Identifying Exact DEs - Differential Equations

1. The differential equation $y dx + (2xy - e^{-2y}) dy = 0$ is not exact. Let $P(y) = \frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})$. If $P(y)$ is a function of $y$ only, then $\mu(y) = e^{\int P(y) dy}$ is an integrating factor. Which statement is true for the given equation?
   Answer: After multiplying by the appropriate integrating factor $\mu(y)$, the resulting equation is exact. First, identify $M(x,y) = y$ and $N(x,y) = 2xy - e^{-2y}$. Compute the partial derivatives: $\frac{\partial M}{\partial y} = 1$ and $\frac{\partial N}{\partial x} = 2y$. Since $1 \neq 2y$, the equation is not exact. Now, let's compute $P(y) = \frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) = \frac{1}{y}(2y - 1) = 2 - \frac{1}{y}$. Since this expression is a function of $y$ only, an integrating factor of the form $\mu(y)$ exists. The theory of integrating factors states that if such a $P(y)$ exists, multiplying the original DE by $\mu(y) = e^{\int P(y) dy}$ will make it exact. Therefore, the resulting equation will be exact. We do not need to find the integrating factor itself to confirm this.
2. For what real 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 condition $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ must hold. First, compute the partial derivative of $M$ with respect to $y$: $\frac{\partial M}{\partial y} = 3y^2 + 4kxy^3$. Next, compute the partial derivative of $N$ with respect to $x$: $\frac{\partial N}{\partial x} = 3y^2 + 40xy^3$. Setting these two expressions equal gives $3y^2 + 4kxy^3 = 3y^2 + 40xy^3$. This simplifies to $4kxy^3 = 40xy^3$, which implies $4k = 40$. Solving for $k$ yields $k = 10$.
3. For the differential equation $(\tan y + 2xy^3)dx + (x\sec^2 y + 3x^2y^2 + h(y))dy = 0$ to be exact, what must be true about the function $h(y)$?
   Answer: $h(y)$ can be any differentiable function of $y$ For exactness, we need $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ where $M = \tan y + 2xy^3$ and $N = x\sec^2 y + 3x^2y^2 + h(y)$. Computing $\frac{\partial M}{\partial y} = \sec^2 y + 6xy^2$ and $\frac{\partial N}{\partial x} = \sec^2 y + 6xy^2 + 0 = \sec^2 y + 6xy^2$ (since $h(y)$ doesn't depend on $x$). These are already equal regardless of the choice of $h(y)$, so the equation is exact for any differentiable function $h(y)$. The function $h(y)$ doesn't affect the exactness condition since it contributes nothing to $\frac{\partial N}{\partial x}$.
4. Consider the differential equation $(3x^2y^2 + 2xy^3)dx + (2x^3y + 3x^2y^2 + e^y)dy = 0$. A student factors the first few terms and writes this as $xy^2(3x + 2y)dx + x^2y(2x + 3y)dy + e^y dy = 0$, then claims the equation cannot be exact because of the "extra" $e^y dy$ term. Evaluate this reasoning.
   Answer: The reasoning is flawed; factoring is irrelevant to exactness testing The student's reasoning is flawed because factoring the terms has no bearing on whether a differential equation is exact. Exactness depends solely on whether $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ for the complete functions $M$ and $N$. Here $M = 3x^2y^2 + 2xy^3$ and $N = 2x^3y + 3x^2y^2 + e^y$. We compute $\frac{\partial M}{\partial y} = 6x^2y + 6xy^2$ and $\frac{\partial N}{\partial x} = 6x^2y + 6xy^2$. Since these are equal, the equation is exact. The presence of the $e^y$ term doesn't "prevent" exactness—it's simply part of $N(x,y)$. The student's approach of factoring and treating terms separately is not how exactness is determined.
5. A student encounters the differential equation $(ye^{xy} + 2x)dx + (xe^{xy} + \cos y)dy = 0$ and immediately concludes it is exact because "the exponential terms match." Without solving the equation, determine whether this reasoning and conclusion are correct.
