Differential Equations Quiz: Classifying Odes

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Classifying OdesQuestion 1 of 20

The differential equation $\frac{dy}{dx} = \frac{2y}{x}$ can be solved using several methods because it fits multiple classifications. Which of the following classifications does NOT apply to this equation?

Separable

First-order linear

Homogeneous

Exact

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Differential Equations Quiz

Differential Equations Quiz: Classifying Odes

Practice Classifying Odes 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 Classifying Odes, 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

Separable
First-order linear
Homogeneous
Exact (correct answer)

Explanation: Let's check each classification. (A) Separable: We can write

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

, so it is separable. (B) Linear: We can write

\frac{dy}{dx} - \frac{2}{x}y = 0

, which is in the form

y' + P(x)y = Q(x)

, so it is linear. (C) Homogeneous: The function

F(x,y) = \frac{2y}{x}

satisfies

F(tx, ty) = \frac{2(ty)}{tx} = \frac{2y}{x} = F(x,y)

, so it is homogeneous. (D) Exact: Rewriting as

2y dx - x dy = 0

, we have

M=2y

and

N=-x

. Then

\frac{\partial M}{\partial y} = 2

and

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

. Since

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

, the equation is not exact.

Question 2

Consider the system of differential equations: $$ \begin{cases} x'(t) = 2x - y^2 \ y'(t) = 3x - 4y + \sin(t) \end{cases}

A linear, autonomous system because the independent variable $t$ does not appear in the coefficients of $x$ and $y$ .
A non-linear, autonomous system because of the $y^2$ term, and the absence of $t$ as a coefficient.
A non-linear, non-autonomous system due to the $y^2$ and $\sin(t)$ terms, respectively. (correct answer)
A linear, non-autonomous system because all terms are linear in $x$ and $y$ except for the forcing function $\sin(t)$ .

Explanation: To classify the system, we examine linearity and autonomy. A system is linear if the dependent variables (

x

and

y

) and their derivatives appear only to the first power and are not part of functions like

\sin(y)

or multiplied together. The term

y^2

in the first equation makes the system non-linear. A system is autonomous if the independent variable (

t

) does not explicitly appear in the equations. The term

\sin(t)

in the second equation makes the system non-autonomous. Therefore, the system is both non-linear and non-autonomous.

Question 3

A chemical reaction rate is modeled by the differential equation $\frac{dC}{dt} = k(C_0 - C)^2$ , where $C(t)$ is the concentration of a product at time $t$ , $C_0$ is the initial concentration of a reactant, and $k$ is a positive rate constant. A modification to the experiment introduces a catalyst that decays over time, making the 'rate constant' a function of time, $k(t) = k_0 e^{-\alpha t}$ . Which classification accurately describes the modified differential equation?

First-order, linear, and autonomous.
First-order, separable, and non-autonomous. (correct answer)
Second-order, non-linear, and autonomous.
First-order, non-linear, and separable.

Explanation: The modified differential equation is

\frac{dC}{dt} = k_0 e^{-\alpha t}(C_0 - C)^2

Order: The highest derivative is $\frac{dC}{dt}$ , so it is a first-order equation.
Linearity: The dependent variable $C$ appears in the term $(C_0 - C)^2 = C_0^2 - 2C_0C + C^2$ . The presence of the $C^2$ term makes the equation non-linear.
Autonomy: The independent variable $t$ appears explicitly in the term $e^{-\alpha t}$ . Therefore, the equation is non-autonomous.
Separability: The equation can be written as $\frac{dC}{(C_0-C)^2} = k_0 e^{-\alpha t} dt$ . Since all terms involving $C$ are on one side and all terms involving $t$ are on the other, the equation is separable. Combining these, the best description is first-order, non-linear, non-autonomous, and separable. Choice B includes 'separable' and 'non-autonomous', which are key features. Choice D incorrectly omits that it is non-autonomous.

Question 4

Consider the differential equation $y = x \frac{dy}{dx} - e^{dy/dx}$ . This is an example of a Clairaut equation. Which of the following statements most accurately describes a key feature of its solution set, derived from its classification?

As a first-order linear equation, its general solution is found using an integrating factor, resulting in a family of exponential curves.
As a non-linear equation, it possesses a general solution which is a family of straight lines, and may also have a singular solution that is not part of this family. (correct answer)
As an exact equation, its solution is an implicit function $F(x,y)=C$ which describes a family of parabolas.
As a homogeneous equation, the substitution $y=vx$ transforms it into a separable equation whose solution is a family of logarithmic curves.

