Differential Equations Quiz: Homogeneous First Order Equations

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Homogeneous First Order EquationsQuestion 1 of 20

The general solution to the differential equation $(x \cos(y/x) + y \sin(y/x))y = (y \sin(y/x) - x \cos(y/x))x y'$ is given by which of the following implicit relations?

xy \sin(y/x) = C

xy \cos(y/x) = C

y\cos(y/x) = C

\cos(y/x) + (y/x)\sin(y/x) = C

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Differential Equations Quiz

Differential Equations Quiz: Homogeneous First Order Equations

Practice Homogeneous First Order Equations 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 Homogeneous First Order Equations, 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

The general solution to the differential equation $(x \cos(y/x) + y \sin(y/x))y = (y \sin(y/x) - x \cos(y/x))x y'$ is given by which of the following implicit relations?

$xy \sin(y/x) = C$
$xy \cos(y/x) = C$ (correct answer)
$y\cos(y/x) = C$
$\cos(y/x) + (y/x)\sin(y/x) = C$

Explanation: When you encounter a differential equation with expressions involving ratios like

y/x

, this signals a homogeneous differential equation that often benefits from substitution methods or recognizing it as an exact equation. Let's rearrange this equation into standard form. Moving terms around, we get:

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

This has the form

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

where:

$M = -(y \sin(y/x) - x \cos(y/x)) = x \cos(y/x) - y \sin(y/x)$
$N = x \cos(y/x) + y \sin(y/x)$

For an exact equation, we need

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

. You can verify this condition holds here, making this an exact differential equation. The solution comes from finding a function

F(x,y)

such that

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

and

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

. By inspection or integration,

F(x,y) = xy\cos(y/x)

, so the general solution is

xy\cos(y/x) = C

. Choice A gives

xy \sin(y/x) = C

, which would correspond to a different differential equation. Choice C,

y\cos(y/x) = C

, is missing the crucial

x

factor. Choice D represents a more complex expression that doesn't match our integrating function. Study tip: For homogeneous differential equations involving

y/x

ratios, always check if the equation is exact first—it's often faster than substitution methods.

Question 2

The differential equation $(2x^2y + y^3)dx + (xy^2 - 2x^3)dy = 0$ is homogeneous. After applying the substitution $v = y/x$ , the equation can be written in the separable form $g(v)dv = dx/x$ . What is the expression for $g(v)$ ?

$\frac{2v^3}{2-v^2}$
$\frac{v^2-2}{4v}$
$\frac{2-v^2}{2v+v^3}$
$\frac{2-v^2}{2v^3}$ (correct answer)

Explanation: When you encounter a homogeneous differential equation, the key strategy is using the substitution

v = y/x

(so

y = vx

) to transform it into a separable equation. This technique works because homogeneous equations have terms where all variables have the same total degree. Starting with

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

, substitute

y = vx

and

dy = v dx + x dv

(2x^2(vx) + (vx)^3)dx + (x(vx)^2 - 2x^3)(v dx + x dv) = 0

Simplifying:

(2x^3v + v^3x^3)dx + (xv^2x^2 - 2x^3)(v dx + x dv) = 0

This becomes:

(2x^3v + v^3x^3)dx + (x^3v^2 - 2x^3)v dx + (x^3v^2 - 2x^3)x dv = 0

Collecting

dx

terms:

[2x^3v + v^3x^3 + x^3v^3 - 2x^3v]dx + x^4(v^2 - 2)dv = 0

The

dx

terms cancel to give

2v^3x^3dx

, so:

2v^3x^3dx + x^4(v^2 - 2)dv = 0

Dividing by

x^3

2v^3dx + x(v^2 - 2)dv = 0

Rearranging:

\frac{dv}{v^2 - 2} = -\frac{2v^3}{x}dx

, which gives us

g(v) = \frac{2-v^2}{2v^3}

when written as

g(v)dv = \frac{dx}{x}

. Answer D is correct. Answer A has the wrong sign in the numerator. Answer B has incorrect powers and missing factors. Answer C has the denominator wrong and doesn't match the algebraic manipulation. Remember: when applying the homogeneous substitution, carefully track all terms through the algebra—small errors in sign or powers will lead you to incorrect separable forms.

Question 3

A homogeneous differential equation has the form $dy/dx = F(y/x)$ . When the substitution $y=vx$ is made, the equation transforms into a separable equation of the form $x \frac{dv}{dx} = f(v)$ . If $F(v) = \frac{1-v}{1+v}$ , which of the following expressions represents the integral $\int \frac{1}{f(v)} dv$ ?

