Solving Exact DEs - Differential Equations

1. A first-order differential equation is given by $\frac{dy}{dx} = -\frac{2xy^3 + \sec^2(x)}{3x^2y^2}$. What is the general solution?
   Answer: $x^2y^3 + \tan(x) = C$ 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 $x^2y^3 + \tan(x) = C$.
2. Consider the differential equation $(2x+y)dx + (x-y)dy = 0$. Which of the following statements is true?
   Answer: The equation is exact, and its general solution is $x^2+xy-\frac{1}{2}y^2=C$. 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.
3. Find the general solution to the differential equation $\frac{dy}{dx} = \frac{x - y\cos(x)}{\sin(x) + y}$.
   Answer: $\frac{x^2}{2} - y\sin(x) - \frac{y^2}{2} = C$ 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.
4. The differential equation $dx + (2y^2 - \frac{x}{y}) dy = 0$ is not exact but can be made exact using an integrating factor $\mu(y)$ that is a function of $y$ alone. Find the general solution of the resulting exact equation.
   Answer: $\frac{x}{y} + y^2 = C$ 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.
5. The differential equation $(2x + y) \, dx + (x - 2y) \, dy = 0$ is exact, with general solution $F(x,y) = x^2 + xy - y^2 = C$. Which of the following differential equations has solution curves that are orthogonal trajectories to the solution curves of the original equation?
   Answer: $(x - 2y) \, dx - (2x + y) \, dy = 0$ 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.
6. The differential equation $(2x + y\sec^2(xy))dx + (x\sec^2(xy) + 3y^2)dy = 0$ is exact. What is the general solution?
   Answer: $x^2 + \tan(xy) + y^3 = C$ When you encounter a differential equation in the form $M(x,y)dx + N(x,y)dy = 0$ and are told it's exact, you need to find a function $F(x,y)$ such that $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$. The general solution will be $F(x,y) = C$. Here, $M(x,y) = 2x + y\sec^2(xy)$ and $N(x,y) = x\sec^2(xy) + 3y^2$. To find $F(x,y)$, integrate $M$ with respect to $x$: $F(x,y) = \int (2x + y\sec^2(xy))dx = x^2 + \tan(xy) + g(y)$ where $g(y)$ is an unknown function of $y$ only. Now differentiate this with respect to $y$: $\frac{\partial F}{\partial y} = x\sec^2(xy) + g'(y)$ This must equal $N(x,y) = x\sec^2(xy) + 3y^2$, so $g'(y) = 3y^2$, which means $g(y) = y^3$. Therefore, $F(x,y) = x^2 + \tan(xy) + y^3$, making the general solution $x^2 + \tan(xy) + y^3 = C$. Choice A includes $3y^3$ instead of $y^3$ - this comes from incorrectly integrating $3y^2$ as $3y^3$ instead of $y^3$. Choice B has $\sec(xy)$ instead of $\tan(xy)$ - a common error when integrating $\sec^2(xy)$. Choice C simply lists the original equation terms without proper integration. Study tip: Always verify your potential function by taking both partial derivatives and checking they match the original $M$ and $N$ terms.
7. The differential equation $(2y^2 + 3x) \, dx + 2xy \, dy = 0$ can be made exact by an integrating factor $\mu(x)$ that is a function of $x$ alone. Find this integrating factor.
   Answer: $\mu(x) = x$ 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.
8. The differential equation $(1+y^2\sin(2x))dx - 2y\cos^2(x)dy = 0$ is exact. Let $F(x,y)$ be its potential function satisfying $F(0,1)=5$. What is the value of $F(\pi/2, 2)$?
   Answer: $\frac{\pi}{2} + 6$ 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.
9. A solution to the exact differential equation $(y e^{xy} + \cos x) \, dx + (x e^{xy} - 1) \, dy = 0$ passes through the point $(\frac{\pi}{2}, 0)$. What is the value of $y$ when $x=0$?
   Answer: $y=-1$ 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.
