Differential Equations Quiz: Solving Exact Des

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Solving Exact DesQuestion 1 of 20

The differential equation $M(x,y)dx + N(x,y)dy = 0$ is known to be exact with potential function $F(x,y)$ . If a new differential equation is formed as $(M(x,y)+f(x))dx + (N(x,y)+g(y))dy = 0$ , where $f(x)$ and $g(y)$ are continuous functions, under what condition is this new equation also exact?

Only if

f'(x) = g'(y)

for all

x, y

in the domain.

Only if

f(x)

and

g(y)

are both constants.

The new equation is always exact for any choice of

f(x)

and

g(y)

Only if

f(x)

and

g(y)

are both identically zero.

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Differential Equations Quiz

Differential Equations Quiz: Solving Exact Des

Practice Solving 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 Solving 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.

All questions

Question 1

Only if $f'(x) = g'(y)$ for all $x, y$ in the domain.
Only if $f(x)$ and $g(y)$ are both constants.
The new equation is always exact for any choice of $f(x)$ and $g(y)$ . (correct answer)
Only if $f(x)$ and $g(y)$ are both identically zero.

Explanation: When you encounter questions about exactness of differential equations, remember that the key criterion is whether the mixed partial derivatives are equal:

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

. Since the original equation

M(x,y)dx + N(x,y)dy = 0

is exact, we know that

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

. For the new equation

(M(x,y)+f(x))dx + (N(x,y)+g(y))dy = 0

, we need to check if

\frac{\partial}{\partial y}[M(x,y)+f(x)] = \frac{\partial}{\partial x}[N(x,y)+g(y)]

. Computing these partial derivatives: the left side gives

\frac{\partial M}{\partial y} + \frac{\partial f(x)}{\partial y} = \frac{\partial M}{\partial y} + 0 = \frac{\partial M}{\partial y}

, since

f(x)

doesn't depend on

y

. The right side gives

\frac{\partial N}{\partial x} + \frac{\partial g(y)}{\partial x} = \frac{\partial N}{\partial x} + 0 = \frac{\partial N}{\partial x}

, since

g(y)

doesn't depend on

x

. Since

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

from the original exact equation, the new equation is also exact regardless of what

f(x)

and

g(y)

are. This makes C correct. A is wrong because the derivatives of

f

and

g

don't appear in the exactness condition. B and D are incorrect because they impose unnecessary restrictions—the functions can be any continuous functions, not just constants or zero. Study tip: Remember that adding functions of only one variable to each term preserves exactness because mixed partials of single-variable functions are always zero.

Question 2

The differential equation $dx + (2y^2 - \frac{x}{y}) dy = 0$ is not exact but can be made exact using an integrating factor $\mu(y)$ that is a function of $y$ alone. Find the general solution of the resulting exact equation.

$\frac{x}{y} - y^2 = C$
$\frac{x}{y} + y^2 = C$ (correct answer)
$x + \frac{2y^3}{3} - x \ln|y| = C$
$x\ln|y| - y^2 = C$

Explanation: When you encounter a non-exact differential equation, your goal is to find an integrating factor that makes it exact, then solve the resulting equation systematically. First, verify this equation isn't exact by checking if

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

where

M = 1

and

N = 2y^2 - \frac{x}{y}

. Since

\frac{\partial M}{\partial y} = 0

but

\frac{\partial N}{\partial x} = -\frac{1}{y}

, the equation is indeed non-exact. To find the integrating factor

\mu(y)

, use the formula:

\frac{d\ln\mu}{dy} = \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{0 - (-\frac{1}{y})}{2y^2 - \frac{x}{y}} = \frac{1}{y(2y^2 - \frac{x}{y})} = \frac{1}{2y^3 - x}

Actually, let's try

\mu(y) = \frac{1}{y}

. Multiplying the original equation by

\frac{1}{y}

\frac{1}{y}dx + (2y - \frac{x}{y^2})dy = 0

Now

M = \frac{1}{y}

and

N = 2y - \frac{x}{y^2}

. Check:

\frac{\partial M}{\partial y} = -\frac{1}{y^2}

and

\frac{\partial N}{\partial x} = -\frac{1}{y^2}

. Perfect—it's exact! For an exact equation,

F(x,y) = C

where

\frac{\partial F}{\partial x} = \frac{1}{y}

and

\frac{\partial F}{\partial y} = 2y - \frac{x}{y^2}

. Integrating the first:

F = \frac{x}{y} + g(y)

. Using the second condition gives

g(y) = y^2

, so

F = \frac{x}{y} + y^2

. The answer is B. Choice A has the wrong sign, while C and D represent different solution approaches that don't match our exact equation's structure. Remember: always verify your integrating factor makes the equation exact before proceeding to solve.

