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.

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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 $x=2$ ?

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.

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Question 1

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 2

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

Question 3

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 4

Given the initial value problem $\frac{dy}{dx} = \frac{y}{x} + \tan(\frac{y}{x})$ with $y(1) = \pi/6$ , what is the value of $y(\sqrt{3})$ ?

$\frac{\pi\sqrt{3}}{6}$
$\frac{\pi}{3}$
$\frac{2\pi\sqrt{3}}{3}$
$\frac{\pi}{\sqrt{3}}$ (correct answer)

Explanation: When you encounter a differential equation of the form $\frac{dy}{dx} = f(\frac{y}{x})$ , you're dealing with a homogeneous equation. The key strategy is to use the substitution $v = \frac{y}{x}$ , which means $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ . Applying this substitution to $\frac{dy}{dx} = \frac{y}{x} + \tan(\frac{y}{x})$ , we get: $v + x\frac{dv}{dx} = v + \tan(v)$ This simplifies to $x\frac{dv}{dx} = \tan(v)$ , or $\frac{dv}{\tan(v)} = \frac{dx}{x}$ . Since $\frac{1}{\tan(v)} = \cot(v)$ , we have $\cot(v)dv = \frac{dx}{x}$ . Integrating both sides: $\ln|\sin(v)| = \ln|x| + C$ . This gives us $\sin(v) = Cx$ , or $\sin(\frac{y}{x}) = Cx$ . Using the initial condition $y(1) = \frac{\pi}{6}$ : $\sin(\frac{\pi/6}{1}) = C \cdot 1$ , so $C = \sin(\frac{\pi}{6}) = \frac{1}{2}$ . Therefore, $\sin(\frac{y}{x}) = \frac{x}{2}$ . At $x = \sqrt{3}$ : $\sin(\frac{y}{\sqrt{3}}) = \frac{\sqrt{3}}{2}$ . Since $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ , we have $\frac{y}{\sqrt{3}} = \frac{\pi}{3}$ , giving us $y = \frac{\pi}{\sqrt{3}}$ . Choice A incorrectly uses the original substitution without proper integration. Choice B forgets to multiply back by $x$ . Choice C appears to involve computational errors in the trigonometric evaluation. Remember: homogeneous differential equations always call for the substitution $v = \frac{y}{x}$ to separate variables effectively.

Question 5

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 6

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 7

Find the particular solution to $x^2 \frac{dy}{dx} = y^2 - xy$ that satisfies the initial condition $y(1)=-1$ , and express $y$ as an explicit function of $x$ .

$y = \frac{2x}{1+x^2}$
$y = \frac{2x^3}{x^2-3}$
$y = \frac{2x}{1-3x^2}$ (correct answer)
$y = 2x(1-3x^2)$

Explanation: When you encounter a differential equation like $x^2 \frac{dy}{dx} = y^2 - xy$ , look for opportunities to use separation of variables or recognize it as a specific type like Bernoulli or homogeneous. Let's rearrange this equation: $x^2 \frac{dy}{dx} = y(y - x)$ , so $\frac{dy}{dx} = \frac{y(y-x)}{x^2}$ . This is a homogeneous differential equation since we can write it as $\frac{dy}{dx} = \frac{y}{x} \cdot \frac{y/x - 1}{x/x} = \frac{y}{x} \cdot (y/x - 1)$ . For homogeneous equations, substitute $v = \frac{y}{x}$ , so $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ . Substituting: $v + x\frac{dv}{dx} = v(v-1) = v^2 - v$ . This simplifies to $x\frac{dv}{dx} = v^2 - 2v = v(v-2)$ . Separating variables: $\frac{dv}{v(v-2)} = \frac{dx}{x}$ . Using partial fractions: $\frac{1}{v(v-2)} = \frac{1/2}{v-2} - \frac{1/2}{v}$ . Integrating both sides and applying the initial condition $y(1) = -1$ (so $v = -1$ when $x = 1$ ), we get $v = \frac{2}{1-3x^2}$ . Since $y = vx$ , we have $y = \frac{2x}{1-3x^2}$ . Answer C is correct. Answer A has the wrong denominator form. Answer B incorrectly includes $x^3$ in the numerator and has a different denominator structure. Answer D isn't even a rational function. Strategy tip: For homogeneous differential equations, always try the substitution $v = y/x$ to convert them into separable equations.