$\frac{1}{2}\ln|1-2v-v^2| + C$
$-\frac{1}{2}\ln|1-2v-v^2| + C$ (correct answer)
$-v - 2\ln|1-v| + C$
$-\ln|1-v| + C$

Explanation: When you encounter a homogeneous differential equation problem involving the substitution

y = vx

, you need to carefully track how the transformation affects the equation structure. Starting with

\frac{dy}{dx} = F(y/x) = \frac{1-v}{1+v}

where

v = y/x

, the substitution

y = vx

gives us

\frac{dy}{dx} = v + x\frac{dv}{dx}

. Setting this equal to

F(v)

v + x\frac{dv}{dx} = \frac{1-v}{1+v}

Solving for

x\frac{dv}{dx}

x\frac{dv}{dx} = \frac{1-v}{1+v} - v = \frac{1-v-v(1+v)}{1+v} = \frac{1-v-v-v^2}{1+v} = \frac{1-2v-v^2}{1+v}

Therefore,

f(v) = \frac{1-2v-v^2}{1+v}

, and we need

\int \frac{1}{f(v)} dv = \int \frac{1+v}{1-2v-v^2} dv

. To evaluate this integral, notice that the derivative of the denominator is

-2-2v

, which is

-2(1+v)

. This means:

\int \frac{1+v}{1-2v-v^2} dv = -\frac{1}{2}\int \frac{-2(1+v)}{1-2v-v^2} dv = -\frac{1}{2}\ln|1-2v-v^2| + C

This confirms answer B is correct. Answer A has the wrong sign on the coefficient. Answer C appears to come from incorrectly attempting partial fractions or making algebraic errors in the substitution. Answer D oversimplifies the denominator structure. Remember: in homogeneous equation problems, always double-check your algebra when computing

f(v) = F(v) - v

, as small errors here cascade through the entire solution.

Question 4

The non-homogeneous differential equation $\frac{dy}{dx} = \frac{y-x+1}{y+x+5}$ can be transformed into a homogeneous equation in variables $u$ and $v$ by a substitution of the form $x=u+h, y=v+k$ . What are the required constant values for $h$ and $k$ ?

$h = 2, k = 3$
$h = -2, k = -3$ (correct answer)
$h = -3, k = -2$
$h = 1, k = 5$

Explanation: When you encounter a differential equation like

\frac{dy}{dx} = \frac{y-x+1}{y+x+5}

, you should recognize this as a non-homogeneous equation that can be made homogeneous through translation. A homogeneous differential equation has the form

\frac{dy}{dx} = f\left(\frac{y}{x}\right)

, but here we have linear terms with constants that prevent this structure. To transform this equation using substitutions

x = u + h

and

y = v + k

, we need the constant terms to vanish. Substituting these expressions gives us:

\frac{dv}{du} = \frac{(v+k)-(u+h)+1}{(v+k)+(u+h)+5} = \frac{v-u+(k-h+1)}{v+u+(k+h+5)}

For this to become homogeneous, both constant terms in the parentheses must equal zero:

$k - h + 1 = 0$
$k + h + 5 = 0$

Solving this system: From the second equation,

k = -h - 5

. Substituting into the first:

(-h - 5) - h + 1 = 0

, which gives

-2h - 4 = 0

, so

h = -2

. Therefore,

k = -(-2) - 5 = -3

. Choice A gives

h = 2, k = 3

, which would leave non-zero constant terms. Choice C reverses the values (

h = -3, k = -2

), failing to satisfy our system. Choice D (

h = 1, k = 5

) also produces non-zero constants when substituted. The answer is B:

h = -2, k = -3

. Study tip: Always set up the system by requiring both constant terms to vanish simultaneously—this is the key to transforming non-homogeneous equations into homogeneous ones.

Question 5

A solution to the differential equation $x dy = (y + \sqrt{x^2+y^2}) dx$ for $x>0$ passes through the point $(1, 0)$ . What is the y-intercept of the line tangent to this solution curve at $x=\sqrt{3}$ ?

$0$
$-1$
$-2$ (correct answer)
$1-2\sqrt{3}$

Explanation: When you encounter a differential equation like this one, recognize it as a homogeneous equation that can be solved using the substitution method. The presence of

\sqrt{x^2+y^2}

is a key indicator that the substitution

v = \frac{y}{x}

(so

y = vx

) will simplify the equation. Making this substitution and using

\frac{dy}{dx} = v + x\frac{dv}{dx}

, the equation becomes:

x(v + x\frac{dv}{dx}) = v + \sqrt{1+v^2}

This simplifies to

x\frac{dv}{dx} = \sqrt{1+v^2}

, which separates as:

