + \sin x - y = 2$. When $x = 0$: $e^{0} + \sin(0) - y = 2$, which gives us $1 + 0 - y = 2$, so $y = -1$. Choice A) $y = 2$ incorrectly uses the constant value. Choice B) $y = 1$ likely comes from forgetting the $-y$ term. Choice C) $y = 0$ might result from confusing the initial condition coordinates. Always verify your potential function by checking both partial derivatives match the original equation's coefficients.

Question 12

1. $\mu(x) = x^2$
2. $\mu(x) = x$ (correct answer)
3. $\mu(x) = e^x$
4. $\mu(x) = 1/x$

Explanation: When you encounter a differential equation that isn't exact, you need to find an integrating factor to make it exact. For the equation $(2y^2 + 3x) \, dx + 2xy \, dy = 0$, first check if it's exact by testing whether $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, where $M = 2y^2 + 3x$ and $N = 2xy$. We get $\frac{\partial M}{\partial y} = 4y$ and $\frac{\partial N}{\partial x} = 2y$. Since $4y \neq 2y$, the equation isn't exact. To find an integrating factor $\mu(x)$ that depends only on $x$, use the formula: $\frac{d\mu}{dx} = \mu \cdot \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N}$ Substituting our values: $\frac{d\mu}{dx} = \mu \cdot \frac{4y - 2y}{2xy} = \mu \cdot \frac{2y}{2xy} = \frac{\mu}{x}$ This gives us $\frac{d\mu}{\mu} = \frac{dx}{x}$, which integrates to $\ln|\mu| = \ln|x| + C$. Therefore, $\mu(x) = x$. Looking at the wrong answers: A) $x^2$ would arise from a different coefficient in our calculation. C) $e^x$ would result if the integrating factor formula yielded a constant instead of $1/x$. D) $1/x$ would be correct if we had mistakenly swapped the numerator terms in our fraction. Study tip: Always verify that $(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})/N$ depends only on $x$ before using this method. If it depends on both variables, try $(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})/M$ for a $\mu(y)$ instead.

Question 13

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

Explanation: When you encounter a non-exact differential equation, your goal is to find an integrating factor that makes it exact, then solve the resulting equation systematically. First, verify this equation isn't exact by checking if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ where $M = 1$ and $N = 2y^2 - \frac{x}{y}$. Since $\frac{\partial M}{\partial y} = 0$ but $\frac{\partial N}{\partial x} = -\frac{1}{y}$, the equation is indeed non-exact. To find the integrating factor $\mu(y)$, use the formula: $\frac{d\ln\mu}{dy} = \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{0 - (-\frac{1}{y})}{2y^2 - \frac{x}{y}} = \frac{1}{y(2y^2 - \frac{x}{y})} = \frac{1}{2y^3 - x}$ Actually, let's try $\mu(y) = \frac{1}{y}$. Multiplying the original equation by $\frac{1}{y}$: $\frac{1}{y}dx + (2y - \frac{x}{y^2})dy = 0$ Now $M = \frac{1}{y}$ and $N = 2y - \frac{x}{y^2}$. Check: $\frac{\partial M}{\partial y} = -\frac{1}{y^2}$ and $\frac{\partial N}{\partial x} = -\frac{1}{y^2}$. Perfect—it's exact! For an exact equation, $F(x,y) = C$ where $\frac{\partial F}{\partial x} = \frac{1}{y}$ and $\frac{\partial F}{\partial y} = 2y - \frac{x}{y^2}$. Integrating the first: $F = \frac{x}{y} + g(y)$. Using the second condition gives $g(y) = y^2$, so $F = \frac{x}{y} + y^2$. The answer is B. Choice A has the wrong sign, while C and D represent different solution approaches that don't match our exact equation's structure. Remember: always verify your integrating factor makes the equation exact before proceeding to solve.

Question 14

1. $(x - 2y) \, dx + (2x + y) \, dy = 0$
2. $(2x + y) \, dx - (x - 2y) \, dy = 0$
3. $(x - 2y) \, dx - (2x + y) \, dy = 0$ (correct answer)
4. $(2x - y) \, dx + (x + 2y) \, dy = 0$

Explanation: The slope of the tangent line to the solution curves of the given equation $M \, dx + N \, dy = 0$ is given by $\frac{dy}{dx} = -\frac{M}{N}$. For the given equation, the slope is $\frac{dy}{dx} = -\frac{2x+y}{x-2y} = \frac{2x+y}{2y-x}$. The slope of the orthogonal trajectories, let's call it $\frac{dy}{dx}_{\perp}$, is the negative reciprocal of the original slope. $\frac{dy}{dx}_{\perp} = -\frac{1}{\frac{2x+y}{2y-x}} = -\frac{2y-x}{2x+y} = \frac{x-2y}{2x+y}$. Now, we convert this slope back into a differential equation of the form $M_{\perp} \, dx + N_{\perp} \, dy = 0$. From $\frac{dy}{dx} = \frac{x-2y}{2x+y}$, we can write $(2x+y) \, dy = (x-2y) \, dx$. Rearranging this into the standard form gives: $(x-2y) \, dx - (2x+y) \, dy = 0$. This matches option C. Option A results from a sign error when finding the negative reciprocal. Option B represents the original family of curves.

Question 15

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

Explanation: First, rewrite the equation in the standard form $M(x, y) \, dx + N(x, y) \, dy = 0$. $(\sin(x) + y) \, dy = (x - y\cos(x)) \, dx$ $(x - y\cos(x)) \, dx - (\sin(x) + y) \, dy = 0$. Here, $M(x, y) = x - y\cos(x)$ and $N(x, y) = -(\sin(x) + y)$. Check for exactness: $\frac{\partial M}{\partial y} = -\cos(x)$. $\frac{\partial N}{\partial x} = -\cos(x)$. The equation is exact. Find the potential function $F(x, y)$. Integrate $M$ with respect to $x$: $F(x, y) = \int (x - y\cos(x)) \, dx = \frac{x^2}{2} - y\sin(x) + g(y)$. Differentiate $F$ with respect to $y$ and set it equal to $N$: $\frac{\partial F}{\partial y} = -\sin(x) + g'(y) = -\sin(x) - y$. This implies $g'(y) = -y$. Integrating gives $g(y) = -\frac{y^2}{2}$. The general solution is $F(x, y) = C$, which is $\frac{x^2}{2} - y\sin(x) - \frac{y^2}{2} = C$. Distractor B results from an incorrect sign during the initial rearrangement of the equation.