Differential Equations Quiz: Identifying Exact Des
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Identifying Exact DesQuestion 1 of 14

A differential equation M(x,y)dx+N(x,y)dy=0M(x,y) dx + N(x,y) dy = 0 is exact. Which of the following must be true for any solution curve y(x)y(x) of the equation in a simply connected region where MM and NN have continuous first partial derivatives?

The curve is a level curve of some potential function F(x,y)F(x,y).
The vector field V⃗=⟨N,M⟩\vec{V} = \langle N, M \rangle is conservative.
The line integral ∫CMdx+Ndy\int_C M dx + N dy depends on the path CC.
Both functions M(x,y)M(x,y) and N(x,y)N(x,y) must be homogeneous.
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Differential Equations Quiz

Differential Equations Quiz: Identifying Exact Des

Practice Identifying Exact Des in Differential Equations 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 Identifying Exact Des, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

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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.

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Question 1

A differential equation M(x,y)dx+N(x,y)dy=0M(x,y) dx + N(x,y) dy = 0 is exact. Which of the following must be true for any solution curve y(x)y(x) of the equation in a simply connected region where MM and NN have continuous first partial derivatives?

  1. The curve is a level curve of some potential function F(x,y)F(x,y). (correct answer)
  2. The vector field V⃗=⟨N,M⟩\vec{V} = \langle N, M \rangle is conservative.
  3. The line integral ∫CMdx+Ndy\int_C M dx + N dy depends on the path CC.
  4. Both functions M(x,y)M(x,y) and N(x,y)N(x,y) must be homogeneous.
Explanation: An exact differential equation Mdx+Ndy=0M dx + N dy = 0 is one for which there exists a potential function F(x,y)F(x,y) such that ∂F∂x=M\frac{\partial F}{\partial x} = M and ∂F∂y=N\frac{\partial F}{\partial y} = N. The solutions to the differential equation are given by the implicit relation F(x,y)=CF(x,y) = C, where CC is a constant. This means the solution curves are the level curves of the potential function FF. The condition for exactness, ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}, is equivalent to the condition for the vector field V⃗=⟨M,N⟩\vec{V} = \langle M, N \rangle to be conservative, not ⟨N,M⟩\langle N, M \rangle. For an exact differential (or a conservative vector field), the line integral ∫CMdx+Ndy\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 2

For the differential equation (eaxcos⁡(y)+y2)dx+(2xy−eaxsin⁡(y))dy=0(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 aa?

  1. a=−1a = -1
  2. a=0a = 0
  3. a=1a = 1 (correct answer)
  4. a=2a = 2
Explanation: Let M(x,y)=eaxcos⁡(y)+y2M(x,y) = e^{ax} \cos(y) + y^2 and N(x,y)=2xy−eaxsin⁡(y)N(x,y) = 2xy - e^{ax} \sin(y). For the equation to be exact, we must have ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. Let's compute these derivatives. ∂M∂y=∂∂y(eaxcos⁡(y)+y2)=−eaxsin⁡(y)+2y\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(e^{ax} \cos(y) + y^2) = -e^{ax} \sin(y) + 2y. ∂N∂x=∂∂x(2xy−eaxsin⁡(y))=2y−aeaxsin⁡(y)\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(2xy - e^{ax} \sin(y)) = 2y - a e^{ax} \sin(y). Setting them equal: −eaxsin⁡(y)+2y=2y−aeaxsin⁡(y)-e^{ax} \sin(y) + 2y = 2y - a e^{ax} \sin(y). This simplifies to −eaxsin⁡(y)=−aeaxsin⁡(y)-e^{ax} \sin(y) = -a e^{ax} \sin(y). For this equality to hold for all xx and yy, we must have a=1a=1.

Question 3

The differential equation ydx+(2xy−e−2y)dy=0y dx + (2xy - e^{-2y}) dy = 0 is not exact. Let P(y)=1M(∂N∂x−∂M∂y)P(y) = \frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}). If P(y)P(y) is a function of yy only, then μ(y)=e∫P(y)dy\mu(y) = e^{\int P(y) dy} is an integrating factor. Which statement is true for the given equation?