Question 3

The differential equation $(2x + y) \, dx + (x - 2y) \, dy = 0$ is exact, with general solution $F(x,y) = x^2 + xy - y^2 = C$ . Which of the following differential equations has solution curves that are orthogonal trajectories to the solution curves of the original equation?

$(x - 2y) \, dx + (2x + y) \, dy = 0$
$(2x + y) \, dx - (x - 2y) \, dy = 0$
$(x - 2y) \, dx - (2x + y) \, dy = 0$ (correct answer)
$(2x - y) \, dx + (x + 2y) \, dy = 0$

Explanation: The slope of the tangent line to the solution curves of the given equation

M \, dx + N \, dy = 0

is given by

\frac{dy}{dx} = -\frac{M}{N}

. For the given equation, the slope is

\frac{dy}{dx} = -\frac{2x+y}{x-2y} = \frac{2x+y}{2y-x}

. The slope of the orthogonal trajectories, let's call it

\frac{dy}{dx}_{\perp}

, is the negative reciprocal of the original slope.

\frac{dy}{dx}_{\perp} = -\frac{1}{\frac{2x+y}{2y-x}} = -\frac{2y-x}{2x+y} = \frac{x-2y}{2x+y}

. Now, we convert this slope back into a differential equation of the form

M_{\perp} \, dx + N_{\perp} \, dy = 0

. From

\frac{dy}{dx} = \frac{x-2y}{2x+y}

, we can write

(2x+y) \, dy = (x-2y) \, dx

. Rearranging this into the standard form gives:

(x-2y) \, dx - (2x+y) \, dy = 0

. This matches option C. Option A results from a sign error when finding the negative reciprocal. Option B represents the original family of curves.

Question 4

Consider the differential equation $M(x,y) \, dx + N(x,y) \, dy = 0$ . If $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ for all $(x,y)$ in a simply connected region $R$ , which statement is guaranteed to be true?

The equation is separable and can be solved by separating variables.
The equation has a unique solution passing through any point in $R$ .
There exists a function $F(x,y)$ such that $M = \frac{\partial F}{\partial y}$ and $N = \frac{\partial F}{\partial x}$ .
The general solution is given by a family of level curves of some potential function $F(x,y)$ . (correct answer)

Explanation: The condition

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

is the test for exactness. If a differential equation is exact on a simply connected region, there exists a potential function

F(x,y)

such that

\frac{\partial F}{\partial x} = M

and

\frac{\partial F}{\partial y} = N

. The general solution to the differential equation is given implicitly by

F(x,y) = C

, where

C

is an arbitrary constant. Each value of

C

defines a specific curve, and the set of these curves for all possible

C

forms a family of level curves of the surface

z = F(x,y)

. Distractor A is incorrect; exact equations are not necessarily separable. Distractor B is incorrect; exactness alone does not guarantee uniqueness for an initial value problem, which requires additional conditions like the Lipschitz condition. Distractor C is incorrect because the partial derivatives are switched; it should be

M = \frac{\partial F}{\partial x}

and

N = \frac{\partial F}{\partial y}

Question 5

Consider the differential equation $(3x^2y + e^x)dx + (x^3 + 2y)dy = 0$ . After verifying that this equation is exact, what is the general solution?

$x^3y + e^x + y^2 = C$ (correct answer)
$x^3y + e^x - y^2 = C$
$3x^2y + e^x + x^3 + 2y = C$
$x^3y + e^x + 2y^2 = C$

Explanation: First verify exactness: ∂M/∂y = ∂(3x²y + eˣ)/∂y = 3x² and ∂N/∂x = ∂(x³ + 2y)/∂x = 3x². Since these are equal, the equation is exact. To find F(x,y), integrate M with respect to x: F = ∫(3x²y + eˣ)dx = x³y + eˣ + g(y). Then ∂F/∂y = x³ + g'(y) = N = x³ + 2y, so g'(y) = 2y and g(y) = y². Therefore F(x,y) = x³y + eˣ + y² = C. Choice B has wrong sign on y². Choice C incorrectly adds all terms. Choice D has coefficient error in y² term.