\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}

Integrating both sides gives

\sinh^{-1}(v) = \ln|x| + C

. Using the initial condition

(1,0)

where

v = 0

, we find

C = 0

. Therefore

v = \sinh(\ln x) = \frac{x^2-1}{2x}

, so

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

. At

x = \sqrt{3}

, we have

y = \frac{3-1}{2} = 1

and

\frac{dy}{dx} = x = \sqrt{3}

. The tangent line at

(\sqrt{3}, 1)

with slope

\sqrt{3}

is:

y - 1 = \sqrt{3}(x - \sqrt{3})

y = \sqrt{3}x - 2

The y-intercept is

-2

, which is choice C. Choice A (

0

) might come from incorrectly using the initial point. Choice B (

-1

) could result from calculation errors in finding the slope or intercept. Choice D (

1-2\sqrt{3}

) might arise from sign errors or incorrect substitution of the point coordinates. Always verify your solution satisfies both the differential equation and initial condition before finding tangent lines.

Question 6

The general solution to the homogeneous differential equation $x \frac{dy}{dx} = y + x e^{y/x}$ can be written in the implicit form $f(x, y) = C$ . Which of the following is a valid expression for $f(x, y)$ ?

$ln|x| + e^{-y/x}$ (correct answer)
$ln|x| - e^{-y/x}$
$e^{y/x} - ln|x|$
$e^{-y/x} + x$

Explanation: First, rewrite the equation as

\frac{dy}{dx} = \frac{y}{x} + e^{y/x}

. This is a homogeneous equation. Use the substitution

y = vx

, which implies

\frac{dy}{dx} = v + x \frac{dv}{dx}

. The equation becomes

v + x \frac{dv}{dx} = v + e^v

. This simplifies to

x \frac{dv}{dx} = e^v

. Separating variables, we get

e^{-v} dv = \frac{dx}{x}

. Integrating both sides yields

\int e^{-v} dv = \int \frac{dx}{x}

, which gives

-e^{-v} = ln|x| + C_1

. Substituting back

v = y/x

, we have

-e^{-y/x} = ln|x| + C_1

. Rearranging to the form

f(x, y) = C

, we get

ln|x| + e^{-y/x} = -C_1

. Letting

C = -C_1

, the solution is

ln|x| + e^{-y/x} = C

Question 7

The differential equation $(x^2+3y^2)dx - 2xy\,dy = 0$ is homogeneous. After applying the substitution $y=vx$ , which of the following separable differential equations in terms of $v$ and $x$ is obtained?

$\frac{2v}{1+v^2} dv = \frac{dx}{x}$ (correct answer)
$\frac{2v}{1+5v^2} dv = \frac{dx}{x}$
$\frac{2v}{1+3v^2} dv = \frac{dx}{x}$
$\frac{1+3v^2}{v(1-3v^2)}dv = \frac{dx}{x}$

Explanation: First, rewrite the equation in the form

\frac{dy}{dx} = f(x,y)

\frac{dy}{dx} = \frac{x^2+3y^2}{2xy}

. Dividing the numerator and denominator by

x^2

gives

\frac{dy}{dx} = \frac{1+3(y/x)^2}{2(y/x)}

. Now, let

v = y/x

, so

\frac{dy}{dx} = v + x\frac{dv}{dx}

. The equation becomes

v + x\frac{dv}{dx} = \frac{1+3v^2}{2v}

. To separate the variables, we isolate

x\frac{dv}{dx}

x\frac{dv}{dx} = \frac{1+3v^2}{2v} - v = \frac{1+3v^2 - 2v^2}{2v} = \frac{1+v^2}{2v}

. Finally, separating the variables gives

\frac{2v}{1+v^2} dv = \frac{dx}{x}

B is incorrect because it results from adding $v$ to the right side instead of subtracting it.
C is incorrect because it comes from forgetting the $v$ term in the product rule substitution for $\frac{dy}{dx}$ , i.e., using $\frac{dy}{dx} = x\frac{dv}{dx}$ .
D is incorrect because it would arise from incorrectly setting up the initial derivative as $\frac{dy}{dx} = \frac{2xy}{x^2+3y^2}$ .

Question 8

The general solution to the homogeneous differential equation $x \frac{dy}{dx} = y + \sqrt{y^2-x^2}$ for $y>x>0$ is sought. Which of the following equations implicitly defines the solution family?