   Answer: The reasoning is flawed, but the conclusion happens to be correct The student's reasoning is flawed because "matching exponential terms" is not a criterion for exactness. The correct test is whether $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Let's check: $M = ye^{xy} + 2x$ and $N = xe^{xy} + \cos y$. We have $\frac{\partial M}{\partial y} = e^{xy} + ye^{xy} \cdot x = e^{xy} + xye^{xy} = e^{xy}(1 + xy)$ and $\frac{\partial N}{\partial x} = e^{xy} + xe^{xy} \cdot y = e^{xy} + xye^{xy} = e^{xy}(1 + xy)$. Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is indeed exact. So the student's conclusion is correct, but the reasoning ("exponential terms match") is not a valid mathematical justification for exactness.
6. The solution to an exact differential equation is given implicitly by the family of curves $F(x,y) = C$. If $F(x,y) = x^2 e^y + y^3 \sin(x)$, which differential equation corresponds to this family of solutions?
   Answer: $(2x e^y + y^3 \cos(x)) dx + (x^2 e^y + 3y^2 \sin(x)) dy = 0$ When you encounter a problem about finding the differential equation from a given family of solutions $F(x,y) = C$, you need to use implicit differentiation. The key insight is that differentiating both sides of $F(x,y) = C$ with respect to $x$ gives you the differential equation. Starting with $F(x,y) = x^2 e^y + y^3 \sin(x) = C$, differentiate both sides with respect to $x$: $\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \cdot \frac{dy}{dx} = 0$ Calculate the partial derivatives: - $\frac{\partial F}{\partial x} = 2x e^y + y^3 \cos(x)$ - $\frac{\partial F}{\partial y} = x^2 e^y + 3y^2 \sin(x)$ Substituting these into our equation: $(2x e^y + y^3 \cos(x)) + (x^2 e^y + 3y^2 \sin(x)) \frac{dy}{dx} = 0$ Rearranging to standard form $M dx + N dy = 0$: $(2x e^y + y^3 \cos(x)) dx + (x^2 e^y + 3y^2 \sin(x)) dy = 0$ This matches answer choice D. Answer choice A has the wrong sign on the second term - it shows subtraction instead of addition. Answer choice B swaps the coefficients of $dx$ and $dy$, which would come from incorrectly identifying which partial derivative goes with which differential. Answer choice C has the wrong sign on the $y^3 \cos(x)$ term in the first coefficient. Remember: when working backwards from $F(x,y) = C$ to find the differential equation, always use implicit differentiation and be careful with signs when rearranging to standard form.
7. The equation $(3xy + y^2) dx + (x^2 + xy) dy = 0$ is not exact. For which function $\mu(x,y)$ does multiplication by $\mu$ make the equation $(\mu(3xy + y^2)) dx + (\mu(x^2 + xy)) dy = 0$ exact?
   Answer: $\mu(x,y) = x$ We need to test each proposed integrating factor by multiplying it through the equation and then checking the exactness condition $\frac{\partial M_{new}}{\partial y} = \frac{\partial N_{new}}{\partial x}$. A) Let $\mu(x,y)=x$. The new equation is $(3x^2y + xy^2)dx + (x^3 + x^2y)dy = 0$. Let $M_1 = 3x^2y + xy^2$ and $N_1 = x^3 + x^2y$. Then $\frac{\partial M_1}{\partial y} = 3x^2 + 2xy$ and $\frac{\partial N_1}{\partial x} = 3x^2 + 2xy$. Since they are equal, the equation becomes exact. This is the correct answer. B) Let $\mu(x,y)=y$. $M_2 = 3xy^2 + y^3$, $N_2 = x^2y + xy^2$. $\frac{\partial M_2}{\partial y} = 6xy + 3y^2$, $\frac{\partial N_2}{\partial x} = 2xy + y^2$. Not exact. C) Let $\mu(x,y)=xy$. $M_3 = 3x^2y^2 + xy^3$, $N_3 = x^3y + x^2y^2$. $\frac{\partial M_3}{\partial y} = 6x^2y + 3xy^2$, $\frac{\partial N_3}{\partial x} = 3x^2y + 2xy^2$. Not exact. D) Let $\mu(x,y)=x^2$. $M_4 = 3x^3y + x^2y^2$, $N_4 = x^4 + x^3y$. $\frac{\partial M_4}{\partial y} = 3x^3 + 2x^2y$, $\frac{\partial N_4}{\partial x} = 4x^3 + 3x^2y$. Not exact.