Question 6

The differential equation $\left(e^y + ye^x\right)dx + \left(xe^y + e^x\right)dy = 0$ is exact. After finding the potential function $F(x,y)$ , what is $\frac{\partial^2 F}{\partial x \partial y}$ ?

$e^x + ye^x$
$e^y + e^x$ (correct answer)
$xe^y + ye^x$
$e^y + xe^y$

Explanation: When you encounter an exact differential equation, you're working with equations where

M(x,y)dx + N(x,y)dy = 0

and

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

. The key insight is that there exists a potential function

F(x,y)

such that

\frac{\partial F}{\partial x} = M

and

\frac{\partial F}{\partial y} = N

. Here,

M(x,y) = e^y + ye^x

and

N(x,y) = xe^y + e^x

. The mixed partial derivative

\frac{\partial^2 F}{\partial x \partial y}

can be found by taking

\frac{\partial M}{\partial y}

\frac{\partial N}{\partial x}

(they're equal for exact equations). Taking

\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(e^y + ye^x) = e^y + e^x

, which is answer B. Let's verify:

\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(xe^y + e^x) = e^y + e^x

. Perfect match. Looking at the wrong answers: A)

e^x + ye^x

appears to be

\frac{\partial M}{\partial x}

, not the mixed partial. C)

xe^y + ye^x

incorrectly combines terms from both

M

and

N

. D)

e^y + xe^y

seems to be

\frac{\partial N}{\partial y}

, which gives the wrong mixed partial. Remember: For exact equations,

\frac{\partial^2 F}{\partial x \partial y} = \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

. This equality is what makes the equation exact in the first place, and either calculation gives you the mixed partial derivative.

Question 7

Find the general solution to the differential equation $(y\sin(x) + xy\cos(x))dx + (x\sin(x) + 1)dy = 0$ .

$y(x\sin(x)) = C$
$y(x\sin(x)+1) = C$ (correct answer)
$y(x\cos(x)-\sin(x)) = C$
$y(x\sin(x)-x\cos(x)+1) = C$

Explanation: When you encounter a differential equation in the form

M(x,y)dx + N(x,y)dy = 0

, you should first check if it's exact. An exact equation has the property that

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

, and its solution comes from finding a function

F(x,y) = C

whose total differential equals the given equation. Here,

M(x,y) = y\sin(x) + xy\cos(x)

and

N(x,y) = x\sin(x) + 1

. Let's verify exactness:

\frac{\partial M}{\partial y} = \sin(x) + x\cos(x)

and

\frac{\partial N}{\partial x} = \sin(x) + x\cos(x)

. Since these are equal, the equation is exact. To find the solution, integrate

M

with respect to

x

F(x,y) = y\int(\sin(x) + x\cos(x))dx = y(x\sin(x) + C_1) = yx\sin(x) + C_1y

. Adding any function of

y

alone, we get

F(x,y) = yx\sin(x) + g(y)

. Using

\frac{\partial F}{\partial y} = N

x\sin(x) + g'(y) = x\sin(x) + 1

, so

g'(y) = 1

and

g(y) = y

. Therefore,

F(x,y) = yx\sin(x) + y = y(x\sin(x) + 1)

, giving us the solution

y(x\sin(x) + 1) = C

. Answer A omits the constant term. Answer C uses cosine instead of sine and has incorrect signs. Answer D includes extra terms that don't belong in this solution. Remember: for exact equations, always verify the exactness condition first, then integrate systematically to build the solution function.

Question 8

Consider the initial value problem $(\frac{y}{x} + 6x)dx + (\ln(x) - 2)dy = 0$ , with the initial condition $y(e)=4$ . What is the value of $y(1)$ ?