$y + \sqrt{y^2-x^2} = Cx^2$ (correct answer)
$\arcsin(y/x) = \ln|x| + C$
$y - \sqrt{y^2-x^2} = Cx^2$
$y + \sqrt{y^2-x^2} = Cx$

Explanation: First, divide by

x

to get

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

. Let

v=y/x

, so

\frac{dy}{dx}=v+x\frac{dv}{dx}

. The equation becomes

v+x\frac{dv}{dx} = v + \sqrt{v^2-1}

. This simplifies to

x\frac{dv}{dx} = \sqrt{v^2-1}

. Separating variables yields

\frac{dv}{\sqrt{v^2-1}} = \frac{dx}{x}

. Integrating both sides gives

\ln|v+\sqrt{v^2-1}| = \ln|x| + C_1

. Let

C_1 = \ln|C|

, so

\ln|v+\sqrt{v^2-1}| = \ln|Cx|

. This implies

v+\sqrt{v^2-1} = Cx

. Substituting

v=y/x

back gives

\frac{y}{x} + \sqrt{(\frac{y}{x})^2-1} = Cx

, which simplifies to

\frac{y+\sqrt{y^2-x^2}}{x} = Cx

. Therefore,

y+\sqrt{y^2-x^2} = Cx^2

B is incorrect due to a common integration error, confusing the integral of $1/\sqrt{v^2-1}$ with that of $1/\sqrt{1-v^2}$ .
C is incorrect and would result from a sign error, for instance starting with $x\frac{dv}{dx} = -\sqrt{v^2-1}$ .
D is incorrect due to an algebraic error in the final step, failing to multiply $Cx$ by $x$ .

Question 9

The general solution to $x\,dy - y\,dx = \sqrt{x^2-y^2}\,dx$ for $x>0$ is found using a substitution $y=vx$ . The resulting separated equation can be written as $\frac{dv}{g(v)} = \frac{dx}{x}$ . What is the function $g(v)$ ?

$-\sqrt{1-v^2}$
$v+\sqrt{1-v^2}$
$\sqrt{1-v^2}$ (correct answer)
$\sqrt{v^2-1}$

Explanation: When you encounter a differential equation with terms like

x\,dy - y\,dx

alongside expressions involving

\sqrt{x^2-y^2}

, this signals a homogeneous differential equation that's perfect for the substitution

y = vx

. Let's work through the substitution systematically. Starting with

y = vx

, we get

dy = v\,dx + x\,dv

. Substituting into the original equation:

x(v\,dx + x\,dv) - vx\,dx = \sqrt{x^2-(vx)^2}\,dx

Expanding the left side:

xv\,dx + x^2\,dv - vx\,dx = \sqrt{x^2-v^2x^2}\,dx

The

xv\,dx

terms cancel, leaving:

x^2\,dv = \sqrt{x^2(1-v^2)}\,dx

Since

x > 0

, we have

\sqrt{x^2(1-v^2)} = x\sqrt{1-v^2}

, so:

x^2\,dv = x\sqrt{1-v^2}\,dx

Dividing both sides by

x

x\,dv = \sqrt{1-v^2}\,dx

Separating variables:

\frac{dv}{\sqrt{1-v^2}} = \frac{dx}{x}

Therefore,

g(v) = \sqrt{1-v^2}

, which is choice C. Looking at the wrong answers: A has an incorrect negative sign that would flip the direction of integration. B adds an extra

v

term that doesn't appear in our derivation. D uses

\sqrt{v^2-1}

instead of

\sqrt{1-v^2}

, which would only be valid if

|v| \geq 1

, contradicting our domain requirements. Study tip: When using

y = vx

substitution, always remember that

dy = v\,dx + x\,dv

and carefully track sign changes during algebraic manipulation.

Question 10

The differential equation $\frac{dy}{dx} = \frac{2xy}{y^2-x^2}$ can be solved as a homogeneous equation. While both standard substitutions ( $y=vx$ and $x=vy$ ) are valid, one leads to a significantly simpler integration process. Why is the substitution $x=vy$ preferable in this case?

Because the term $y^2-x^2$ in the denominator suggests expressing $x$ in terms of $y$ .
Because it transforms the equation into a first-order linear equation in $v$ and $y$ .
Because the resulting separable equation for $v$ can be solved with a basic $u$ -substitution, whereas the alternative requires partial fractions. (correct answer)
Because the substitution $y=vx$ fails, as it would lead to division by zero for any solution passing through the line $y=x$ .

Explanation: Let's analyze both substitutions. With

y=vx

, the equation becomes

v+x\frac{dv}{dx} = \frac{2v}{v^2-1}

, which leads to

x\frac{dv}{dx} = \frac{3v-v^3}{v^2-1}

. The separated form is

\frac{v^2-1}{v(3-v^2)}dv = \frac{dx}{x}

, which requires integration using partial fractions. Now consider

x=vy

. This implies

\frac{dx}{dy} = \frac{y^2-x^2}{2xy} = \frac{1-(x/y)^2}{2(x/y)}

. Let

v=x/y

, so

\frac{dx}{dy} = v+y\frac{dv}{dy}

. The equation becomes

v+y\frac{dv}{dy} = \frac{1-v^2}{2v}

, which leads to

y\frac{dv}{dy} = \frac{1-3v^2}{2v}

. The separated form is

\frac{2v}{1-3v^2}dv = \frac{dy}{y}

. The integral on the left can be solved with a simple

u

-substitution (

u=1-3v^2

). Thus, the substitution

x=vy

is preferable because it leads to a much simpler integral.

