Differential Equations Quiz: Integrating Factors Non Exact
14 questions · exam conditions
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Integrating Factors Non ExactQuestion 1 of 14

For a non-exact differential equation M(x,y)dx+N(x,y)dy=0M(x,y)dx + N(x,y)dy = 0, an analyst seeks an integrating factor. Under which of the following conditions is the analyst guaranteed to find a simplified integrating factor μ(x)\mu(x) that depends only on xx?

When the expression 1N(∂M∂y−∂N∂x)\frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right) simplifies to a function of xx alone.
When the expression 1M(∂N∂x−∂M∂y)\frac{1}{M} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) simplifies to a function of yy alone.
When N(x,y)N(x,y) is independent of yy, i.e., N(x,y)=N(x)N(x,y)=N(x).
When M(x,y)M(x,y) and N(x,y)N(x,y) are both homogeneous functions of the same degree.
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Differential Equations Quiz

Differential Equations Quiz: Integrating Factors Non Exact

Practice Integrating Factors Non Exact 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 Integrating Factors Non Exact, 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

For a non-exact differential equation M(x,y)dx+N(x,y)dy=0M(x,y)dx + N(x,y)dy = 0, an analyst seeks an integrating factor. Under which of the following conditions is the analyst guaranteed to find a simplified integrating factor μ(x)\mu(x) that depends only on xx?

  1. When the expression 1N(∂M∂y−∂N∂x)\frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right) simplifies to a function of xx alone. (correct answer)
  2. When the expression 1M(∂N∂x−∂M∂y)\frac{1}{M} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) simplifies to a function of yy alone.
  3. When N(x,y)N(x,y) is independent of yy, i.e., N(x,y)=N(x)N(x,y)=N(x).
  4. When M(x,y)M(x,y) and N(x,y)N(x,y) are both homogeneous functions of the same degree.
Explanation: If a non-exact differential equation Mdx+Ndy=0Mdx+Ndy=0 has an integrating factor μ(x)\mu(x) that depends only on xx, then the equation μ(x)Mdx+μ(x)Ndy=0\mu(x)Mdx + \mu(x)Ndy=0 must be exact. The condition for exactness is ∂∂y(μM)=∂∂x(μN)\frac{\partial}{\partial y}(\mu M) = \frac{\partial}{\partial x}(\mu N). Applying the product rule gives μ∂M∂y=dμdxN+μ∂N∂x\mu \frac{\partial M}{\partial y} = \frac{d\mu}{dx}N + \mu \frac{\partial N}{\partial x}. Rearranging this gives dμdxN=μ(∂M∂y−∂N∂x)\frac{d\mu}{dx}N = \mu (\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}), which can be written as 1μdμdx=1N(∂M∂y−∂N∂x)\frac{1}{\mu}\frac{d\mu}{dx} = \frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}). For this to be solvable for μ(x)\mu(x), the left side, which is a function of xx alone, must equal the right side. Therefore, the right side must also be a function of xx alone. This is the condition stated in choice A. Choice B is the condition for an integrating factor μ(y)\mu(y). Choice C is not sufficient, as ∂M∂y\frac{\partial M}{\partial y} may still depend on yy. Choice D describes a homogeneous equation, which is solved by a different method.

Question 2

The differential equation (ay3+bxy)dx+(2x2+3xy2)dy=0(ay^3 + bxy)dx + (2x^2 + 3xy^2)dy = 0 is made exact by the integrating factor μ(x)=x\mu(x) = x. What is the value of a+ba+b?