$2$
$\frac{7 - 3e^2}{2}$ (correct answer)
$\frac{3e^2 - 7}{2}$
$-\frac{3e^2 + 1}{2}$

Explanation: Let

M = y/x + 6x

and

N = \ln(x) - 2

. The equation is defined for

x>0

. Check for exactness:

\frac{\partial M}{\partial y} = 1/x

and

\frac{\partial N}{\partial x} = 1/x

. The equation is exact. Find the potential function

F(x,y)

by integrating

N

with respect to

y

F(x,y) = \int (\ln(x)-2)dy = y(\ln(x)-2) + h(x)

. Differentiate with respect to

x

\frac{\partial F}{\partial x} = y(1/x) + h'(x)

. Set this equal to

M

y/x + h'(x) = y/x + 6x

. So,

h'(x) = 6x

, which gives

h(x) = 3x^2

. The general solution is

y(\ln(x)-2) + 3x^2 = C

. Apply the initial condition

y(e)=4

4(\ln(e)-2) + 3e^2 = C \implies 4(1-2) + 3e^2 = C \implies C = 3e^2 - 4

. The particular solution is

y(\ln(x)-2) + 3x^2 = 3e^2 - 4

. To find

y(1)

, substitute

x=1

y(1)(\ln(1)-2) + 3(1)^2 = 3e^2 - 4 \implies y(1)(0-2) + 3 = 3e^2 - 4 \implies -2y(1) = 3e^2 - 7 \implies y(1) = \frac{7-3e^2}{2}

Question 9

Consider the differential equation $(2x+y)dx + (x-y)dy = 0$ . Which of the following statements is true?

The equation becomes exact after multiplying by the integrating factor $\mu(x)=e^x$ .
The equation is not exact, because $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$ .
The equation is exact, and its general solution is $x^2+y^2=C$ .
The equation is exact, and its general solution is $x^2+xy-\frac{1}{2}y^2=C$ . (correct answer)

Explanation: When you encounter a differential equation in the form

M(x,y)dx + N(x,y)dy = 0

, your first step should be checking if it's exact by testing whether

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

. Here,

M(x,y) = 2x+y

and

N(x,y) = x-y

. Computing the partial derivatives:

\frac{\partial M}{\partial y} = 1

and

\frac{\partial N}{\partial x} = 1

. Since these are equal, the equation is exact. For an exact equation, there exists a function

F(x,y)

such that

\frac{\partial F}{\partial x} = M

and

\frac{\partial F}{\partial y} = N

. To find

F

, integrate

M

with respect to

x

F(x,y) = \int(2x+y)dx = x^2 + xy + g(y)

To find

g(y)

, use the condition

\frac{\partial F}{\partial y} = N

\frac{\partial F}{\partial y} = x + g'(y) = x - y

Therefore,

g'(y) = -y

, so

g(y) = -\frac{1}{2}y^2

This gives us

F(x,y) = x^2 + xy - \frac{1}{2}y^2

, and the general solution is

x^2 + xy - \frac{1}{2}y^2 = C

. Option A is wrong because no integrating factor is needed—the equation is already exact. Option B incorrectly claims the equation isn't exact when our calculation shows it is. Option C has the wrong solution form entirely. Always verify exactness first before seeking integrating factors. This systematic approach prevents unnecessary work and ensures you don't miss when an equation is already in its simplest exact form.

Question 10

For what values of the constants $a$ and $b$ is the differential equation $(6xy^3 + b\cos(y))dx + (ax^2y^2 - x\sin(y))dy = 0$ exact?

$a=9, b=1$ (correct answer)
$a=9, b=-1$
$a=3, b=1$
$a=12, b=-1$

Explanation: For the differential equation

M(x,y)dx + N(x,y)dy = 0

to be exact, the condition

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

must hold. Here,

M(x,y) = 6xy^3 + b\cos(y)

and

N(x,y) = ax^2y^2 - x\sin(y)

. We compute the partial derivatives:

\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(6xy^3 + b\cos(y)) = 18xy^2 - b\sin(y)

\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(ax^2y^2 - x\sin(y)) = 2axy^2 - \sin(y)

. For the equation to be exact, these two expressions must be identical for all

x

and

y

. This means we can equate the coefficients of like terms:

18xy^2 = 2axy^2 \implies 18 = 2a \implies a=9

-b\sin(y) = -\sin(y) \implies b=1

. Thus,

a=9

and

b=1

Question 11

The differential equation $(1+y^2\sin(2x))dx - 2y\cos^2(x)dy = 0$ is exact. Let $F(x,y)$ be its potential function satisfying $F(0,1)=5$ . What is the value of $F(\pi/2, 2)$ ?