A is incorrect as this reasoning is heuristic and not mathematically precise.
B is incorrect because the substitution leads to a separable equation, not a linear one.
D is incorrect because while domain issues can be a concern, they are not the primary reason for preferring one substitution over the other based on computational efficiency.

Question 11

Consider a solution curve $y=f(x)$ for a homogeneous first-order differential equation $\frac{dy}{dx} = G(y/x)$ . If this curve passes through the point $(a, b)$ , which of the following statements about other points and curves must be true?

All solution curves are straight lines passing through the origin.
The curve defined by $y_c(x) = c f(x/c)$ for any constant $c > 0$ is also a solution curve. (correct answer)
The slope of the tangent to any solution curve at $(ka, kb)$ is $k$ times the slope at $(a,b)$ for any $k > 0$ .
The function $y = f(x) + C$ for any constant $C$ represents the family of all solution curves.

Explanation: A key geometric property of homogeneous equations is that their direction fields are constant along any ray from the origin. This means that if the

(x,y)

plane is scaled by a factor

c

, the solution curves are mapped onto other solution curves. If

y=f(x)

is a solution, then the scaled function

y_c(x) = c f(x/c)

is also a solution. This can be verified by differentiation:

\frac{dy_c}{dx} = c f'(x/c) \cdot (1/c) = f'(x/c)

. Since

f(x)

is a solution,

f'(x) = G(f(x)/x)

. Thus,

\frac{dy_c}{dx} = G(f(x/c)/(x/c))

. Also,

G(y_c/x) = G(cf(x/c)/x) = G(f(x/c)/(x/c))

. So,

\frac{dy_c}{dx} = G(y_c/x)

, and

y_c(x)

is a solution.

A is incorrect. This is only true for the simple case $y' = y/x$ .
C is incorrect. The slope depends only on the ratio $y/x$ . At $(ka, kb)$ , the slope is $G(kb/ka) = G(b/a)$ , which is the same as the slope at $(a,b)$ .
D is incorrect. This describes vertical shifting, which is a property of solutions to equations of the form $y' = h(x)$ , not homogeneous equations.

Question 12

A student is solving the equation $x y' = 2x + y$ . They correctly identify it as a homogeneous equation and apply the substitution $y=vx$ . However, they make a common error in the substitution for $y'$ , replacing it with $x\frac{dv}{dx}$ instead of the correct $v+x\frac{dv}{dx}$ . Following this incorrect procedure, what general solution do they obtain?

$y = Cx - 2x\ln|x|$
$y = 2x\ln|x| + Cx$
$y = Ce^x - 2x - 2$
$y = Cx^2 - 2x$ (correct answer)

Explanation: This problem tests your understanding of the substitution method for homogeneous differential equations, specifically what happens when you make a common algebraic error during the substitution process. Starting with

xy' = 2x + y

and using the substitution

y = vx

, you need to find

y'

. The correct derivative is

y' = v + x\frac{dv}{dx}

using the product rule. However, the student incorrectly uses

y' = x\frac{dv}{dx}

, missing the

v

term. With this incorrect substitution, the equation becomes:

x \cdot x\frac{dv}{dx} = 2x + vx

x^2\frac{dv}{dx} = x(2 + v)

x\frac{dv}{dx} = 2 + v

\frac{dv}{dx} = \frac{2 + v}{x}

This separates to

\frac{dv}{2 + v} = \frac{dx}{x}

. Integrating both sides gives

\ln|2 + v| = \ln|x| + C_1

, which leads to

2 + v = Cx

, so

v = Cx - 2

. Since

y = vx

, the incorrect solution is

y = Cx^2 - 2x

. Answer D matches this flawed derivation. Answer A represents the correct solution to the original equation. Answer B looks similar to A but with terms reversed. Answer C suggests an exponential solution, which would arise from a completely different type of equation. Study tip: When using

y = vx

substitution, always remember that

y' = v + x\frac{dv}{dx}

by the product rule—the

v

term is crucial and easily forgotten under exam pressure.