  1. 5
  2. 2
  3. 8 (correct answer)
  4. 6
Explanation: Multiplying the given differential equation by the integrating factor μ(x)=x\mu(x) = x, we obtain a new equation: (axy3+bx2y)dx+(2x3+3x2y2)dy=0(axy^3 + bx^2y)dx + (2x^3 + 3x^2y^2)dy = 0. Let M∗(x,y)=axy3+bx2yM^*(x,y) = axy^3 + bx^2y and N∗(x,y)=2x3+3x2y2N^*(x,y) = 2x^3 + 3x^2y^2. For this new equation to be exact, we must have ∂M∗∂y=∂N∗∂x\frac{\partial M^*}{\partial y} = \frac{\partial N^*}{\partial x}. Calculating the partial derivatives: ∂M∗∂y=3axy2+bx2\frac{\partial M^*}{\partial y} = 3axy^2 + bx^2 ∂N∗∂x=6x2+6xy2\frac{\partial N^*}{\partial x} = 6x^2 + 6xy^2 Setting them equal: 3axy2+bx2=6x2+6xy23axy^2 + bx^2 = 6x^2 + 6xy^2. For this equality to hold for all xx and yy, the coefficients of like terms must be equal. Equating coefficients of xy2xy^2: 3a=6  ⟹  a=23a = 6 \implies a = 2. Equating coefficients of x2x^2: b=6b = 6. The question asks for the value of a+ba+b, which is 2+6=82+6=8.

Question 3

The equation y′−2xy=x2cos⁡(x)y' - \frac{2}{x}y = x^2 \cos(x) is a first-order linear equation. It can also be written in the form M(x,y)dx+N(x,y)dy=0M(x,y)dx + N(x,y)dy = 0 and made exact using an integrating factor. What is the integrating factor μ(x)\mu(x) found by treating it as a non-exact equation?

  1. x2x^2
  2. x−2x^{-2} (correct answer)
  3. cos⁡(x)\cos(x)
  4. e−2xe^{-2x}
Explanation: First, we rewrite the linear equation in differential form Mdx+Ndy=0Mdx+Ndy=0. dydx=2xy+x2cos⁡(x)\frac{dy}{dx} = \frac{2}{x}y + x^2\cos(x) dy=(2yx+x2cos⁡(x))dxdy = (\frac{2y}{x} + x^2\cos(x))dx (2yx+x2cos⁡(x))dx−dy=0(\frac{2y}{x} + x^2\cos(x))dx - dy = 0. Here, M(x,y)=2yx+x2cos⁡(x)M(x,y) = \frac{2y}{x} + x^2\cos(x) and N(x,y)=−1N(x,y)=-1. We compute the partial derivatives: ∂M∂y=2x\frac{\partial M}{\partial y} = \frac{2}{x} and ∂N∂x=0\frac{\partial N}{\partial x} = 0. The equation is not exact. We check for an integrating factor of xx: 1N(∂M∂y−∂N∂x)=1−1(2x−0)=−2x\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{1}{-1}(\frac{2}{x} - 0) = -\frac{2}{x}. This is a function of xx alone. The integrating factor is μ(x)=e∫−2xdx=e−2ln⁡∣x∣=eln⁡(x−2)=x−2\mu(x) = e^{\int -\frac{2}{x} dx} = e^{-2\ln|x|} = e^{\ln(x^{-2})} = x^{-2}. This matches the integrating factor found using the standard formula for first-order linear equations, e∫P(x)dxe^{\int P(x)dx}, where P(x)=−2/xP(x)=-2/x.

Question 4

A student attempts to solve (2x+y2)dx+2xydy=0(2x + y^2)dx + 2xy dy = 0 by finding an integrating factor. They compute ∂M∂y−∂N∂xM=2y−2y2x+y2=0\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{M} = \frac{2y - 2y}{2x + y^2} = 0 and conclude no integrating factor is needed. What error did the student make?

  1. They computed ∂M∂y\frac{\partial M}{\partial y} incorrectly as 2y2y instead of 22
  2. They used the wrong formula; they should compute ∂N∂x−∂M∂yN\frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{N}
  3. They computed ∂N∂x\frac{\partial N}{\partial x} correctly but their conclusion is wrong since the equation is exact (correct answer)
  4. They should have computed ∂M∂y−∂N∂xN\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} instead of dividing by MM
Explanation: Let's check the student's work: M=2x+y2M = 2x + y^2, N=2xyN = 2xy. We have ∂M∂y=2y\frac{\partial M}{\partial y} = 2y and ∂N∂x=2y\frac{\partial N}{\partial x} = 2y. The student computed these correctly. Since ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}, the equation is exact and no integrating factor is needed. The student's conclusion that no integrating factor is needed is actually correct, but their reasoning about the formula is confused. The correct interpretation is that when ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}, the equation is already exact. The student happened to get the right answer but for partially wrong reasons - they were checking for an integrating factor when they should have recognized exactness.