$\frac{\pi}{2}$
$\frac{\pi}{2} + 5$
$\frac{\pi}{2} + 4$
$\frac{\pi}{2} + 6$ (correct answer)

Explanation: When you encounter an exact differential equation, you're looking for a potential function

F(x,y)

such that

\frac{\partial F}{\partial x} = M

and

\frac{\partial F}{\partial y} = N

, where the equation is written as

M dx + N dy = 0

. Here,

M = 1 + y^2\sin(2x)

and

N = -2y\cos^2(x)

. To find

F(x,y)

, integrate

M

with respect to

x

F(x,y) = \int (1 + y^2\sin(2x)) dx = x - \frac{y^2\cos(2x)}{2} + g(y)

To find

g(y)

, use the condition

\frac{\partial F}{\partial y} = N

\frac{\partial F}{\partial y} = -y\cos(2x) + g'(y) = -2y\cos^2(x)

Since

\cos(2x) = 2\cos^2(x) - 1

, we have

-y\cos(2x) = -y(2\cos^2(x) - 1) = -2y\cos^2(x) + y

. Therefore:

g'(y) = y

, so

g(y) = \frac{y^2}{2} + C

Thus:

F(x,y) = x - \frac{y^2\cos(2x)}{2} + \frac{y^2}{2} + C

Using

F(0,1) = 5

0 - \frac{1}{2} + \frac{1}{2} + C = 5

, so

C = 5

. Therefore:

F(x,y) = x + \frac{y^2(1-\cos(2x))}{2} + 5

(\pi/2, 2)

F(\pi/2, 2) = \frac{\pi}{2} + \frac{4(1-(-1))}{2} + 5 = \frac{\pi}{2} + 4 + 5 = \frac{\pi}{2} + 6

Choice A omits the constant and

y^2

term. Choice B includes only the original constant. Choice C includes the

y^2

contribution but misses part of the constant adjustment. Key strategy: Always verify your potential function by checking both partial derivative conditions, and don't forget to apply initial conditions to determine all constants.

Question 12

Find the implicit solution to the initial value problem $(y^2 e^{xy^2} + 4x^3) dx + (2xye^{xy^2} - 3y^2) dy = 0$ , with $y(1)=0$ .

$x e^{xy^2} + x^4 - y^3 = 1$
$e^{xy^2} + x^4 - y^3 = 1$
$e^{xy^2} + x^4 - y^3 = 2$ (correct answer)
$e^{xy^2} + 4x^4 - 3y^3 = 1$

Explanation: When you encounter a differential equation of the form

M(x,y)dx + N(x,y)dy = 0

, check if it's exact by verifying whether

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

. Here,

M = y^2 e^{xy^2} + 4x^3

and

N = 2xye^{xy^2} - 3y^2

. Taking partial derivatives:

\frac{\partial M}{\partial y} = 2ye^{xy^2} + y^2 \cdot e^{xy^2} \cdot 2xy = 2ye^{xy^2}(1 + xy^2)

\frac{\partial N}{\partial x} = 2ye^{xy^2} + 2xy \cdot e^{xy^2} \cdot y^2 = 2ye^{xy^2}(1 + xy^2)

Since these are equal, the equation is exact. The solution has the form

F(x,y) = C

where

\frac{\partial F}{\partial x} = M

and

\frac{\partial F}{\partial y} = N

. Integrating

\frac{\partial F}{\partial x} = y^2 e^{xy^2} + 4x^3

F = e^{xy^2} + x^4 + g(y)

To find

g(y)

, use

\frac{\partial F}{\partial y} = N

2xye^{xy^2} + g'(y) = 2xye^{xy^2} - 3y^2

This gives

g'(y) = -3y^2

, so

g(y) = -y^3

. Therefore:

F(x,y) = e^{xy^2} + x^4 - y^3

Applying the initial condition

y(1) = 0

e^{1 \cdot 0^2} + 1^4 - 0^3 = e^0 + 1 = 2

The solution is

e^{xy^2} + x^4 - y^3 = 2

, which is choice C. Choice A has an extra factor of

x

in the exponential term. Choice B uses the wrong constant (1 instead of 2). Choice D has incorrect coefficients (4 and 3 instead of 1). Remember: always check if a differential equation is exact first, then integrate systematically and apply initial conditions carefully to determine the constant.