  1. P(y)P(y) is not a function of yy only, so an integrating factor of the form μ(y)\mu(y) does not exist.
  2. An integrating factor is μ(y)=y\mu(y) = y.
  3. After multiplying by the appropriate integrating factor μ(y)\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)=yM(x,y) = y and N(x,y)=2xy−e−2yN(x,y) = 2xy - e^{-2y}. Compute the partial derivatives: ∂M∂y=1\frac{\partial M}{\partial y} = 1 and ∂N∂x=2y\frac{\partial N}{\partial x} = 2y. Since 1≠2y1 \neq 2y, the equation is not exact. Now, let's compute P(y)=1M(∂N∂x−∂M∂y)=1y(2y−1)=2−1yP(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 yy only, an integrating factor of the form μ(y)\mu(y) exists. The theory of integrating factors states that if such a P(y)P(y) exists, multiplying the original DE by μ(y)=e∫P(y)dy\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 4

The slope of the tangent line to a curve at any point (x,y)(x,y) is given by y′=−M(x,y)N(x,y)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=0M(x,y)dx + N(x,y)dy = 0. For which of the following pairs of functions MM and NN does the corresponding family of curves represent the level sets of a potential function F(x,y)F(x,y)?

  1. M(x,y)=x2−yM(x,y) = x^2 - y, N(x,y)=x+y2N(x,y) = x + y^2
  2. M(x,y)=arctan⁡(y)M(x,y) = \arctan(y), N(x,y)=x1+y2N(x,y) = \frac{x}{1+y^2} (correct answer)
  3. M(x,y)=yln⁡xM(x,y) = y \ln x, N(x,y)=xln⁡yN(x,y) = x \ln y
  4. M(x,y)=2xeyM(x,y) = 2x e^y, N(x,y)=xey+yN(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=0M(x,y)dx + N(x,y)dy = 0 is exact. We must check the condition ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} for each pair of functions. A) My=−1M_y = -1, Nx=1N_x = 1. Not exact. B) My=∂∂y(arctan⁡(y))=11+y2M_y = \frac{\partial}{\partial y}(\arctan(y)) = \frac{1}{1+y^2}. Nx=∂∂x(x1+y2)=11+y2N_x = \frac{\partial}{\partial x}(\frac{x}{1+y^2}) = \frac{1}{1+y^2}. Since My=NxM_y = N_x, this equation is exact. C) My=ln⁡xM_y = \ln x, Nx=ln⁡yN_x = \ln y. Not exact. D) My=2xeyM_y = 2x e^y, Nx=eyN_x = e^y. Not exact.

Question 5

The solution to an exact differential equation is given implicitly by the family of curves F(x,y)=CF(x,y) = C. If F(x,y)=x2ey+y3sin⁡(x)F(x,y) = x^2 e^y + y^3 \sin(x), which differential equation corresponds to this family of solutions?