Question 5

Consider the non-exact equation M(x,y)dx+N(x,y)dy=0M(x,y)dx + N(x,y)dy = 0 where ∂M∂y−∂N∂xxN−yM=2x2+y2\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{xN - yM} = \frac{2}{x^2 + y^2}. What type of integrating factor should be sought?

  1. An integrating factor of the form μ(x)\mu(x) only
  2. An integrating factor of the form μ(y)\mu(y) only
  3. An integrating factor of the form μ(xy)\mu(xy)
  4. An integrating factor of the form μ(x2+y2)\mu(x^2 + y^2) (correct answer)
Explanation: The given expression ∂M∂y−∂N∂xxN−yM=2x2+y2\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{xN - yM} = \frac{2}{x^2 + y^2} suggests looking for an integrating factor that depends on x2+y2x^2 + y^2. This is because the denominator xN−yMxN - yM appears in the standard formula for integrating factors of the form μ(x2+y2)\mu(x^2 + y^2). When the expression ∂M∂y−∂N∂xxN−yM\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{xN - yM} depends only on x2+y2x^2 + y^2, then an integrating factor μ(x2+y2)\mu(x^2 + y^2) exists where dln⁡μd(x2+y2)=12⋅2x2+y2=1x2+y2\frac{d\ln\mu}{d(x^2 + y^2)} = \frac{1}{2} \cdot \frac{2}{x^2 + y^2} = \frac{1}{x^2 + y^2}. The other options (μ(x)\mu(x), μ(y)\mu(y), μ(xy)\mu(xy)) would require different expressions in their respective formulas.

Question 6

The differential equation (y4+2y)dx+(xy3+2y4−4x)dy=0(y^4 + 2y) dx + (xy^3 + 2y^4 - 4x) dy = 0 is not exact. An integrating factor of the form μ(y)=yk\mu(y) = y^k makes the equation exact. What is the value of kk?

  1. k=3k = 3
  2. k=−3k = -3 (correct answer)
  3. k=−2k = -2
  4. k=−4k = -4
Explanation: Let M(x,y)=y4+2yM(x,y) = y^4 + 2y and N(x,y)=xy3+2y4−4xN(x,y) = xy^3 + 2y^4 - 4x. We calculate the partial derivatives: ∂M∂y=4y3+2\frac{\partial M}{\partial y} = 4y^3 + 2 and ∂N∂x=y3−4\frac{\partial N}{\partial x} = y^3 - 4. Since ∂M∂y≠∂N∂x\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}, the equation is not exact. For an integrating factor that is a function of yy alone, μ(y)\mu(y), the expression 1M(∂N∂x−∂M∂y)\frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) must be a function of yy only. Let's compute this: (y3−4)−(4y3+2)y4+2y=−3y3−6y(y3+2)=−3(y3+2)y(y3+2)=−3y\frac{(y^3 - 4) - (4y^3 + 2)}{y^4 + 2y} = \frac{-3y^3 - 6}{y(y^3 + 2)} = \frac{-3(y^3 + 2)}{y(y^3 + 2)} = -\frac{3}{y}. This is a function of yy only. The integrating factor is μ(y)=e∫−3ydy=e−3ln⁡∣y∣=y−3\mu(y) = e^{\int -\frac{3}{y} dy} = e^{-3\ln|y|} = y^{-3}. Comparing this to the form μ(y)=yk\mu(y)=y^k, we find that k=−3k=-3.

Question 7

Consider the differential equation dydx=x2y−y32x3−4xy2\frac{dy}{dx} = \frac{x^2y - y^3}{2x^3 - 4xy^2}. An integrating factor for this equation is of the form μ(y)=yk\mu(y) = y^k. Determine the value of kk.