Question 13

A first-order differential equation is given by $\frac{dy}{dx} = -\frac{2xy^3 + \sec^2(x)}{3x^2y^2}$ . What is the general solution?

$x^2y^3 + \tan(x) = C$ (correct answer)
$x^2y^3 - \tan(x) = C$
$y^3 + \frac{\tan(x)}{x^2} = C$
The equation is not exact and cannot be solved this way.

Explanation: First, rewrite the equation in the standard form

M(x,y)dx + N(x,y)dy = 0

. From

\frac{dy}{dx} = -\frac{2xy^3 + \sec^2(x)}{3x^2y^2}

, we get

(3x^2y^2)dy = -(2xy^3 + \sec^2(x))dx

, which is

(2xy^3 + \sec^2(x))dx + (3x^2y^2)dy = 0

. Let

M(x,y) = 2xy^3 + \sec^2(x)

and

N(x,y) = 3x^2y^2

. Check for exactness:

\frac{\partial M}{\partial y} = 6xy^2

and

\frac{\partial N}{\partial x} = 6xy^2

. The equation is exact. We find a potential function

F(x,y)

. Integrating

N

with respect to

y

F(x,y) = \int 3x^2y^2 dy = x^2y^3 + h(x)

. Differentiating with respect to

x

\frac{\partial F}{\partial x} = 2xy^3 + h'(x)

. Setting this equal to

M

2xy^3 + h'(x) = 2xy^3 + \sec^2(x)

. This gives

h'(x) = \sec^2(x)

, so

h(x) = \tan(x)

. The potential function is

F(x,y) = x^2y^3 + \tan(x)

. The general solution is

F(x,y) = C

, which is

x^2y^3 + \tan(x) = C

Question 14

For what value of the constant $k$ is the differential equation $(y^3 + kxy^4 - 2x) \, dx + (3xy^2 + 20x^2y^3) \, dy = 0$ exact?

$k=5$
$k=8$
$k=10$ (correct answer)
$k=20$

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 partial derivative of

M

with respect to

y

must equal the partial derivative of

N

with respect to

x

. First, compute

\frac{\partial M}{\partial y}

\frac{\partial}{\partial y}(y^3 + kxy^4 - 2x) = 3y^2 + 4kxy^3

. Next, compute

\frac{\partial N}{\partial x}

\frac{\partial}{\partial x}(3xy^2 + 20x^2y^3) = 3y^2 + 40xy^3

. Set the two partial derivatives equal to each other:

3y^2 + 4kxy^3 = 3y^2 + 40xy^3

. Subtracting

3y^2

from both sides gives:

4kxy^3 = 40xy^3

. Assuming

x

and

y

are not identically zero, we can divide by

4xy^3

to find

k

k = \frac{40}{4} = 10

Question 15

Find the particular solution to the exact differential equation $(2xy^2 + \cos x) \, dx + (2x^2y - \sin y) \, dy = 0$ subject to the initial condition $y(0) = \pi$ .

$x^2y^2 + \sin x + \cos y = -1$ (correct answer)
$x^2y^2 + \sin x + \cos y = 1$
$x^2y^2 - \sin x + \cos y = -1$
$x^2y^2 + \sin x - \cos y = 1$

Explanation: The equation is exact because

\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2xy^2 + \cos x) = 4xy

and

\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(2x^2y - \sin y) = 4xy

. To find the solution

F(x, y) = C

, we integrate

M(x, y)

with respect to

x

F(x, y) = \int (2xy^2 + \cos x) \, dx = x^2y^2 + \sin x + g(y)

. To find

g(y)

, we differentiate

F(x, y)

with respect to

y

and set it equal to

N(x, y)

\frac{\partial F}{\partial y} = 2x^2y + g'(y) = 2x^2y - \sin y

. This implies

g'(y) = -\sin y

. Integrating with respect to

y

gives

g(y) = \cos y

. The general solution is

F(x, y) = x^2y^2 + \sin x + \cos y = C

. Apply the initial condition

y(0) = \pi

0^2\pi^2 + \sin(0) + \cos(\pi) = C

0 + 0 + (-1) = C \implies C = -1

. Thus, the particular solution is

x^2y^2 + \sin x + \cos y = -1

. Distractor B arises from the common error

\cos(\pi)=1

Question 16

A solution to the exact differential equation $(y e^{xy} + \cos x) \, dx + (x e^{xy} - 1) \, dy = 0$ passes through the point $(\frac{\pi}{2}, 0)$ . What is the value of $y$ when $x=0$ ?