  1. (2xey+y3cos⁡(x))dx−(x2ey+3y2sin⁡(x))dy=0(2x e^y + y^3 \cos(x)) dx - (x^2 e^y + 3y^2 \sin(x)) dy = 0
  2. (x2ey+3y2sin⁡(x))dx+(2xey+y3cos⁡(x))dy=0(x^2 e^y + 3y^2 \sin(x)) dx + (2x e^y + y^3 \cos(x)) dy = 0
  3. (2xey−y3cos⁡(x))dx+(x2ey+3y2sin⁡(x))dy=0(2x e^y - y^3 \cos(x)) dx + (x^2 e^y + 3y^2 \sin(x)) dy = 0
  4. (2xey+y3cos⁡(x))dx+(x2ey+3y2sin⁡(x))dy=0(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)=CF(x,y) = C, you need to use implicit differentiation. The key insight is that differentiating both sides of F(x,y)=CF(x,y) = C with respect to xx gives you the differential equation. Starting with F(x,y)=x2ey+y3sin⁡(x)=CF(x,y) = x^2 e^y + y^3 \sin(x) = C, differentiate both sides with respect to xx: ∂F∂x+∂F∂y⋅dydx=0\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \cdot \frac{dy}{dx} = 0 Calculate the partial derivatives:
  • ∂F∂x=2xey+y3cos⁡(x)\frac{\partial F}{\partial x} = 2x e^y + y^3 \cos(x)
  • ∂F∂y=x2ey+3y2sin⁡(x)\frac{\partial F}{\partial y} = x^2 e^y + 3y^2 \sin(x)
Substituting these into our equation: (2xey+y3cos⁡(x))+(x2ey+3y2sin⁡(x))dydx=0(2x e^y + y^3 \cos(x)) + (x^2 e^y + 3y^2 \sin(x)) \frac{dy}{dx} = 0 Rearranging to standard form Mdx+Ndy=0M dx + N dy = 0: (2xey+y3cos⁡(x))dx+(x2ey+3y2sin⁡(x))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 dxdx and dydy, which would come from incorrectly identifying which partial derivative goes with which differential. Answer choice C has the wrong sign on the y3cos⁡(x)y^3 \cos(x) term in the first coefficient. Remember: when working backwards from F(x,y)=CF(x,y) = C to find the differential equation, always use implicit differentiation and be careful with signs when rearranging to standard form.

Question 6

If the equation M(x,y)dx+N(x,y)dy=0M(x,y)dx + N(x,y)dy = 0 is exact, and the equation M(x,y)dx−N(x,y)dy=0M(x,y)dx - N(x,y)dy = 0 is also exact, which of the following must be true about M(x,y)M(x,y) and N(x,y)N(x,y) on a simply connected domain with continuous first partial derivatives?

  1. The only solution to Mdx+Ndy=0Mdx+Ndy=0 is y(x)=Cy(x)=C.
  2. MM and NN must both be constant functions.
  3. M(x,y)=−N(x,y)M(x,y) = -N(x,y) for all x,yx,y.
  4. MM is a function of xx only and NN is a function of yy 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=0M(x,y)dx + N(x,y)dy = 0 is exact if and only if ∂M∂y=∂N∂x\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=0M(x,y)dx + N(x,y)dy = 0 to be exact, we need: ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} For M(x,y)dx−N(x,y)dy=0M(x,y)dx - N(x,y)dy = 0 to be exact, we need: ∂M∂y=∂(−N)∂x=−∂N∂x\frac{\partial M}{\partial y} = \frac{\partial(-N)}{\partial x} = -\frac{\partial N}{\partial x} Setting these equal: ∂N∂x=−∂N∂x\frac{\partial N}{\partial x} = -\frac{\partial N}{\partial x}, which means ∂N∂x=0\frac{\partial N}{\partial x} = 0. Therefore, NN depends only on yy. Similarly, from the first condition: ∂M∂y=∂N∂x=0\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 0, so MM depends only on xx. This confirms answer choice D. Let's examine why the other options fail: A) This claims the solution is always y=Cy = C, but this would only be true if N=0N = 0, which isn't required. B) Constant functions are too restrictive. For example, M(x)=xM(x) = x and N(y)=yN(y) = y satisfy our conditions but aren't constant. C) The relationship M=−NM = -N isn't required. Functions like M(x)=xM(x) = x and N(y)=2yN(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 7

A student encounters the differential equation (yexy+2x)dx+(xexy+cos⁡y)dy=0(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.

  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 ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. Let's check: M=yexy+2xM = ye^{xy} + 2x and N=xexy+cos⁡yN = xe^{xy} + \cos y. We have ∂M∂y=exy+yexy⋅x=exy+xyexy=exy(1+xy)\frac{\partial M}{\partial y} = e^{xy} + ye^{xy} \cdot x = e^{xy} + xye^{xy} = e^{xy}(1 + xy) and ∂N∂x=exy+xexy⋅y=exy+xyexy=exy(1+xy)\frac{\partial N}{\partial x} = e^{xy} + xe^{xy} \cdot y = e^{xy} + xye^{xy} = e^{xy}(1 + xy). Since ∂M∂y=∂N∂x\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 8

Consider the differential equation (3x2y2+2xy3)dx+(2x3y+3x2y2+ey)dy=0(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 xy2(3x+2y)dx+x2y(2x+3y)dy+eydy=0xy^2(3x + 2y)dx + x^2y(2x + 3y)dy + e^y dy = 0, then claims the equation cannot be exact because of the "extra" eydye^y dy term. Evaluate this reasoning.