10. For the differential equation $\left(\frac{y}{x} + 2x\right)dx + \left(\ln x + y\right)dy = 0$ where $x > 0$, what is the general solution after confirming exactness?
    Answer: $y\ln x + x^2 + \frac{y^2}{2} = C$ 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 verifying that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Here, $M(x,y) = \frac{y}{x} + 2x$ and $N(x,y) = \ln x + y$. Taking partial derivatives: $\frac{\partial M}{\partial y} = \frac{1}{x}$ and $\frac{\partial N}{\partial x} = \frac{1}{x}$. Since these are equal, the equation is exact. For exact equations, you find a function $F(x,y)$ where $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$. Integrating $\frac{\partial F}{\partial x} = \frac{y}{x} + 2x$ with respect to $x$ gives $F = y\ln x + x^2 + g(y)$. To find $g(y)$, use the condition $\frac{\partial F}{\partial y} = N$: $\frac{\partial F}{\partial y} = \ln x + g'(y) = \ln x + y$, so $g'(y) = y$ and $g(y) = \frac{y^2}{2}$. Therefore, the general solution is $y\ln x + x^2 + \frac{y^2}{2} = C$, which is choice C. Choice A is missing the factor of $\frac{1}{2}$ in the $y^2$ term. Choice B incorrectly adds the original equation's terms rather than integrating them. Choice D contains $\ln(xy)$ instead of $y\ln x$, which comes from mishandling the integration of $\frac{y}{x}$. Remember: for exact equations, always verify exactness first, then integrate systematically to build your potential function. Double-check by differentiating your final answer.
11. Find the implicit solution to the initial value problem $(y^2 e^{xy^2} + 4x^3) dx + (2xye^{xy^2} - 3y^2) dy = 0$, with $y(1)=0$.
    Answer: $e^{xy^2} + x^4 - y^3 = 2$ 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.
12. The differential equation $\left(e^y + ye^x\right)dx + \left(xe^y + e^x\right)dy = 0$ is exact. After finding the potential function $F(x,y)$, what is $\frac{\partial^2 F}{\partial x \partial y}$?
    Answer: $e^y + e^x$ When you encounter an exact differential equation, you're working with equations where $M(x,y)dx + N(x,y)dy = 0$ and $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. The key insight is that there exists a potential function $F(x,y)$ such that $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$. Here, $M(x,y) = e^y + ye^x$ and $N(x,y) = xe^y + e^x$. The mixed partial derivative $\frac{\partial^2 F}{\partial x \partial y}$ can be found by taking $\frac{\partial M}{\partial y}$ or $\frac{\partial N}{\partial x}$ (they're equal for exact equations). Taking $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(e^y + ye^x) = e^y + e^x$, which is answer B. Let's verify: $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(xe^y + e^x) = e^y + e^x$. Perfect match. Looking at the wrong answers: A) $e^x + ye^x$ appears to be $\frac{\partial M}{\partial x}$, not the mixed partial. C) $xe^y + ye^x$ incorrectly combines terms from both $M$ and $N$. D) $e^y + xe^y$ seems to be $\frac{\partial N}{\partial y}$, which gives the wrong mixed partial. Remember: For exact equations, $\frac{\partial^2 F}{\partial x \partial y} = \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. This equality is what makes the equation exact in the first place, and either calculation gives you the mixed partial derivative.
13. Find the general solution to the differential equation $(y\sin(x) + xy\cos(x))dx + (x\sin(x) + 1)dy = 0$.
    Answer: $y(x\sin(x)+1) = C$ 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.
14. For the differential equation $(ye^{xy} + 2x)dx + (xe^{xy} + 3y^2)dy = 0$, which of the following represents the correct potential function $F(x,y)$ such that $dF = 0$?
