Differential Equations Quiz: Integrating Factors Non Exact

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.

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. $k = 3$
2. $k = -3$ (correct answer)
3. $k = -2$
4. $k = -4$

Explanation: Let $M(x,y) = y^4 + 2y$ and $N(x,y) = xy^3 + 2y^4 - 4x$. We calculate the partial derivatives: $\frac{\partial M}{\partial y} = 4y^3 + 2$ and $\frac{\partial N}{\partial x} = y^3 - 4$. Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact. For an integrating factor that is a function of $y$ alone, $\mu(y)$, the expression $\frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)$ must be a function of $y$ only. Let's compute this: $\frac{(y^3 - 4) - (4y^3 + 2)}{y^4 + 2y} = \frac{-3y^3 - 6}{y(y^3 + 2)} = \frac{-3(y^3 + 2)}{y(y^3 + 2)} = -\frac{3}{y}$. This is a function of $y$ only. The integrating factor is $\mu(y) = e^{\int -\frac{3}{y} dy} = e^{-3\ln|y|} = y^{-3}$. Comparing this to the form $\mu(y)=y^k$, we find that $k=-3$.

Question 2

1. When the expression $\frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right)$ simplifies to a function of $x$ alone. (correct answer)
2. When the expression $\frac{1}{M} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right)$ simplifies to a function of $y$ alone.
3. When $N(x,y)$ is independent of $y$, i.e., $N(x,y)=N(x)$.
4. When $M(x,y)$ and $N(x,y)$ are both homogeneous functions of the same degree.

Explanation: If a non-exact differential equation $Mdx+Ndy=0$ has an integrating factor $\mu(x)$ that depends only on $x$, then the equation $\mu(x)Mdx + \mu(x)Ndy=0$ must be exact. The condition for exactness is $\frac{\partial}{\partial y}(\mu M) = \frac{\partial}{\partial x}(\mu N)$. Applying the product rule gives $\mu \frac{\partial M}{\partial y} = \frac{d\mu}{dx}N + \mu \frac{\partial N}{\partial x}$. Rearranging this gives $\frac{d\mu}{dx}N = \mu (\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$, which can be written as $\frac{1}{\mu}\frac{d\mu}{dx} = \frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$. For this to be solvable for $\mu(x)$, the left side, which is a function of $x$ alone, must equal the right side. Therefore, the right side must also be a function of $x$ alone. This is the condition stated in choice A. Choice B is the condition for an integrating factor $\mu(y)$. Choice C is not sufficient, as $\frac{\partial M}{\partial y}$ may still depend on $y$. Choice D describes a homogeneous equation, which is solved by a different method.

Question 3

1. 5
2. 2
3. 8 (correct answer)
4. 6

Explanation: Multiplying the given differential equation by the integrating factor $\mu(x) = x$, we obtain a new equation: $(axy^3 + bx^2y)dx + (2x^3 + 3x^2y^2)dy = 0$. Let $M^*(x,y) = axy^3 + bx^2y$ and $N^*(x,y) = 2x^3 + 3x^2y^2$. For this new equation to be exact, we must have $\frac{\partial M^*}{\partial y} = \frac{\partial N^*}{\partial x}$. Calculating the partial derivatives: $\frac{\partial M^*}{\partial y} = 3axy^2 + bx^2$ $\frac{\partial N^*}{\partial x} = 6x^2 + 6xy^2$ Setting them equal: $3axy^2 + bx^2 = 6x^2 + 6xy^2$. For this equality to hold for all $x$ and $y$, the coefficients of like terms must be equal. Equating coefficients of $xy^2$: $3a = 6 \implies a = 2$. Equating coefficients of $x^2$: $b = 6$. The question asks for the value of $a+b$, which is $2+6=8$.