  1. The reasoning is correct; the eydye^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 eye^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 ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} for the complete functions MM and NN. Here M=3x2y2+2xy3M = 3x^2y^2 + 2xy^3 and N=2x3y+3x2y2+eyN = 2x^3y + 3x^2y^2 + e^y. We compute ∂M∂y=6x2y+6xy2\frac{\partial M}{\partial y} = 6x^2y + 6xy^2 and ∂N∂x=6x2y+6xy2\frac{\partial N}{\partial x} = 6x^2y + 6xy^2. Since these are equal, the equation is exact. The presence of the eye^y term doesn't "prevent" exactness—it's simply part of N(x,y)N(x,y). The student's approach of factoring and treating terms separately is not how exactness is determined.

Question 9

Consider the differential equation dydx=exsin⁡y−2xyx2−excos⁡y\frac{dy}{dx} = \frac{e^x \sin y - 2xy}{x^2 - e^x \cos y}. Which of the following statements is true?

  1. The equation is exact. (correct answer)
  2. The equation is separable.
  3. The equation is linear in yy.
  4. The equation becomes exact when multiplied by μ(x)=x\mu(x) = x.
Explanation: First, rewrite the equation in the standard form M(x,y)dx+N(x,y)dy=0M(x,y)dx + N(x,y)dy = 0. From dydx=exsin⁡y−2xyx2−excos⁡y\frac{dy}{dx} = \frac{e^x \sin y - 2xy}{x^2 - e^x \cos y}, we get (x2−excos⁡y)dy=(exsin⁡y−2xy)dx(x^2 - e^x \cos y)dy = (e^x \sin y - 2xy)dx. Rearranging gives (exsin⁡y−2xy)dx−(x2−excos⁡y)dy=0(e^x \sin y - 2xy)dx - (x^2 - e^x \cos y)dy = 0, which is (exsin⁡y−2xy)dx+(excos⁡y−x2)dy=0(e^x \sin y - 2xy)dx + (e^x \cos y - x^2)dy = 0. Let M(x,y)=exsin⁡y−2xyM(x,y) = e^x \sin y - 2xy and N(x,y)=excos⁡y−x2N(x,y) = e^x \cos y - x^2. To check for exactness, we compute the partial derivatives: ∂M∂y=excos⁡y−2x\frac{\partial M}{\partial y} = e^x \cos y - 2x and ∂N∂x=excos⁡y−2x\frac{\partial N}{\partial x} = e^x \cos y - 2x. Since ∂M∂y=∂N∂x\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 xx and only yy. It is not linear due to the sin⁡y\sin y and cos⁡y\cos y terms. Multiplying by an integrating factor is unnecessary as it's already exact.

Question 10

The equation (3xy+y2)dx+(x2+xy)dy=0(3xy + y^2) dx + (x^2 + xy) dy = 0 is not exact. For which function μ(x,y)\mu(x,y) does multiplication by μ\mu make the equation (μ(3xy+y2))dx+(μ(x2+xy))dy=0(\mu(3xy + y^2)) dx + (\mu(x^2 + xy)) dy = 0 exact?