Explanation: The equation

y = xy' - e^{y'}

is in the form of a Clairaut equation,

y = xy' + f(y')

. Clairaut equations are a special type of first-order, non-linear ODE. A key feature of their solutions is that the general solution is a family of straight lines, given by

y = Cx - e^C

, obtained by replacing

y'

with a constant

C

. Additionally, Clairaut equations often have a singular solution that forms an envelope to the family of lines. This singular solution cannot be obtained by specifying a value for

C

in the general solution. The other classifications are incorrect: it is non-linear, not exact, and not homogeneous.

Question 5

The temperature $T$ of an object in an environment with a varying ambient temperature $T_a(t) = A\cos(\omega t)$ is modeled by Newton's law of cooling, $\frac{dT}{dt} = -k(T - T_a(t))$ , where $k$ , $A$ , and $\omega$ are positive constants. How is this differential equation for $T(t)$ best classified?

First-order, linear, homogeneous, and autonomous.
First-order, linear, non-homogeneous, and non-autonomous. (correct answer)
First-order, non-linear, because of the trigonometric term.
Second-order, linear, because the forcing function is a cosine.

Explanation: Let's write out the full equation:

\frac{dT}{dt} = -k(T - A\cos(\omega t))

. Rearranging this gives

\frac{dT}{dt} + kT = kA\cos(\omega t)

Order: The highest derivative is the first, so it is first-order.
Linearity: The equation is of the form $T'(t) + p(t)T(t) = g(t)$ , with $p(t)=k$ and $g(t) = kA\cos(\omega t)$ . The dependent variable $T$ and its derivative $T'$ appear to the first power. The coefficients are functions of $t$ (or constants). Thus, the equation is linear. The trigonometric term $\cos(\omega t)$ is a function of the independent variable $t$ , not the dependent variable $T$ , so it does not make the equation non-linear.
Homogeneity: The term on the right-hand side, $g(t) = kA\cos(\omega t)$ , is not zero. Therefore, the equation is non-homogeneous.
Autonomy: The independent variable $t$ appears explicitly in the term $\cos(\omega t)$ . Therefore, the equation is non-autonomous. Combining these facts, the equation is first-order, linear, non-homogeneous, and non-autonomous.

Question 6

A student attempts to solve $(3xy+y^2)dx + (x^2+xy)dy=0$ and finds it is not exact. They correctly determine that multiplying by an integrating factor $\mu(x)$ will make it exact. Which property of the equation's coefficients, $M(x,y)$ and $N(x,y)$ , is the necessary and sufficient condition that justifies this specific approach?

The functions $M(x,y)$ and $N(x,y)$ are both homogeneous polynomials of degree 2, which allows for simplification.
The expression $\frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})$ simplifies to a function of $y$ only.
The expression $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$ simplifies to a function of $x$ only. (correct answer)
The equation can be rewritten in the form of a first-order linear equation after an algebraic rearrangement.

Explanation: When you encounter a differential equation that's not exact, you need to find an integrating factor to make it exact. The key is identifying which type of integrating factor exists based on specific conditions involving the coefficients. For the equation

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

, we have

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

and

N(x,y) = x^2+xy

. Let's check if an integrating factor

\mu(x)

(depending only on

x

) exists. First, calculate the partial derivatives:

\frac{\partial M}{\partial y} = 3x+2y

and

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

. Now compute

\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{1}{x^2+xy}[(3x+2y)-(2x+y)] = \frac{x+y}{x(x+y)} = \frac{1}{x}

. Since this expression depends only on

x

, an integrating factor

\mu(x)

exists, confirming answer C is correct. Answer A is wrong because while both functions are homogeneous of degree 2, this property alone doesn't guarantee the existence of an integrating factor depending only on

x

. Answer B incorrectly reverses the roles of

M

and

N

in the formula - this expression would indicate an integrating factor depending only on

y

. Answer D is incorrect because the equation cannot be easily rearranged into standard first-order linear form. Remember: when checking for integrating factors, if

\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})

is a function of

x

only, then

\mu(x)

exists. If

\frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})

is a function of

y

only, then

\mu(y)

exists.

Question 7

The Riccati equation $y' = -y^2 + xy + 1$ has a particular solution $y_1(x) = x$ . To find the general solution, the substitution $y = y_1 + u = x+u$ is used. Which of the following best describes the resulting differential equation in the variable $u$ ?