  1. k=7k = 7
  2. k=−6k = -6
  3. k=−7k = -7 (correct answer)
  4. k=6k = 6
Explanation: First, we rewrite the equation in the form M(x,y)dx+N(x,y)dy=0M(x,y)dx + N(x,y)dy = 0. (2x3−4xy2)dy=(x2y−y3)dx(2x^3 - 4xy^2)dy = (x^2y - y^3)dx (y3−x2y)dx+(2x3−4xy2)dy=0(y^3 - x^2y)dx + (2x^3 - 4xy^2)dy = 0. So, M=y3−x2yM = y^3 - x^2y and N=2x3−4xy2N = 2x^3 - 4xy^2. We find the partial derivatives: ∂M∂y=3y2−x2\frac{\partial M}{\partial y} = 3y^2 - x^2 ∂N∂x=6x2−4y2\frac{\partial N}{\partial x} = 6x^2 - 4y^2 The equation is not exact. Since the integrating factor depends on yy, we compute the expression 1M(∂N∂x−∂M∂y)\frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}): (6x2−4y2)−(3y2−x2)y3−x2y=7x2−7y2y(y2−x2)=−7(y2−x2)y(y2−x2)=−7y\frac{(6x^2 - 4y^2) - (3y^2 - x^2)}{y^3 - x^2y} = \frac{7x^2 - 7y^2}{y(y^2 - x^2)} = \frac{-7(y^2 - x^2)}{y(y^2 - x^2)} = -\frac{7}{y}. This is a function of yy only. The integrating factor is μ(y)=e∫−7ydy=e−7ln⁡∣y∣=y−7\mu(y) = e^{\int -\frac{7}{y} dy} = e^{-7\ln|y|} = y^{-7}. Comparing this to μ(y)=yk\mu(y) = y^k, we have k=−7k=-7.

Question 8

For which of the following non-exact differential equations can an integrating factor μ(x)\mu(x), which depends only on xx, be found?

  1. (y)dx+(2xy−e−2y)dy=0(y)dx + (2xy-e^{-2y})dy=0
  2. (2x+y2)dx+(2xy)dy=0(2x+y^2)dx + (2xy)dy = 0
  3. (xy)dx+(y2+1)dy=0(xy)dx + (y^2+1)dy=0
  4. (y2+x)dx+xy dy=0(y^2+x)dx + xy \, dy = 0 (correct answer)
Explanation: An integrating factor μ(x)\mu(x) exists if the expression P(x)=1N(∂M∂y−∂N∂x)P(x) = \frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) is a function of xx alone. We test each option: (A) M=y,N=2xy−e−2yM=y, N=2xy-e^{-2y}. My=1,Nx=2yM_y=1, N_x=2y. P(x)=1−2y2xy−e−2yP(x) = \frac{1-2y}{2xy-e^{-2y}}, not a function of xx alone. (B) M=2x+y2,N=2xyM=2x+y^2, N=2xy. My=2y,Nx=2yM_y=2y, N_x=2y. My=NxM_y=N_x, so the equation is already exact. (C) M=xy,N=y2+1M=xy, N=y^2+1. My=x,Nx=0M_y=x, N_x=0. P(x)=x−0y2+1=xy2+1P(x) = \frac{x-0}{y^2+1} = \frac{x}{y^2+1}, not a function of xx alone. (D) M=y2+x,N=xyM=y^2+x, N=xy. My=2y,Nx=yM_y=2y, N_x=y. P(x)=2y−yxy=yxy=1xP(x) = \frac{2y-y}{xy} = \frac{y}{xy} = \frac{1}{x}. This is a function of xx alone, so an integrating factor μ(x)\mu(x) can be found.

Question 9

The differential equation (ytan⁡x)dx+dy=0(y \tan x) dx + dy = 0 is made exact by an integrating factor of the form μ(x)=sec⁡kx\mu(x) = \sec^k x. What is the value of kk?