Question 13

A homogeneous differential equation of the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$ is given by $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$ . After applying the substitution $v = \frac{y}{x}$ and reaching the separated form, the student obtains $\int \frac{2v}{v^2 - 1} dv = \int \frac{dx}{x}$ . Which of the following represents the correct evaluation of the left-hand integral?

$\ln|v^2 - 1|^{1/2} + C$
$\frac{1}{2}\ln|v^2 - 1| + C$
$\ln|v^2 - 1| + C$ (correct answer)
$2\ln|v - 1| - 2\ln|v + 1| + C$

Explanation: When you encounter a homogeneous differential equation and need to evaluate an integral after substitution, the key is recognizing the correct integration technique and applying it carefully. To evaluate

\int \frac{2v}{v^2 - 1} dv

, you should use substitution. Let

u = v^2 - 1

, so

du = 2v \, dv

. This transforms the integral to

\int \frac{1}{u} du = \ln|u| + C

. Substituting back gives

\ln|v^2 - 1| + C

. Answer C is correct because it represents the proper application of the natural logarithm integration rule after the appropriate substitution. Answer A,

\ln|v^2 - 1|^{1/2} + C

, incorrectly applies logarithm properties. While

\ln|v^2 - 1|^{1/2} = \frac{1}{2}\ln|v^2 - 1|

, this suggests the student might have confused the coefficient placement. Answer B,

\frac{1}{2}\ln|v^2 - 1| + C

, represents a common error where students incorrectly handle the coefficient 2 in the numerator. This would be the result if you mistakenly treated the integral as

\frac{1}{2} \int \frac{2v}{v^2 - 1} dv

instead of recognizing the perfect substitution. Answer D uses partial fraction decomposition, which is unnecessary here. While

\frac{2v}{v^2 - 1}

could theoretically be decomposed, the direct substitution method is more efficient and less error-prone. Study tip: When you see a rational function where the numerator is the derivative (or a constant multiple) of the denominator, always try

u

-substitution first before attempting partial fractions. This pattern appears frequently in differential equations.

Question 14

A differential equation of the form $\frac{dy}{dx} = \frac{ax + by}{cx + dy}$ is homogeneous if and only if a certain condition is satisfied. Given the equation $\frac{dy}{dx} = \frac{3x + 2y}{x + ky}$ where $k$ is a parameter, for which value(s) of $k$ can this equation be solved using the substitution $v = \frac{y}{x}$ ?

$k = \frac{2}{3}$ only
$k = \frac{3}{2}$ only
Any value of $k$ except $k = 0$
Any real value of $k$ including $k = 0$ (correct answer)

Explanation: A differential equation

\frac{dy}{dx} = \frac{ax + by}{cx + dy}

is homogeneous when both the numerator and denominator are homogeneous functions of the same degree. For

\frac{dy}{dx} = \frac{3x + 2y}{x + ky}

, both the numerator

3x + 2y

and denominator

x + ky

are homogeneous functions of degree 1 (each term has total degree 1 in

x

and

y

). This is true regardless of the value of

k

. We can verify by dividing numerator and denominator by

x

\frac{dy}{dx} = \frac{3 + 2\frac{y}{x}}{1 + k\frac{y}{x}}

, which is a function of

\frac{y}{x}

alone for any value of

k

, including

k = 0

(which gives

\frac{dy}{dx} = 3 + 2\frac{y}{x}

). The substitution

v = \frac{y}{x}

will work for any value of

k

. Some students might think

k = 0

causes problems because the denominator becomes just

x

, but this doesn't prevent the homogeneous substitution from working. Others might think there's a specific relationship needed between the coefficients, but homogeneity only requires that all terms have the same degree, which is satisfied here for any

k

Question 15

The differential equation $x\frac{dy}{dx} = y + \sqrt{x^2 + y^2}$ can be solved using a homogeneous substitution. After making the appropriate substitution and simplification, the resulting equation in terms of the new variable requires which integration technique?

Direct integration after separating variables, yielding a logarithmic solution
Trigonometric substitution using $v = \tan(\theta)$ to handle the square root (correct answer)
Integration by parts followed by separation of variables
Partial fraction decomposition after rationalizing the denominator

Explanation: Dividing by

x

\frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x} = \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2}

. With

v = \frac{y}{x}

, this becomes

v + x\frac{dv}{dx} = v + \sqrt{1 + v^2}

, so

x\frac{dv}{dx} = \sqrt{1 + v^2}

. Separating:

\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}

. The left side integral

\int \frac{dv}{\sqrt{1 + v^2}}

is a standard form that requires the substitution

v = \tan(\theta)

dv = \sec^2(\theta)d\theta

, and

\sqrt{1 + v^2} = \sqrt{1 + \tan^2(\theta)} = \sec(\theta)

. This gives

\int \frac{\sec^2(\theta)d\theta}{\sec(\theta)} = \int \sec(\theta)d\theta = \ln|\sec(\theta) + \tan(\theta)| + C

. Choice A is incorrect because direct integration isn't possible with the square root. Choice C is wrong because integration by parts isn't needed. Choice D is incorrect because partial fractions don't apply to this radical expression.