Question 4

1. $k = 7$
2. $k = -6$
3. $k = -7$ (correct answer)
4. $k = 6$

Explanation: First, we rewrite the equation in the form $M(x,y)dx + N(x,y)dy = 0$. $(2x^3 - 4xy^2)dy = (x^2y - y^3)dx$ $(y^3 - x^2y)dx + (2x^3 - 4xy^2)dy = 0$. So, $M = y^3 - x^2y$ and $N = 2x^3 - 4xy^2$. We find the partial derivatives: $\frac{\partial M}{\partial y} = 3y^2 - x^2$ $\frac{\partial N}{\partial x} = 6x^2 - 4y^2$ The equation is not exact. Since the integrating factor depends on $y$, we compute the expression $\frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})$: $\frac{(6x^2 - 4y^2) - (3y^2 - x^2)}{y^3 - x^2y} = \frac{7x^2 - 7y^2}{y(y^2 - x^2)} = \frac{-7(y^2 - x^2)}{y(y^2 - x^2)} = -\frac{7}{y}$. This is a function of $y$ only. The integrating factor is $\mu(y) = e^{\int -\frac{7}{y} dy} = e^{-7\ln|y|} = y^{-7}$. Comparing this to $\mu(y) = y^k$, we have $k=-7$.

Question 5

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

Explanation: An integrating factor $\mu(x)$ exists if the expression $P(x) = \frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$ is a function of $x$ alone. We test each option: (A) $M=y, N=2xy-e^{-2y}$. $M_y=1, N_x=2y$. $P(x) = \frac{1-2y}{2xy-e^{-2y}}$, not a function of $x$ alone. (B) $M=2x+y^2, N=2xy$. $M_y=2y, N_x=2y$. $M_y=N_x$, so the equation is already exact. (C) $M=xy, N=y^2+1$. $M_y=x, N_x=0$. $P(x) = \frac{x-0}{y^2+1} = \frac{x}{y^2+1}$, not a function of $x$ alone. (D) $M=y^2+x, N=xy$. $M_y=2y, N_x=y$. $P(x) = \frac{2y-y}{xy} = \frac{y}{xy} = \frac{1}{x}$. This is a function of $x$ alone, so an integrating factor $\mu(x)$ can be found.

Question 6

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

Explanation: Given the equation $(y \tan x) dx + 1 dy = 0$, we have $M(x,y) = y \tan x$ and $N(x,y) = 1$. We calculate the partial derivatives: $\frac{\partial M}{\partial y} = \tan x$ and $\frac{\partial N}{\partial x} = 0$. The equation is not exact. To find an integrating factor that is a function of $x$ only, we compute the expression $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{\tan x - 0}{1} = \tan x$. Since this is a function of $x$ alone, the integrating factor is $\mu(x) = e^{\int \tan x dx}$. The integral of $\tan x$ is $\ln|\sec x|$. Thus, $\mu(x) = e^{\ln|\sec x|} = |\sec x|$. We can choose the positive value, so $\mu(x) = \sec x$. Comparing this with the given form $\mu(x) = \sec^k x$, we see that $k=1$.

Question 7

1. The function $\mu(x,y)$ 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 $\mu(x,y)$ must be a function of either $x$ alone or $y$ alone.

Explanation: The purpose of an integrating factor $\mu(x,y)$ is to transform a non-exact equation into an exact one without altering the solution set. If $y(x)$ is a solution to the original equation, it satisfies $M+Ny'=0$. Multiplying by $\mu$ gives $\mu M + \mu N y' = 0$, so $y(x)$ is also a solution to the new equation. Conversely, since $\mu$ 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; $\mu$ is a factor, not a solution curve $y(x)$. 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 $x$ and $y$ exist.

Question 8

1. $3x^4 + 6x^2y^2 + 4x^3 = C$ (correct answer)
2. $3x^4 + 12x^2y^2 + 4x^3 = C$
3. $x^4 + 2x^2y^2 + x^3 = C$
4. $4x^4 + 6x^2y^2 + 3x^3 = C$