  1. k=−1k = -1
  2. k=1k = 1 (correct answer)
  3. k=2k = 2
  4. k=−2k = -2
Explanation: Given the equation (ytan⁡x)dx+1dy=0(y \tan x) dx + 1 dy = 0, we have M(x,y)=ytan⁡xM(x,y) = y \tan x and N(x,y)=1N(x,y) = 1. We calculate the partial derivatives: ∂M∂y=tan⁡x\frac{\partial M}{\partial y} = \tan x and ∂N∂x=0\frac{\partial N}{\partial x} = 0. The equation is not exact. To find an integrating factor that is a function of xx only, we compute the expression 1N(∂M∂y−∂N∂x)=tan⁡x−01=tan⁡x\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{\tan x - 0}{1} = \tan x. Since this is a function of xx alone, the integrating factor is μ(x)=e∫tan⁡xdx\mu(x) = e^{\int \tan x dx}. The integral of tan⁡x\tan x is ln⁡∣sec⁡x∣\ln|\sec x|. Thus, μ(x)=eln⁡∣sec⁡x∣=∣sec⁡x∣\mu(x) = e^{\ln|\sec x|} = |\sec x|. We can choose the positive value, so μ(x)=sec⁡x\mu(x) = \sec x. Comparing this with the given form μ(x)=sec⁡kx\mu(x) = \sec^k x, we see that k=1k=1.

Question 10

If multiplying the non-exact differential equation M(x,y)dx+N(x,y)dy=0M(x,y)dx + N(x,y)dy=0 by a non-zero function μ(x,y)\mu(x,y) results in an exact equation, which of the following statements must be true?

  1. The function μ(x,y)\mu(x,y) is a particular solution to the original differential equation.
  2. The solutions to the new exact equation are also the solutions to the original equation. (correct answer)
  3. The original equation must have been a first-order linear differential equation.
  4. The function μ(x,y)\mu(x,y) must be a function of either xx alone or yy alone.
Explanation: The purpose of an integrating factor μ(x,y)\mu(x,y) is to transform a non-exact equation into an exact one without altering the solution set. If y(x)y(x) is a solution to the original equation, it satisfies M+Ny′=0M+Ny'=0. Multiplying by μ\mu gives μM+μNy′=0\mu M + \mu N y' = 0, so y(x)y(x) is also a solution to the new equation. Conversely, since μ\mu is assumed to be non-zero, any solution to the new equation also solves the original. Therefore, the solutions are the same. A is incorrect; μ\mu is a factor, not a solution curve y(x)y(x). C is incorrect; this method applies to a wide class of non-linear equations. D is incorrect; while we often seek integrating factors that depend on a single variable for simplicity, more general integrating factors depending on both xx and yy exist.

Question 11

What is the general solution of the differential equation (x2+y2+x)dx+(xy)dy=0(x^2+y^2+x)dx + (xy)dy=0?

  1. 3x4+6x2y2+4x3=C3x^4 + 6x^2y^2 + 4x^3 = C (correct answer)
  2. 3x4+12x2y2+4x3=C3x^4 + 12x^2y^2 + 4x^3 = C
  3. x4+2x2y2+x3=Cx^4 + 2x^2y^2 + x^3 = C
  4. 4x4+6x2y2+3x3=C4x^4 + 6x^2y^2 + 3x^3 = C
Explanation: Let M=x2+y2+xM=x^2+y^2+x and N=xyN=xy. We find ∂M∂y=2y\frac{\partial M}{\partial y} = 2y and ∂N∂x=y\frac{\partial N}{\partial x} = y. The equation is not exact. We look for an integrating factor: 1N(∂M∂y−∂N∂x)=2y−yxy=1x\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{2y-y}{xy} = \frac{1}{x}. So, the integrating factor is μ(x)=e∫1xdx=x\mu(x) = e^{\int \frac{1}{x} dx} = x. Multiplying the DE by xx yields (x3+xy2+x2)dx+(x2y)dy=0(x^3+xy^2+x^2)dx + (x^2y)dy = 0. This new equation is exact. Let M∗=x3+xy2+x2M^*=x^3+xy^2+x^2 and N∗=x2yN^*=x^2y. We find the solution f(x,y)=Cf(x,y)=C by integrating N∗N^* with respect to yy: f(x,y)=∫x2y dy=12x2y2+h(x)f(x,y) = \int x^2y \, dy = \frac{1}{2}x^2y^2 + h(x). Differentiating with respect to xx: ∂f∂x=xy2+h′(x)\frac{\partial f}{\partial x} = xy^2 + h'(x). We set this equal to M∗M^*: xy2+h′(x)=x3+xy2+x2xy^2 + h'(x) = x^3+xy^2+x^2. This implies h′(x)=x3+x2h'(x) = x^3+x^2. Integrating gives h(x)=14x4+13x3h(x) = \frac{1}{4}x^4 + \frac{1}{3}x^3. The general solution is 12x2y2+14x4+13x3=C1\frac{1}{2}x^2y^2 + \frac{1}{4}x^4 + \frac{1}{3}x^3 = C_1. To clear the fractions, we can multiply by 12, giving 6x2y2+3x4+4x3=C6x^2y^2 + 3x^4 + 4x^3 = C.