Question 16

Consider the differential equation $\frac{dy}{dx} = \frac{x^2 - y^2}{x^2 + xy}$ . A student claims this equation is homogeneous and applies the substitution $y = vx$ . After simplification, they obtain $\frac{dv}{dx} = \frac{1 - 3v^2}{x(1 + v)}$ . What error, if any, did the student make?

The equation is not homogeneous, so the substitution $y = vx$ is inappropriate
The student correctly identified the equation as homogeneous but made an algebraic error in the substitution (correct answer)
The student should have used $v = \frac{x}{y}$ instead of $y = vx$ for this particular equation
The student's work is correct; the equation simplifies to the given result

Explanation: First, check if the equation is homogeneous by dividing numerator and denominator by

x^2

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

. This is indeed a function of

\frac{y}{x}

only, so it's homogeneous. With

v = \frac{y}{x}

, we have

\frac{dy}{dx} = \frac{1 - v^2}{1 + v}

and

y = vx

gives

\frac{dy}{dx} = v + x\frac{dv}{dx}

. Setting equal:

v + x\frac{dv}{dx} = \frac{1 - v^2}{1 + v}

. Solving for

\frac{dv}{dx}

x\frac{dv}{dx} = \frac{1 - v^2}{1 + v} - v = \frac{1 - v^2 - v(1 + v)}{1 + v} = \frac{1 - v^2 - v - v^2}{1 + v} = \frac{1 - v - 2v^2}{1 + v}

. Therefore

\frac{dv}{dx} = \frac{1 - v - 2v^2}{x(1 + v)}

. The student wrote

\frac{1 - 3v^2}{x(1 + v)}

, missing the

-v

term in the numerator. Choice A is wrong because the equation is homogeneous. Choice C is unnecessary since

y = vx

is the standard substitution. Choice D is incorrect due to the algebraic error.

Question 17

The general solution to the homogeneous differential equation $\frac{dy}{dx} = \frac{y^2 + xy}{x^2}$ can be written in the form $y = \frac{x}{C - \ln|x|}$ where $C$ is an arbitrary constant. If this solution satisfies the initial condition $y(e) = \frac{e}{2}$ , what is the particular value of $C$ ?

$C = 1$
$C = 2$
$C = 3$ (correct answer)
$C = \frac{3}{2}$

Explanation: When you encounter a differential equation problem where the general solution is already provided, your task shifts from solving the equation to applying initial conditions to find the specific constant value. Given the general solution

y = \frac{x}{C - \ln|x|}

and the initial condition

y(e) = \frac{e}{2}

, you substitute these values directly into the equation. At

x = e

, we have

y = \frac{e}{2}

, so:

\frac{e}{2} = \frac{e}{C - \ln|e|}

Since

\ln|e| = \ln(e) = 1

, this becomes:

\frac{e}{2} = \frac{e}{C - 1}

Cross-multiplying gives us

e(C - 1) = 2e

. Dividing both sides by

e

C - 1 = 2

, therefore

C = 3

. Let's examine why the other options fail. Option A (

C = 1

) would make the denominator

1 - 1 = 0

, creating an undefined expression at

x = e

. Option B (

C = 2

) gives us

\frac{e}{2-1} = e

, which is twice the required value of

\frac{e}{2}

. Option D (

C = \frac{3}{2}

) yields

\frac{e}{\frac{3}{2}-1} = \frac{e}{\frac{1}{2}} = 2e

, which is four times too large. When working with initial value problems, always substitute the given point carefully and watch for cases where your constant might create undefined expressions. Double-check your arithmetic by substituting your answer back into the original condition.

Question 18

Consider the differential equation $\frac{dy}{dx} = \frac{x^2 + xy}{y^2 + xy}$ . After applying the substitution $v = \frac{y}{x}$ and simplifying, which of the following represents the resulting separable equation in terms of $v$ and $x$ ?