  1. μ(x,y)=x\mu(x,y) = x (correct answer)
  2. μ(x,y)=y\mu(x,y) = y
  3. μ(x,y)=xy\mu(x,y) = xy
  4. μ(x,y)=x2\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 ∂Mnew∂y=∂Nnew∂x\frac{\partial M_{new}}{\partial y} = \frac{\partial N_{new}}{\partial x}. A) Let μ(x,y)=x\mu(x,y)=x. The new equation is (3x2y+xy2)dx+(x3+x2y)dy=0(3x^2y + xy^2)dx + (x^3 + x^2y)dy = 0. Let M1=3x2y+xy2M_1 = 3x^2y + xy^2 and N1=x3+x2yN_1 = x^3 + x^2y. Then ∂M1∂y=3x2+2xy\frac{\partial M_1}{\partial y} = 3x^2 + 2xy and ∂N1∂x=3x2+2xy\frac{\partial N_1}{\partial x} = 3x^2 + 2xy. Since they are equal, the equation becomes exact. This is the correct answer. B) Let μ(x,y)=y\mu(x,y)=y. M2=3xy2+y3M_2 = 3xy^2 + y^3, N2=x2y+xy2N_2 = x^2y + xy^2. ∂M2∂y=6xy+3y2\frac{\partial M_2}{\partial y} = 6xy + 3y^2, ∂N2∂x=2xy+y2\frac{\partial N_2}{\partial x} = 2xy + y^2. Not exact. C) Let μ(x,y)=xy\mu(x,y)=xy. M3=3x2y2+xy3M_3 = 3x^2y^2 + xy^3, N3=x3y+x2y2N_3 = x^3y + x^2y^2. ∂M3∂y=6x2y+3xy2\frac{\partial M_3}{\partial y} = 6x^2y + 3xy^2, ∂N3∂x=3x2y+2xy2\frac{\partial N_3}{\partial x} = 3x^2y + 2xy^2. Not exact. D) Let μ(x,y)=x2\mu(x,y)=x^2. M4=3x3y+x2y2M_4 = 3x^3y + x^2y^2, N4=x4+x3yN_4 = x^4 + x^3y. ∂M4∂y=3x3+2x2y\frac{\partial M_4}{\partial y} = 3x^3 + 2x^2y, ∂N4∂x=4x3+3x2y\frac{\partial N_4}{\partial x} = 4x^3 + 3x^2y. Not exact.

Question 11

For what real value of the constant kk is the differential equation (y3+kxy4−2x)dx+(3xy2+20x2y3)dy=0(y^3 + kxy^4 - 2x) dx + (3xy^2 + 20x^2y^3) dy = 0 exact?

  1. k=5k=5
  2. k=10k=10 (correct answer)
  3. k=20k=20
  4. k=40k=40
Explanation: Let M(x,y)=y3+kxy4−2xM(x,y) = y^3 + kxy^4 - 2x and N(x,y)=3xy2+20x2y3N(x,y) = 3xy^2 + 20x^2y^3. For the equation to be exact, the condition ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} must hold. First, compute the partial derivative of MM with respect to yy: ∂M∂y=3y2+4kxy3\frac{\partial M}{\partial y} = 3y^2 + 4kxy^3. Next, compute the partial derivative of NN with respect to xx: ∂N∂x=3y2+40xy3\frac{\partial N}{\partial x} = 3y^2 + 40xy^3. Setting these two expressions equal gives 3y2+4kxy3=3y2+40xy33y^2 + 4kxy^3 = 3y^2 + 40xy^3. This simplifies to 4kxy3=40xy34kxy^3 = 40xy^3, which implies 4k=404k = 40. Solving for kk yields k=10k = 10.

Question 12

Consider the differential equation (f(x)+2xy2)dx+(2x2y−cos⁡(y))dy=0(f(x) + 2xy^2) dx + (2x^2y - \cos(y)) dy = 0, where f(x)f(x) is any differentiable function. Which statement about the exactness of this equation is true?

  1. The equation is exact only if f(x)f(x) is a constant.
  2. The equation is exact only if f(x)=cx4f(x) = cx^4 for some constant cc.
  3. The equation is exact for any differentiable function f(x)f(x). (correct answer)
  4. The equation is never exact, regardless of the choice of f(x)f(x).
Explanation: Let M(x,y)=f(x)+2xy2M(x,y) = f(x) + 2xy^2 and N(x,y)=2x2y−cos⁡(y)N(x,y) = 2x^2y - \cos(y). We check the condition for exactness, ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. The partial derivative of MM with respect to yy is ∂M∂y=∂∂y(f(x)+2xy2)=0+4xy=4xy\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(f(x) + 2xy^2) = 0 + 4xy = 4xy. The partial derivative of NN with respect to xx is ∂N∂x=∂∂x(2x2y−cos⁡(y))=4xy−0=4xy\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(2x^2y - \cos(y)) = 4xy - 0 = 4xy. Since ∂M∂y=4xy\frac{\partial M}{\partial y} = 4xy and ∂N∂x=4xy\frac{\partial N}{\partial x} = 4xy, the condition for exactness is satisfied for any differentiable function f(x)f(x). The choice of f(x)f(x) does not affect the outcome of the test.