Question 12

The differential equation 2sin⁡(y2)dx+xycos⁡(y2)dy=02\sin(y^2)dx + xy\cos(y^2)dy = 0 is made exact by an integrating factor μ(x)=xk\mu(x) = x^k. What is the value of kk?

  1. k=1k = 1
  2. k=4k = 4
  3. k=−3k = -3
  4. k=3k = 3 (correct answer)
Explanation: Let M(x,y)=2sin⁡(y2)M(x,y) = 2\sin(y^2) and N(x,y)=xycos⁡(y2)N(x,y) = xy\cos(y^2). We calculate the partial derivatives using the chain rule: ∂M∂y=2cos⁡(y2)⋅(2y)=4ycos⁡(y2)\frac{\partial M}{\partial y} = 2\cos(y^2) \cdot (2y) = 4y\cos(y^2). ∂N∂x=ycos⁡(y2)\frac{\partial N}{\partial x} = y\cos(y^2). The equation is not exact. Since the integrating factor depends on xx, we compute the expression 1N(∂M∂y−∂N∂x)\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}): 4ycos⁡(y2)−ycos⁡(y2)xycos⁡(y2)=3ycos⁡(y2)xycos⁡(y2)=3x\frac{4y\cos(y^2) - y\cos(y^2)}{xy\cos(y^2)} = \frac{3y\cos(y^2)}{xy\cos(y^2)} = \frac{3}{x}. Since this is a function of xx alone, the integrating factor is μ(x)=e∫3xdx=e3ln⁡∣x∣=∣x∣3\mu(x) = e^{\int \frac{3}{x} dx} = e^{3\ln|x|} = |x|^3. We can take μ(x)=x3\mu(x) = x^3. Comparing this to the form μ(x)=xk\mu(x) = x^k, we have k=3k=3.

Question 13

Consider the initial value problem given by (3xy+y2)+(x2+xy)y′=0(3xy + y^2) + (x^2 + xy)y' = 0, with the initial condition y(1)=−4y(1) = -4. What is the value of y(2)y(2)?