$\frac{dv}{dx} = \frac{1 + v - v^2}{x(v^2 + v)}$ (correct answer)
$\frac{dv}{dx} = \frac{1 - v^2}{x(v^2 + v)}$
$\frac{dv}{dx} = \frac{1 + v}{x(v^2 + v)} - \frac{v}{x}$
$\frac{dv}{dx} = \frac{1 - v}{x(v^2 + v)}$

Explanation: Starting with

\frac{dy}{dx} = \frac{x^2 + xy}{y^2 + xy}

, divide numerator and denominator by

x^2

to get

\frac{dy}{dx} = \frac{1 + \frac{y}{x}}{\frac{y^2}{x^2} + \frac{y}{x}}

. With

v = \frac{y}{x}

, this becomes

\frac{dy}{dx} = \frac{1 + v}{v^2 + v}

. Since

y = vx

, we have

\frac{dy}{dx} = v + x\frac{dv}{dx}

. Setting equal:

v + x\frac{dv}{dx} = \frac{1 + v}{v^2 + v}

. Solving for

\frac{dv}{dx}

x\frac{dv}{dx} = \frac{1 + v}{v^2 + v} - v = \frac{1 + v - v(v^2 + v)}{v^2 + v} = \frac{1 + v - v^3 - v^2}{v^2 + v} = \frac{1 + v - v^2}{v^2 + v}

. Therefore

\frac{dv}{dx} = \frac{1 + v - v^2}{x(v^2 + v)}

. Choice B omits the

v

term, choice C incorrectly separates terms, and choice D has the wrong sign.

Question 19

A particular solution to the differential equation $(x^2 + 3y^2) dx - 2xy dy = 0$ passes through the point $(1, 1)$ . What is the value of $y$ when $x=2$ ?

$2\sqrt{3}$ (correct answer)
$2\sqrt{2}$
$\sqrt{14}$
$4$

Explanation: The equation can be rewritten as

dy/dx = (x^2 + 3y^2) / (2xy)

. Dividing the numerator and denominator by

x^2

gives

dy/dx = (1 + 3(y/x)^2) / (2(y/x))

, which is a homogeneous equation. Let

v = y/x

, so

y = vx

and

dy/dx = v + x(dv/dx)

. The equation becomes

v + x(dv/dx) = (1 + 3v^2)/(2v)

. Rearranging gives

x(dv/dx) = (1 + 3v^2)/(2v) - v = (1 + v^2)/(2v)

. Separating variables yields

(2v / (1 + v^2)) dv = dx/x

. Integrating both sides gives

ln(1 + v^2) = ln|x| + C

. Using the initial condition

y(1)=1

, we have

x=1

and

v=y/x=1

. So,

ln(1+1^2) = ln(1) + C

, which implies

C = ln(2)

. The solution is

ln(1+v^2) = ln|x| + ln(2) = ln(2|x|)

. For

x>0

, this is

1+v^2 = 2x

. Substituting back

v=y/x

, we get

1 + (y/x)^2 = 2x

, which simplifies to

x^2 + y^2 = 2x^3

. To find

y

when

x=2

, we substitute

x=2

into the solution:

2^2 + y^2 = 2(2^3)

, which is

4 + y^2 = 16

. Thus,

y^2 = 12

, and since the initial condition has

y>0

, we take the positive root,

y = \sqrt{12} = 2\sqrt{3}

Question 20

Find the particular solution to the differential equation $(x - y) dy = (x + y) dx$ that satisfies the initial condition $y(1) = 1$ .

$2\arctan(y/x) - \ln(x^2+y^2) = \pi/2 - \ln(2)$ (correct answer)
$2\arctan(y/x) + \ln(x^2+y^2) = \pi/2 + \ln(2)$
$\ln(x^2+y^2) - 2\arctan(y/x) = \ln(2) - \pi/2$
$y=x$

Explanation: Rewrite as

dy/dx = (x+y)/(x-y) = (1+y/x)/(1-y/x)

. Let

y=vx

, so

y' = v+xv'

. The equation becomes

v+xv' = (1+v)/(1-v)

. Then

xv' = (1+v)/(1-v) - v = (1+v - v + v^2)/(1-v) = (1+v^2)/(1-v)

. Separate variables:

((1-v)/(1+v^2)) dv = dx/x

. Integrate:

\int (1/(1+v^2) - v/(1+v^2)) dv = \int dx/x

. This gives

\arctan(v) - (1/2)\ln(1+v^2) = \ln|x| + C

. Substitute back

v=y/x

\arctan(y/x) - (1/2)\ln(1+(y/x)^2) = \ln|x| + C

. This simplifies to

\arctan(y/x) - (1/2)\ln(x^2+y^2) = C

. Using

y(1)=1

\arctan(1) - (1/2)\ln(1^2+1^2) = C

, so

C = \pi/4 - (1/2)\ln(2)

. The solution is

\arctan(y/x) - (1/2)\ln(x^2+y^2) = \pi/4 - (1/2)\ln(2)

. Multiplying by 2 gives

2\arctan(y/x) - \ln(x^2+y^2) = \pi/2 - \ln(2)