Differential Equations Quiz: Identifying Exact Des

What this quiz covers

This quiz focuses on Identifying 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. The equation is exact. (correct answer)
2. The equation is separable.
3. The equation is linear in $y$.
4. The equation becomes exact when multiplied by $\mu(x) = x$.

Explanation: 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.

Question 2

1. The curve is a level curve of some potential function $F(x,y)$. (correct answer)
2. The vector field $\vec{V} = \langle N, M \rangle$ is conservative.
3. The line integral $\int_C M dx + N dy$ depends on the path $C$.
4. Both functions $M(x,y)$ and $N(x,y)$ must be homogeneous.

Explanation: 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.

Question 3

1. $\mu(x,y) = x$ (correct answer)
2. $\mu(x,y) = y$
3. $\mu(x,y) = xy$
4. $\mu(x,y) = x^2$

Explanation: 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.

Question 4

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

Explanation: 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 $a=1$.

Question 5

1. $P(y)$ is not a function of $y$ only, so an integrating factor of the form $\mu(y)$ does not exist.
2. An integrating factor is $\mu(y) = y$.
3. After multiplying by the appropriate integrating factor $\mu(y)$, the resulting equation is exact. (correct answer)
4. The equation is already exact, so no integrating factor is needed.

Explanation: 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.

Question 6

1. $M(x,y) = x^2 - y$, $N(x,y) = x + y^2$
2. $M(x,y) = \arctan(y)$, $N(x,y) = \frac{x}{1+y^2}$ (correct answer)
3. $M(x,y) = y \ln x$, $N(x,y) = x \ln y$
4. $M(x,y) = 2x e^y$, $N(x,y) = x e^y + y$

Explanation: 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.

Question 7

1. $k=5$
2. $k=10$ (correct answer)
3. $k=20$
4. $k=40$

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 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 $k = 10$.

Question 8

1. The equation is exact only if $f(x)$ is a constant.
2. The equation is exact only if $f(x) = cx^4$ for some constant $c$.
3. The equation is exact for any differentiable function $f(x)$. (correct answer)
4. The equation is never exact, regardless of the choice of $f(x)$.

Explanation: 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.

Question 9

1. $(2x + y^2) dx + 2xy dy = 0$
2. $(\cos(x+y) - x\sin(x+y)) dx - x\sin(x+y) dy = 0$
3. $(y e^{xy} + 2x) dx + (x e^{xy} - 2y) dy = 0$
4. $(x^2 + y^2) dx + xy dy = 0$ (correct answer)

Explanation: 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.

Question 10

1. $(2x e^y + y^3 \cos(x)) dx - (x^2 e^y + 3y^2 \sin(x)) dy = 0$
2. $(x^2 e^y + 3y^2 \sin(x)) dx + (2x e^y + y^3 \cos(x)) dy = 0$
3. $(2x e^y - y^3 \cos(x)) dx + (x^2 e^y + 3y^2 \sin(x)) dy = 0$
4. $(2x e^y + y^3 \cos(x)) dx + (x^2 e^y + 3y^2 \sin(x)) dy = 0$ (correct answer)

Explanation: 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:

• $\frac{\partial F}{\partial x} = 2x e^y + y^3 \cos(x)$
• $\frac{\partial F}{\partial y} = x^2 e^y + 3y^2 \sin(x)$

Question 11

1. The only solution to $Mdx+Ndy=0$ is $y(x)=C$.
2. $M$ and $N$ must both be constant functions.
3. $M(x,y) = -N(x,y)$ for all $x,y$.
4. $M$ is a function of $x$ only and $N$ is a function of $y$ only. (correct answer)

Explanation: 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.

Question 12

1. Both the reasoning and conclusion are correct
2. The reasoning is flawed, but the conclusion happens to be correct (correct answer)
3. The reasoning is correct, but the conclusion is wrong due to a calculation error
4. Both the reasoning and conclusion are incorrect

Explanation: 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.

Question 13

1. The reasoning is correct; the $e^y dy$ term prevents exactness
2. The reasoning is flawed; factoring is irrelevant to exactness testing (correct answer)
3. The reasoning is partially correct; factoring helps, but the conclusion about the $e^y$ term is wrong
4. The factoring is incorrect, making the entire analysis invalid

Explanation: 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.

Question 14

1. $h(y)$ can be any differentiable function of $y$ (correct answer)
2. $h(y)$ must be a constant function
3. $h(y)$ must satisfy $h'(y) = \sec^2 y$
4. No such function $h(y)$ exists that makes the equation exact

Explanation: 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 $\frac{\partial N}{\partial x}$.