  1. −2+6-2 + \sqrt{6}
  2. −3-3
  3. −2−2-2 - \sqrt{2}
  4. −2−6-2 - \sqrt{6} (correct answer)
Explanation: First, write the equation in differential form: (3xy+y2)dx+(x2+xy)dy=0(3xy + y^2)dx + (x^2 + xy)dy = 0. Let M=3xy+y2M = 3xy + y^2 and N=x2+xyN = x^2 + xy. Then ∂M∂y=3x+2y\frac{\partial M}{\partial y} = 3x + 2y and ∂N∂x=2x+y\frac{\partial N}{\partial x} = 2x + y. The equation is not exact. We check for an integrating factor μ(x)\mu(x): 1N(∂M∂y−∂N∂x)=(3x+2y)−(2x+y)x2+xy=x+yx(x+y)=1x\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{(3x+2y)-(2x+y)}{x^2+xy} = \frac{x+y}{x(x+y)} = \frac{1}{x}. The integrating factor is μ(x)=e∫1xdx=x\mu(x) = e^{\int \frac{1}{x} dx} = x. Multiplying the DE by xx gives (3x2y+xy2)dx+(x3+x2y)dy=0(3x^2y + xy^2)dx + (x^3 + x^2y)dy = 0. This equation is exact. The solution f(x,y)=Cf(x,y)=C is found by integrating ∂f∂y=x3+x2y\frac{\partial f}{\partial y} = x^3 + x^2y, which gives f(x,y)=x3y+12x2y2+h(x)f(x,y) = x^3y + \frac{1}{2}x^2y^2 + h(x). Differentiating with respect to xx gives ∂f∂x=3x2y+xy2+h′(x)\frac{\partial f}{\partial x} = 3x^2y + xy^2 + h'(x). Setting this equal to the new MM, we get h′(x)=0h'(x)=0. So, the general solution is x3y+12x2y2=Cx^3y + \frac{1}{2}x^2y^2 = C. Using y(1)=−4y(1)=-4: (1)3(−4)+12(1)2(−4)2=−4+8=4(1)^3(-4) + \frac{1}{2}(1)^2(-4)^2 = -4 + 8 = 4. So C=4C=4. The particular solution is x3y+12x2y2=4x^3y + \frac{1}{2}x^2y^2 = 4. For x=2x=2, we have 8y+2y2=48y + 2y^2 = 4, or y2+4y−2=0y^2+4y-2=0. The quadratic formula gives y=−4±16−4(−2)2=−2±6y = \frac{-4 \pm \sqrt{16-4(-2)}}{2} = -2 \pm \sqrt{6}. Since the initial condition is y(1)=−4y(1)=-4, we need the branch of the solution that is near −4-4. The value −2−6≈−4.45-2-\sqrt{6} \approx -4.45 is on this branch, while −2+6≈0.45-2+\sqrt{6} \approx 0.45 is not.

Question 14

Consider the equation (2xy2+y)dx+(2x2y+3x)dy=0(2xy^2 + y)dx + (2x^2y + 3x)dy = 0. After determining it's not exact, you find that multiplying by 1xy\frac{1}{xy} makes it exact. What is the resulting exact equation?

  1. (2y+1x)dx+(2x+3y)dy=0(2y + \frac{1}{x})dx + (2x + \frac{3}{y})dy = 0 (correct answer)
  2. (2y+1x)dx+(2x+3)dy=0(2y + \frac{1}{x})dx + (2x + 3)dy = 0
  3. (2xy+1)dx+(2xy+3)dy=0(2xy + 1)dx + (2xy + 3)dy = 0
  4. 2y+1xdx+2x+3ydy=0\frac{2y + 1}{x}dx + \frac{2x + 3}{y}dy = 0
Explanation: First verify the original equation is not exact: M=2xy2+yM = 2xy^2 + y, N=2x2y+3xN = 2x^2y + 3x. We have ∂M∂y=4xy+1\frac{\partial M}{\partial y} = 4xy + 1 and ∂N∂x=4xy+3\frac{\partial N}{\partial x} = 4xy + 3. Since 4xy+1≠4xy+34xy + 1 \neq 4xy + 3, it's not exact. Multiplying by μ=1xy\mu = \frac{1}{xy}: 2xy2+yxydx+2x2y+3xxydy=0\frac{2xy^2 + y}{xy}dx + \frac{2x^2y + 3x}{xy}dy = 0, which simplifies to y(2xy+1)xydx+x(2xy+3)xydy=0\frac{y(2xy + 1)}{xy}dx + \frac{x(2xy + 3)}{xy}dy = 0, giving (2xy+1x)dx+(2xy+3y)dy=0(\frac{2xy + 1}{x})dx + (\frac{2xy + 3}{y})dy = 0, or (2y+1x)dx+(2x+3y)dy=0(2y + \frac{1}{x})dx + (2x + \frac{3}{y})dy = 0. Check exactness: ∂∂y(2y+1x)=2\frac{\partial}{\partial y}(2y + \frac{1}{x}) = 2 and ∂∂x(2x+3y)=2\frac{\partial}{\partial x}(2x + \frac{3}{y}) = 2. Now it's exact.