Question 8

The general solution to the differential equation $2xy y' = y^2-x^2$ represents a family of curves. Which of the following provides the best geometric description of this family for $x > 0$ ?

Parabolas with vertices at the origin.
Circles centered at the origin.
Circles passing through the origin with centers on the x-axis. (correct answer)
Circles passing through the origin with centers on the y-axis.

Explanation: When you encounter a differential equation that asks about the geometric description of its solution family, your goal is to solve the equation and identify the resulting curve type. This differential equation $2xy y' = y^2 - x^2$ can be solved using substitution. Rearranging gives $y' = \frac{y^2 - x^2}{2xy}$ . This suggests trying the substitution $v = \frac{y}{x}$ , so $y = vx$ and $y' = v + xv'$ . Substituting: $v + xv' = \frac{(vx)^2 - x^2}{2x(vx)} = \frac{x^2(v^2 - 1)}{2x^2v} = \frac{v^2 - 1}{2v}$ This simplifies to $xv' = \frac{v^2 - 1}{2v} - v = \frac{v^2 - 1 - 2v^2}{2v} = \frac{-v^2 - 1}{2v}$ Separating variables and integrating leads to the general solution: $x^2 + y^2 = 2Cx$ , where $C$ is an arbitrary constant. Completing the square gives $(x - C)^2 + y^2 = C^2$ , which represents circles with centers at $(C, 0)$ and radius $|C|$ . Since the radius equals the x-coordinate of the center, these circles all pass through the origin. Choice A is wrong because these aren't parabolas—they're clearly circular equations. Choice B is incorrect because the centers are at $(C, 0)$ , not at the origin. Choice D places centers on the y-axis, but our centers are at $(C, 0)$ on the x-axis. Study tip: When solving differential equations geometrically, always complete the square on the final implicit equation to identify the standard form and geometric properties of the curve family.

Question 9

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 10

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 11

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 12

Find the particular solution to the initial value problem $x^2 \frac{dy}{dx} = y^2 + xy$ , with $y(1) = -2$ .

$y = \frac{-2x}{2\ln x + 1}$ (correct answer)
$y = \frac{2x^2}{1-2x}$
$y = \frac{2x}{1 - 2\ln x}$
$y = -2x^{3/2}$

Explanation: The equation can be rewritten as $\frac{dy}{dx} = (\frac{y}{x})^2 + \frac{y}{x}$ , which is a homogeneous equation. Let $v = y/x$ , so $y=vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ . Substituting gives $v + x\frac{dv}{dx} = v^2 + v$ , which simplifies to $x\frac{dv}{dx} = v^2$ . This is a separable equation: $\frac{dv}{v^2} = \frac{dx}{x}$ . Integrating both sides gives $-\frac{1}{v} = \ln|x| + C$ . Substituting back $v=y/x$ gives $-\frac{x}{y} = \ln|x| + C$ . Using the initial condition $y(1)=-2$ : $-\frac{1}{-2} = \ln|1| + C$ , which implies $C = 1/2$ . The implicit solution is $-\frac{x}{y} = \ln x + \frac{1}{2}$ for $x>0$ . Solving for $y$ gives $y = \frac{-x}{\ln x + 1/2} = \frac{-2x}{2\ln x + 1}$ .

B is incorrect because it results from the common error of replacing $\frac{dy}{dx}$ with just $x\frac{dv}{dx}$ instead of $v+x\frac{dv}{dx}$ .
C is incorrect due to a sign error when solving for the constant of integration or manipulating the final equation.
D is incorrect due to a fundamental error in integration, such as treating $\int v^{-2}dv$ as $\ln(v^2)$ .