$y=2$
$y=1$
$y=0$
$y=-1$ (correct answer)

Explanation: When you encounter a differential equation like this, you need to recognize it's exact and solve by finding a potential function whose partial derivatives match the given coefficients. For an exact equation

M \, dx + N \, dy = 0

, there exists a function

F(x,y)

where

\frac{\partial F}{\partial x} = M = y e^{xy} + \cos x

and

\frac{\partial F}{\partial y} = N = x e^{xy} - 1

. To find

F

, integrate the first equation with respect to

x

F(x,y) = \int (y e^{xy} + \cos x) \, dx = e^{xy} + \sin x + g(y)

To find

g(y)

, take the partial derivative with respect to

y

and set it equal to

N

\frac{\partial F}{\partial y} = x e^{xy} + g'(y) = x e^{xy} - 1

This gives us

g'(y) = -1

, so

g(y) = -y + C

. Therefore,

F(x,y) = e^{xy} + \sin x - y = C

. Using the initial condition

(\frac{\pi}{2}, 0)

e^{(\pi/2)(0)} + \sin(\frac{\pi}{2}) - 0 = 1 + 1 - 0 = 2

So our solution is

e^{xy} + \sin x - y = 2

. When

x = 0

e^{0} + \sin(0) - y = 2

, which gives us

1 + 0 - y = 2

, so

y = -1

. Choice A)

y = 2

incorrectly uses the constant value. Choice B)

y = 1

likely comes from forgetting the

-y

term. Choice C)

y = 0

might result from confusing the initial condition coordinates. Always verify your potential function by checking both partial derivatives match the original equation's coefficients.

Question 17

The differential equation $(2y^2 + 3x) \, dx + 2xy \, dy = 0$ can be made exact by an integrating factor $\mu(x)$ that is a function of $x$ alone. Find this integrating factor.

$\mu(x) = x^2$
$\mu(x) = x$ (correct answer)
$\mu(x) = e^x$
$\mu(x) = 1/x$

Explanation: When you encounter a differential equation that isn't exact, you need to find an integrating factor to make it exact. For the equation

(2y^2 + 3x) \, dx + 2xy \, dy = 0

, first check if it's exact by testing whether

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

, where

M = 2y^2 + 3x

and

N = 2xy

. We get

\frac{\partial M}{\partial y} = 4y

and

\frac{\partial N}{\partial x} = 2y

. Since

4y \neq 2y

, the equation isn't exact. To find an integrating factor

\mu(x)

that depends only on

x

, use the formula:

\frac{d\mu}{dx} = \mu \cdot \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N}

Substituting our values:

\frac{d\mu}{dx} = \mu \cdot \frac{4y - 2y}{2xy} = \mu \cdot \frac{2y}{2xy} = \frac{\mu}{x}

This gives us

\frac{d\mu}{\mu} = \frac{dx}{x}

, which integrates to

\ln|\mu| = \ln|x| + C

. Therefore,

\mu(x) = x

. Looking at the wrong answers: A)

x^2

would arise from a different coefficient in our calculation. C)

e^x

would result if the integrating factor formula yielded a constant instead of

1/x

. D)

1/x

would be correct if we had mistakenly swapped the numerator terms in our fraction. Study tip: Always verify that

(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})/N

depends only on

x

before using this method. If it depends on both variables, try

(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})/M

for a

\mu(y)

instead.

Question 18

Find the general solution to the differential equation $\frac{dy}{dx} = \frac{x - y\cos(x)}{\sin(x) + y}$ .