    Answer: $F(x,y) = e^{xy} + x^2 + y^3$ First check exactness: ∂M/∂y = ∂(yeˣʸ + 2x)/∂y = eˣʸ + xyeˣʸ and ∂N/∂x = ∂(xeˣʸ + 3y²)/∂x = eˣʸ + xyeˣʸ. Since these are equal, the equation is exact. To find F, integrate M with respect to x: F = ∫(yeˣʸ + 2x)dx = eˣʸ + x² + g(y). Then ∂F/∂y = xeˣʸ + g'(y) = N = xeˣʸ + 3y², so g'(y) = 3y² and g(y) = y³. Therefore F(x,y) = eˣʸ + x² + y³. Choice B incorrectly lists the coefficients M and N. Choice C has wrong coefficient on y³. Choice D incorrectly includes xy as a factor of eˣʸ.
15. Consider the differential equation $M(x,y) \, dx + N(x,y) \, dy = 0$. If $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ for all $(x,y)$ in a simply connected region $R$, which statement is guaranteed to be true?
    Answer: The general solution is given by a family of level curves of some potential function $F(x,y)$. The condition $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ is the test for exactness. If a differential equation is exact on a simply connected region, there exists a potential function $F(x,y)$ such that $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$. The general solution to the differential equation is given implicitly by $F(x,y) = C$, where $C$ is an arbitrary constant. Each value of $C$ defines a specific curve, and the set of these curves for all possible $C$ forms a family of level curves of the surface $z = F(x,y)$. Distractor A is incorrect; exact equations are not necessarily separable. Distractor B is incorrect; exactness alone does not guarantee uniqueness for an initial value problem, which requires additional conditions like the Lipschitz condition. Distractor C is incorrect because the partial derivatives are switched; it should be $M = \frac{\partial F}{\partial x}$ and $N = \frac{\partial F}{\partial y}$.
16. For what value of the constant $k$ is the differential equation $(y^3 + kxy^4 - 2x) \, dx + (3xy^2 + 20x^2y^3) \, dy = 0$ exact?
    Answer: $k=10$ 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 $k$: $k = \frac{40}{4} = 10$.
17. Find the particular solution to the exact differential equation $(2xy^2 + \cos x) \, dx + (2x^2y - \sin y) \, dy = 0$ subject to the initial condition $y(0) = \pi$.
    Answer: $x^2y^2 + \sin x + \cos y = -1$ 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 $\cos(\pi)=1$.
18. The differential equation $M(x,y)dx + N(x,y)dy = 0$ is known to be exact with potential function $F(x,y)$. If a new differential equation is formed as $(M(x,y)+f(x))dx + (N(x,y)+g(y))dy = 0$, where $f(x)$ and $g(y)$ are continuous functions, under what condition is this new equation also exact?
    Answer: The new equation is always exact for any choice of $f(x)$ and $g(y)$. 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.
19. Consider the initial value problem $(\frac{y}{x} + 6x)dx + (\ln(x) - 2)dy = 0$, with the initial condition $y(e)=4$. What is the value of $y(1)$?
    Answer: $\frac{7 - 3e^2}{2}$ 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 $x=1$: $y(1)(\ln(1)-2) + 3(1)^2 = 3e^2 - 4 \implies y(1)(0-2) + 3 = 3e^2 - 4 \implies -2y(1) = 3e^2 - 7 \implies y(1) = \frac{7-3e^2}{2}$.
20. Consider the differential equation $(3x^2y + e^x)dx + (x^3 + 2y)dy = 0$. After verifying that this equation is exact, what is the general solution?
    Answer: $x^3y + e^x + y^2 = C$ First verify exactness: ∂M/∂y = ∂(3x²y + eˣ)/∂y = 3x² and ∂N/∂x = ∂(x³ + 2y)/∂x = 3x². Since these are equal, the equation is exact. To find F(x,y), integrate M with respect to x: F = ∫(3x²y + eˣ)dx = x³y + eˣ + g(y). Then ∂F/∂y = x³ + g'(y) = N = x³ + 2y, so g'(y) = 2y and g(y) = y². Therefore F(x,y) = x³y + eˣ + y² = C. Choice B has wrong sign on y². Choice C incorrectly adds all terms. Choice D has coefficient error in y² term.

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