Explanation: Let $M=x^2+y^2+x$ and $N=xy$. We find $\frac{\partial M}{\partial y} = 2y$ and $\frac{\partial N}{\partial x} = y$. The equation is not exact. We look for an integrating factor: $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{2y-y}{xy} = \frac{1}{x}$. So, the integrating factor is $\mu(x) = e^{\int \frac{1}{x} dx} = x$. Multiplying the DE by $x$ yields $(x^3+xy^2+x^2)dx + (x^2y)dy = 0$. This new equation is exact. Let $M^*=x^3+xy^2+x^2$ and $N^*=x^2y$. We find the solution $f(x,y)=C$ by integrating $N^*$ with respect to $y$: $f(x,y) = \int x^2y \, dy = \frac{1}{2}x^2y^2 + h(x)$. Differentiating with respect to $x$: $\frac{\partial f}{\partial x} = xy^2 + h'(x)$. We set this equal to $M^*$: $xy^2 + h'(x) = x^3+xy^2+x^2$. This implies $h'(x) = x^3+x^2$. Integrating gives $h(x) = \frac{1}{4}x^4 + \frac{1}{3}x^3$. The general solution is $\frac{1}{2}x^2y^2 + \frac{1}{4}x^4 + \frac{1}{3}x^3 = C_1$. To clear the fractions, we can multiply by 12, giving $6x^2y^2 + 3x^4 + 4x^3 = C$.

Question 9

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

Explanation: Let $M(x,y) = 2\sin(y^2)$ and $N(x,y) = xy\cos(y^2)$. We calculate the partial derivatives using the chain rule: $\frac{\partial M}{\partial y} = 2\cos(y^2) \cdot (2y) = 4y\cos(y^2)$. $\frac{\partial N}{\partial x} = y\cos(y^2)$. The equation is not exact. Since the integrating factor depends on $x$, we compute the expression $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$: $\frac{4y\cos(y^2) - y\cos(y^2)}{xy\cos(y^2)} = \frac{3y\cos(y^2)}{xy\cos(y^2)} = \frac{3}{x}$. Since this is a function of $x$ alone, the integrating factor is $\mu(x) = e^{\int \frac{3}{x} dx} = e^{3\ln|x|} = |x|^3$. We can take $\mu(x) = x^3$. Comparing this to the form $\mu(x) = x^k$, we have $k=3$.

Question 10

1. $x^2$
2. $x^{-2}$ (correct answer)
3. $\cos(x)$
4. $e^{-2x}$

Explanation: First, we rewrite the linear equation in differential form $Mdx+Ndy=0$. $\frac{dy}{dx} = \frac{2}{x}y + x^2\cos(x)$ $dy = (\frac{2y}{x} + x^2\cos(x))dx$ $(\frac{2y}{x} + x^2\cos(x))dx - dy = 0$. Here, $M(x,y) = \frac{2y}{x} + x^2\cos(x)$ and $N(x,y)=-1$. We compute the partial derivatives: $\frac{\partial M}{\partial y} = \frac{2}{x}$ and $\frac{\partial N}{\partial x} = 0$. The equation is not exact. We check for an integrating factor of $x$: $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{1}{-1}(\frac{2}{x} - 0) = -\frac{2}{x}$. This is a function of $x$ alone. The integrating factor is $\mu(x) = e^{\int -\frac{2}{x} dx} = e^{-2\ln|x|} = e^{\ln(x^{-2})} = x^{-2}$. This matches the integrating factor found using the standard formula for first-order linear equations, $e^{\int P(x)dx}$, where $P(x)=-2/x$.

Question 11

1. $-2 + \sqrt{6}$
2. $-3$
3. $-2 - \sqrt{2}$
4. $-2 - \sqrt{6}$ (correct answer)