8. Consider the differential equation $\frac{dy}{dx} = \frac{e^x \sin y - 2xy}{x^2 - e^x \cos y}$. Which of the following statements is true?
   Answer: The equation is exact. First, rewrite the equation in the standard form $M(x,y)dx + N(x,y)dy = 0$. From $\frac{dy}{dx} = \frac{e^x \sin y - 2xy}{x^2 - e^x \cos y}$, we get $(x^2 - e^x \cos y)dy = (e^x \sin y - 2xy)dx$. Rearranging gives $(e^x \sin y - 2xy)dx - (x^2 - e^x \cos y)dy = 0$, which is $(e^x \sin y - 2xy)dx + (e^x \cos y - x^2)dy = 0$. Let $M(x,y) = e^x \sin y - 2xy$ and $N(x,y) = e^x \cos y - x^2$. To check for exactness, we compute the partial derivatives: $\frac{\partial M}{\partial y} = e^x \cos y - 2x$ and $\frac{\partial N}{\partial x} = e^x \cos y - 2x$. Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact. It is not separable because the terms cannot be factored into functions of only $x$ and only $y$. It is not linear due to the $\sin y$ and $\cos y$ terms. Multiplying by an integrating factor is unnecessary as it's already exact.
9. Consider the differential equation $(f(x) + 2xy^2) dx + (2x^2y - \cos(y)) dy = 0$, where $f(x)$ is any differentiable function. Which statement about the exactness of this equation is true?
   Answer: The equation is exact for any differentiable function $f(x)$. Let $M(x,y) = f(x) + 2xy^2$ and $N(x,y) = 2x^2y - \cos(y)$. We check the condition for exactness, $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. The partial derivative of $M$ with respect to $y$ is $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(f(x) + 2xy^2) = 0 + 4xy = 4xy$. The partial derivative of $N$ with respect to $x$ is $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(2x^2y - \cos(y)) = 4xy - 0 = 4xy$. Since $\frac{\partial M}{\partial y} = 4xy$ and $\frac{\partial N}{\partial x} = 4xy$, the condition for exactness is satisfied for any differentiable function $f(x)$. The choice of $f(x)$ does not affect the outcome of the test.
10. The slope of the tangent line to a curve at any point $(x,y)$ is given by $y' = -\frac{M(x,y)}{N(x,y)}$. The set of all such curves forms the solution to $M(x,y)dx + N(x,y)dy = 0$. For which of the following pairs of functions $M$ and $N$ does the corresponding family of curves represent the level sets of a potential function $F(x,y)$?
    Answer: $M(x,y) = \arctan(y)$, $N(x,y) = \frac{x}{1+y^2}$ The family of curves represents the level sets of a potential function if and only if the differential equation $M(x,y)dx + N(x,y)dy = 0$ is exact. We must check the condition $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ for each pair of functions. A) $M_y = -1$, $N_x = 1$. Not exact. B) $M_y = \frac{\partial}{\partial y}(\arctan(y)) = \frac{1}{1+y^2}$. $N_x = \frac{\partial}{\partial x}(\frac{x}{1+y^2}) = \frac{1}{1+y^2}$. Since $M_y = N_x$, this equation is exact. C) $M_y = \ln x$, $N_x = \ln y$. Not exact. D) $M_y = 2x e^y$, $N_x = e^y$. Not exact.
11. A differential equation $M(x,y) dx + N(x,y) dy = 0$ is exact. Which of the following must be true for any solution curve $y(x)$ of the equation in a simply connected region where $M$ and $N$ have continuous first partial derivatives?
    Answer: The curve is a level curve of some potential function $F(x,y)$. An exact differential equation $M dx + N dy = 0$ is one for which there exists a potential function $F(x,y)$ such that $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$. The solutions to the differential equation are given by the implicit relation $F(x,y) = C$, where $C$ is a constant. This means the solution curves are the level curves of the potential function $F$. The condition for exactness, $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, is equivalent to the condition for the vector field $\vec{V} = \langle M, N \rangle$ to be conservative, not $\langle N, M \rangle$. For an exact differential (or a conservative vector field), the line integral $\int_C M dx + N dy$ is path-independent. The condition for a homogeneous equation is different from the condition for an exact equation.