Question 13

Which of the following differential equations is NOT exact?

  1. (2x+y2)dx+2xydy=0(2x + y^2) dx + 2xy dy = 0
  2. (cos⁡(x+y)−xsin⁡(x+y))dx−xsin⁡(x+y)dy=0(\cos(x+y) - x\sin(x+y)) dx - x\sin(x+y) dy = 0
  3. (yexy+2x)dx+(xexy−2y)dy=0(y e^{xy} + 2x) dx + (x e^{xy} - 2y) dy = 0
  4. (x2+y2)dx+xydy=0(x^2 + y^2) dx + xy dy = 0 (correct answer)
Explanation: We test each equation for exactness using the condition ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. A) M=2x+y2,N=2xyM=2x+y^2, N=2xy. ∂M∂y=2y\frac{\partial M}{\partial y} = 2y. ∂N∂x=2y\frac{\partial N}{\partial x} = 2y. Exact. B) M=cos⁡(x+y)−xsin⁡(x+y),N=−xsin⁡(x+y)M=\cos(x+y) - x\sin(x+y), N=-x\sin(x+y). ∂M∂y=−sin⁡(x+y)−xcos⁡(x+y)\frac{\partial M}{\partial y} = -\sin(x+y) - x\cos(x+y). ∂N∂x=−(1⋅sin⁡(x+y)+xcos⁡(x+y))=−sin⁡(x+y)−xcos⁡(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=yexy+2x,N=xexy−2yM=y e^{xy} + 2x, N=x e^{xy} - 2y. ∂M∂y=exy+y(xexy)=exy(1+xy)\frac{\partial M}{\partial y} = e^{xy} + y(xe^{xy}) = e^{xy}(1+xy). ∂N∂x=exy+x(yexy)=exy(1+xy)\frac{\partial N}{\partial x} = e^{xy} + x(ye^{xy}) = e^{xy}(1+xy). Exact. D) M=x2+y2,N=xyM=x^2+y^2, N=xy. ∂M∂y=2y\frac{\partial M}{\partial y} = 2y. ∂N∂x=y\frac{\partial N}{\partial x} = y. Since 2y≠y2y \neq y, this equation is not exact.

Question 14

For the differential equation (tan⁡y+2xy3)dx+(xsec⁡2y+3x2y2+h(y))dy=0(\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)h(y)?

  1. h(y)h(y) can be any differentiable function of yy (correct answer)
  2. h(y)h(y) must be a constant function
  3. h(y)h(y) must satisfy h′(y)=sec⁡2yh'(y) = \sec^2 y
  4. No such function h(y)h(y) exists that makes the equation exact
Explanation: For exactness, we need ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} where M=tan⁡y+2xy3M = \tan y + 2xy^3 and N=xsec⁡2y+3x2y2+h(y)N = x\sec^2 y + 3x^2y^2 + h(y). Computing ∂M∂y=sec⁡2y+6xy2\frac{\partial M}{\partial y} = \sec^2 y + 6xy^2 and ∂N∂x=sec⁡2y+6xy2+0=sec⁡2y+6xy2\frac{\partial N}{\partial x} = \sec^2 y + 6xy^2 + 0 = \sec^2 y + 6xy^2 (since h(y)h(y) doesn't depend on xx). These are already equal regardless of the choice of h(y)h(y), so the equation is exact for any differentiable function h(y)h(y). The function h(y)h(y) doesn't affect the exactness condition since it contributes nothing to ∂N∂x\frac{\partial N}{\partial x}.