$\frac{x^2}{2} - y\sin(x) - \frac{y^2}{2} = C$ (correct answer)
$y\sin(x) - \frac{x^2}{2} + \frac{y^2}{2} = C$
$\frac{x^2}{2} + y\sin(x) + \frac{y^2}{2} = C$
$y\cos(x) - x^2 - y^2 = C$

Explanation: First, rewrite the equation in the standard form

M(x, y) \, dx + N(x, y) \, dy = 0

(\sin(x) + y) \, dy = (x - y\cos(x)) \, dx

(x - y\cos(x)) \, dx - (\sin(x) + y) \, dy = 0

. Here,

M(x, y) = x - y\cos(x)

and

N(x, y) = -(\sin(x) + y)

. Check for exactness:

\frac{\partial M}{\partial y} = -\cos(x)

\frac{\partial N}{\partial x} = -\cos(x)

. The equation is exact. Find the potential function

F(x, y)

. Integrate

M

with respect to

x

F(x, y) = \int (x - y\cos(x)) \, dx = \frac{x^2}{2} - y\sin(x) + g(y)

. Differentiate

F

with respect to

y

and set it equal to

N

\frac{\partial F}{\partial y} = -\sin(x) + g'(y) = -\sin(x) - y

. This implies

g'(y) = -y

. Integrating gives

g(y) = -\frac{y^2}{2}

. The general solution is

F(x, y) = C

, which is

\frac{x^2}{2} - y\sin(x) - \frac{y^2}{2} = C

. Distractor B results from an incorrect sign during the initial rearrangement of the equation.

Question 19

The implicit solution to an exact first-order differential equation is given by $F(x, y) = x^2 \sin(y) + y^2 e^x = C$ . Which of the following is the differential equation?

$(2x \sin(y) + y^2 e^x) \, dx + (x^2 \cos(y) + 2y e^x) \, dy = 0$ (correct answer)
$(x^2 \cos(y) + 2y e^x) \, dx + (2x \sin(y) + y^2 e^x) \, dy = 0$
$(2x \sin(y) + y^2 e^x) \, dx - (x^2 \cos(y) + 2y e^x) \, dy = 0$
$(2x \sin(y) + 2y e^x) \, dx + (x^2 \cos(y) + y^2 e^x) \, dy = 0$

Explanation: An exact differential equation is of the form

M(x, y) \, dx + N(x, y) \, dy = 0

, where

M = \frac{\partial F}{\partial x}

and

N = \frac{\partial F}{\partial y}

for some function

F(x, y)

. The solution is given by

F(x, y) = C

. Given

F(x, y) = x^2 \sin(y) + y^2 e^x

, we need to find its partial derivatives.

M(x, y) = \frac{\partial F}{\partial x} = \frac{\partial}{\partial x}(x^2 \sin(y) + y^2 e^x) = 2x \sin(y) + y^2 e^x

N(x, y) = \frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(x^2 \sin(y) + y^2 e^x) = x^2 \cos(y) + 2y e^x

. The corresponding differential equation is

M \, dx + N \, dy = 0

, which is:

(2x \sin(y) + y^2 e^x) \, dx + (x^2 \cos(y) + 2y e^x) \, dy = 0

. Distractor B incorrectly swaps

M

and

N

. Distractor C introduces an incorrect sign. Distractor D contains errors in partial differentiation.

Question 20

For the differential equation $(ye^{xy} + 2x)dx + (xe^{xy} + 3y^2)dy = 0$ , which of the following represents the correct potential function $F(x,y)$ such that $dF = 0$ ?

$F(x,y) = e^{xy} + x^2 + y^3$ (correct answer)
$F(x,y) = ye^{xy} + 2x + xe^{xy} + 3y^2$
$F(x,y) = e^{xy} + x^2 + 3y^3$
$F(x,y) = xye^{xy} + x^2 + y^3$

Explanation: First check exactness: ∂M/∂y = ∂(yeˣʸ + 2x)/∂y = eˣʸ + xyeˣʸ and ∂N/∂x = ∂(xeˣʸ + 3y²)/∂x = eˣʸ + xyeˣʸ. Since these are equal, the equation is exact. To find F, integrate M with respect to x: F = ∫(yeˣʸ + 2x)dx = eˣʸ + x² + g(y). Then ∂F/∂y = xeˣʸ + g'(y) = N = xeˣʸ + 3y², so g'(y) = 3y² and g(y) = y³. Therefore F(x,y) = eˣʸ + x² + y³. Choice B incorrectly lists the coefficients M and N. Choice C has wrong coefficient on y³. Choice D incorrectly includes xy as a factor of eˣʸ.