12. For the differential equation $(e^{ax} \cos(y) + y^2) dx + (2xy - e^{ax} \sin(y)) dy = 0$ to be exact, what must be the value of the constant $a$?
    Answer: $a = 1$ Let $M(x,y) = e^{ax} \cos(y) + y^2$ and $N(x,y) = 2xy - e^{ax} \sin(y)$. For the equation to be exact, we must have $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Let's compute these derivatives. $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(e^{ax} \cos(y) + y^2) = -e^{ax} \sin(y) + 2y$. $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(2xy - e^{ax} \sin(y)) = 2y - a e^{ax} \sin(y)$. Setting them equal: $-e^{ax} \sin(y) + 2y = 2y - a e^{ax} \sin(y)$. This simplifies to $-e^{ax} \sin(y) = -a e^{ax} \sin(y)$. For this equality to hold for all $x$ and $y$, we must have $a=1$.
13. Which of the following differential equations is NOT exact?
    Answer: $(x^2 + y^2) dx + xy dy = 0$ We test each equation for exactness using the condition $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. A) $M=2x+y^2, N=2xy$. $\frac{\partial M}{\partial y} = 2y$. $\frac{\partial N}{\partial x} = 2y$. Exact. B) $M=\cos(x+y) - x\sin(x+y), N=-x\sin(x+y)$. $\frac{\partial M}{\partial y} = -\sin(x+y) - x\cos(x+y)$. $\frac{\partial N}{\partial x} = -(1\cdot\sin(x+y) + x\cos(x+y)) = -\sin(x+y) - x\cos(x+y)$. Exact. C) $M=y e^{xy} + 2x, N=x e^{xy} - 2y$. $\frac{\partial M}{\partial y} = e^{xy} + y(xe^{xy}) = e^{xy}(1+xy)$. $\frac{\partial N}{\partial x} = e^{xy} + x(ye^{xy}) = e^{xy}(1+xy)$. Exact. D) $M=x^2+y^2, N=xy$. $\frac{\partial M}{\partial y} = 2y$. $\frac{\partial N}{\partial x} = y$. Since $2y \neq y$, this equation is not exact.
14. If the equation $M(x,y)dx + N(x,y)dy = 0$ is exact, and the equation $M(x,y)dx - N(x,y)dy = 0$ is also exact, which of the following must be true about $M(x,y)$ and $N(x,y)$ on a simply connected domain with continuous first partial derivatives?
    Answer: $M$ is a function of $x$ only and $N$ is a function of $y$ only. When you encounter problems about exact differential equations, you need to recall the exactness condition: an equation $M(x,y)dx + N(x,y)dy = 0$ is exact if and only if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Let's apply this condition to both given equations. For $M(x,y)dx + N(x,y)dy = 0$ to be exact, we need: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ For $M(x,y)dx - N(x,y)dy = 0$ to be exact, we need: $\frac{\partial M}{\partial y} = \frac{\partial(-N)}{\partial x} = -\frac{\partial N}{\partial x}$ Setting these equal: $\frac{\partial N}{\partial x} = -\frac{\partial N}{\partial x}$, which means $\frac{\partial N}{\partial x} = 0$. Therefore, $N$ depends only on $y$. Similarly, from the first condition: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 0$, so $M$ depends only on $x$. This confirms answer choice D. Let's examine why the other options fail: A) This claims the solution is always $y = C$, but this would only be true if $N = 0$, which isn't required. B) Constant functions are too restrictive. For example, $M(x) = x$ and $N(y) = y$ satisfy our conditions but aren't constant. C) The relationship $M = -N$ isn't required. Functions like $M(x) = x$ and $N(y) = 2y$ work perfectly. Study tip: When you see "both equations exact" problems, immediately write out both exactness conditions and look for what constraints this places on the partial derivatives. The intersection of these constraints often reveals the answer.

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