Explanation: First, write the equation in differential form: $(3xy + y^2)dx + (x^2 + xy)dy = 0$. Let $M = 3xy + y^2$ and $N = x^2 + xy$. Then $\frac{\partial M}{\partial y} = 3x + 2y$ and $\frac{\partial N}{\partial x} = 2x + y$. The equation is not exact. We check for an integrating factor $\mu(x)$: $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{(3x+2y)-(2x+y)}{x^2+xy} = \frac{x+y}{x(x+y)} = \frac{1}{x}$. The integrating factor is $\mu(x) = e^{\int \frac{1}{x} dx} = x$. Multiplying the DE by $x$ gives $(3x^2y + xy^2)dx + (x^3 + x^2y)dy = 0$. This equation is exact. The solution $f(x,y)=C$ is found by integrating $\frac{\partial f}{\partial y} = x^3 + x^2y$, which gives $f(x,y) = x^3y + \frac{1}{2}x^2y^2 + h(x)$. Differentiating with respect to $x$ gives $\frac{\partial f}{\partial x} = 3x^2y + xy^2 + h'(x)$. Setting this equal to the new $M$, we get $h'(x)=0$. So, the general solution is $x^3y + \frac{1}{2}x^2y^2 = C$. Using $y(1)=-4$: $(1)^3(-4) + \frac{1}{2}(1)^2(-4)^2 = -4 + 8 = 4$. So $C=4$. The particular solution is $x^3y + \frac{1}{2}x^2y^2 = 4$. For $x=2$, we have $8y + 2y^2 = 4$, or $y^2+4y-2=0$. The quadratic formula gives $y = \frac{-4 \pm \sqrt{16-4(-2)}}{2} = -2 \pm \sqrt{6}$. Since the initial condition is $y(1)=-4$, we need the branch of the solution that is near $-4$. The value $-2-\sqrt{6} \approx -4.45$ is on this branch, while $-2+\sqrt{6} \approx 0.45$ is not.

Question 12

1. They computed $\frac{\partial M}{\partial y}$ incorrectly as $2y$ instead of $2$
2. They used the wrong formula; they should compute $\frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{N}$
3. They computed $\frac{\partial N}{\partial x}$ correctly but their conclusion is wrong since the equation is exact (correct answer)
4. They should have computed $\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N}$ instead of dividing by $M$

Explanation: Let's check the student's work: $M = 2x + y^2$, $N = 2xy$. We have $\frac{\partial M}{\partial y} = 2y$ and $\frac{\partial N}{\partial x} = 2y$. The student computed these correctly. Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, 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 $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, 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 13

1. An integrating factor of the form $\mu(x)$ only
2. An integrating factor of the form $\mu(y)$ only
3. An integrating factor of the form $\mu(xy)$
4. An integrating factor of the form $\mu(x^2 + y^2)$ (correct answer)

Explanation: The given expression $\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{xN - yM} = \frac{2}{x^2 + y^2}$ suggests looking for an integrating factor that depends on $x^2 + y^2$. This is because the denominator $xN - yM$ appears in the standard formula for integrating factors of the form $\mu(x^2 + y^2)$. When the expression $\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{xN - yM}$ depends only on $x^2 + y^2$, then an integrating factor $\mu(x^2 + y^2)$ exists where $\frac{d\ln\mu}{d(x^2 + y^2)} = \frac{1}{2} \cdot \frac{2}{x^2 + y^2} = \frac{1}{x^2 + y^2}$. The other options ($\mu(x)$, $\mu(y)$, $\mu(xy)$) would require different expressions in their respective formulas.

Question 14

1. $(2y + \frac{1}{x})dx + (2x + \frac{3}{y})dy = 0$ (correct answer)
2. $(2y + \frac{1}{x})dx + (2x + 3)dy = 0$
3. $(2xy + 1)dx + (2xy + 3)dy = 0$
4. $\frac{2y + 1}{x}dx + \frac{2x + 3}{y}dy = 0$

Explanation: First verify the original equation is not exact: $M = 2xy^2 + y$, $N = 2x^2y + 3x$. We have $\frac{\partial M}{\partial y} = 4xy + 1$ and $\frac{\partial N}{\partial x} = 4xy + 3$. Since $4xy + 1 \neq 4xy + 3$, it's not exact. Multiplying by $\mu = \frac{1}{xy}$: $\frac{2xy^2 + y}{xy}dx + \frac{2x^2y + 3x}{xy}dy = 0$, which simplifies to $\frac{y(2xy + 1)}{xy}dx + \frac{x(2xy + 3)}{xy}dy = 0$, giving $(\frac{2xy + 1}{x})dx + (\frac{2xy + 3}{y})dy = 0$, or $(2y + \frac{1}{x})dx + (2x + \frac{3}{y})dy = 0$. Check exactness: $\frac{\partial}{\partial y}(2y + \frac{1}{x}) = 2$ and $\frac{\partial}{\partial x}(2x + \frac{3}{y}) = 2$. Now it's exact.