Question 13

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 14

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 15

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 16

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 17

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 18

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 19

Find the explicit solution to the initial value problem $y' = \frac{y}{x} + \cos^2(\frac{y}{x})$ , with $y(1) = \pi/4$ .

$y = x \arctan(\ln x)$
$y = x \arctan(\ln x + 1)$ (correct answer)
$y = x (\ln x + \pi/4)$
$y = x \,\mathrm{arccot}(1-\ln x)$

Explanation: When you encounter a differential equation of the form $y' = f\left(\frac{y}{x}\right)$ , this signals a homogeneous equation that requires the substitution $v = \frac{y}{x}$ , making $y = vx$ and $y' = v'x + v$ . Applying this substitution to $y' = \frac{y}{x} + \cos^2\left(\frac{y}{x}\right)$ : $v'x + v = v + \cos^2(v)$ $v'x = \cos^2(v)$ $\frac{dv}{\cos^2(v)} = \frac{dx}{x}$ Integrating both sides: $\int \sec^2(v) \, dv = \int \frac{dx}{x}$ This gives us $\tan(v) = \ln|x| + C$ , so $v = \arctan(\ln|x| + C)$ . Since $y = vx$ , we have $y = x \arctan(\ln|x| + C)$ . Using the initial condition $y(1) = \pi/4$ : $\frac{\pi}{4} = 1 \cdot \arctan(\ln(1) + C) = \arctan(0 + C) = \arctan(C)$ Therefore $C = \tan(\pi/4) = 1$ , giving us $y = x \arctan(\ln x + 1)$ , which is choice B. Choice A omits the constant of integration. Choice C incorrectly assumes a linear relationship without proper integration of the trigonometric term. Choice D uses arccotangent instead of arctangent and has the wrong sign pattern. For homogeneous differential equations, always remember the key substitution $v = \frac{y}{x}$ and don't forget to apply your initial conditions to find the integration constant.

Question 20

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

$-2 + \sqrt{10}$ (correct answer)
$-2 - \sqrt{10}$
$1$
$0$

Explanation: The equation is homogeneous. Let $v=y/x$ . Then $\frac{dy}{dx} = \frac{1-y/x}{1+y/x} = \frac{1-v}{1+v}$ . Using $\frac{dy}{dx} = v+x\frac{dv}{dx}$ , we get $v+x\frac{dv}{dx} = \frac{1-v}{1+v}$ . This leads to $x\frac{dv}{dx} = \frac{1-v}{1+v} - v = \frac{1-v-v-v^2}{1+v} = -\frac{v^2+2v-1}{v+1}$ . Separating variables: $\frac{v+1}{v^2+2v-1}dv = -\frac{dx}{x}$ . Integrating gives $\frac{1}{2}\ln|v^2+2v-1| = -\ln|x|+C$ . Substituting back $v=y/x$ and multiplying by $x^2$ leads to the implicit solution $y^2+2xy-x^2=k$ . Using the initial condition $y(1)=1$ , we find $1^2+2(1)(1)-1^2 = k$ , so $k=2$ . The particular solution is $y^2+2xy-x^2=2$ . At $x=2$ , we have $y^2+4y-4=2$ , or $y^2+4y-6=0$ . Using the quadratic formula, $y = \frac{-4 \pm \sqrt{16-4(-6)}}{2} = -2 \pm \sqrt{10}$ . Since the solution starts at $y(1)=1$ (positive), and the function is continuous, we choose the positive root, $y(2) = -2+\sqrt{10} \approx 1.16$ .

B is incorrect as it is the other root from the quadratic formula, which does not match the trajectory of the solution curve starting at $(1,1)$ .
C is incorrect and could arise from an error in calculating the constant $k$ , for example, finding $k=1$ , which leads to the equation $y^2+4y-5=0$ at $x=2$ .
D is incorrect and would result from a sign error during separation leading to the implicit solution $y^2+2xy+x^2=k$ .

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