A first-order, separable equation.
A first-order, linear equation.
A first-order, homogeneous equation.
A first-order, Bernoulli equation. (correct answer)

Explanation: Given

y' = -y^2 + xy + 1

and the substitution

y = x+u

. First, find

y'

in terms of

u

y' = 1 + u'

. Now substitute

y

and

y'

into the original equation:

1 + u' = -(x+u)^2 + x(x+u) + 1

. Expand the terms:

1 + u' = -(x^2 + 2xu + u^2) + x^2 + xu + 1

. Simplify:

1 + u' = -x^2 - 2xu - u^2 + x^2 + xu + 1

. This reduces to

u' = -xu - u^2

. Rearranging gives

u' + xu = -u^2

. This is a Bernoulli equation of the form

u' + P(x)u = Q(x)u^n

with

n=2

P(x)=x

, and

Q(x)=-1

Question 8

The substitution $v = y^{-2}$ is applied to the Bernoulli differential equation $y' + p(x)y = q(x)y^3$ . Which of the following accurately describes the resulting differential equation in the variable $v$ and its properties?

The resulting equation is a first-order linear equation in $v$ , which is solvable using an integrating factor. (correct answer)
The resulting equation is a second-order linear equation in $v$ because the original exponent was greater than one.
The resulting equation remains a non-linear Bernoulli equation, as the substitution only changes the dependent variable.
The resulting equation becomes a separable equation in $v$ , provided that $p(x)$ and $q(x)$ are constants.

Explanation: The substitution for a Bernoulli equation of the form

y' + p(x)y = q(x)y^n

v = y^{1-n}

. In this case,

n=3

, so the substitution is

v = y^{1-3} = y^{-2}

. We differentiate

v

with respect to

x

v' = -2y^{-3}y'

. We can write

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

. Substitute this into the original equation:

(-\frac{1}{2}y^3 v') + p(x)y = q(x)y^3

. Now, divide by

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

v' - 2p(x)y^{-2} = -2q(x)

. Since

v = y^{-2}

, the equation becomes

v' - 2p(x)v = -2q(x)

. This is a first-order linear differential equation in the variable

v

, which can be solved using an integrating factor. The general form is

v' + P(x)v = Q(x)

Question 9

The differential equation $y' - \frac{1}{x}y = x$ is given. Which of the following pairs of classifications both correctly apply to this equation, and which of the associated solution methods is generally considered more direct?

Homogeneous and linear; the homogeneous method is more direct.
Separable and linear; the separation of variables method is more direct.
Exact and Bernoulli; the method for exact equations is more direct.
Linear and non-homogeneous; the method of integrating factors is the most direct approach. (correct answer)

Explanation: The equation is

y' - \frac{1}{x}y = x

. This is in the standard form for a first-order linear equation,

y' + p(x)y = g(x)

, with

p(x) = -1/x

and

g(x) = x

. Since

g(x) \neq 0

, it is non-homogeneous. The standard solution method for this form is using an integrating factor. Let's check other classifications. It is not separable due to the

x

term on the right. It is not homogeneous because the right side, written as

y/x + x

, cannot be expressed as a function of

y/x

. It is not a Bernoulli equation in its current form. It is not exact if written as

(x-y/x)dx - dy = 0

. Therefore, the most accurate and useful classification is first-order linear, non-homogeneous, solved by an integrating factor.

Question 10

To solve the differential equation $(x+y)dx - x dy = 0$ , a standard strategy is to use a substitution that transforms it into a separable equation. Which substitution is most appropriate for this purpose?

The substitution $v = y/x$ , because the equation is homogeneous. (correct answer)
The substitution $v = x+y$ , because the right-hand side is a function of $x+y$ .
An integrating factor, because the equation can be made exact but is not linear.
The substitution $v=y^{-1}$ , because the equation is a form of Bernoulli equation.

Explanation: First, classify the equation. We can rewrite it as

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

. Since the right-hand side can be expressed as a function of the ratio

y/x

, the equation is homogeneous. The standard substitution for homogeneous equations is

v = y/x

(or

y=vx

). This substitution is guaranteed to transform the equation into one that is separable in the variables

v

and

x

Question 11

An ordinary differential equation of the form $M(x, y)dx + N(x, y)dy = 0$ is not exact. However, it is discovered that multiplying the equation by the function $\mu(x) = e^{\int \frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) dx}$ renders it exact. This fact implies which of the following about the term inside the integral?

The original equation must have been linear in the variable $y$ .
The expression $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$ must be a function of $x$ only. (correct answer)
The original equation must have been homogeneous.
The expression $\frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})$ must be a function of $x$ only.

Explanation: For a non-exact differential equation, an integrating factor

\mu

can sometimes be found. If the expression

g(x) = \frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})

depends only on

x

, then an integrating factor is

\mu(x) = e^{\int g(x) dx}

. The fact that such a

\mu(x)

exists and is given in this form means that the condition for its existence must be met. Therefore, the expression

\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})

must be a function of

x

only. Distractor D describes a mixed condition; the expression shown is part of the test for an integrating factor that depends on

y

only,

\mu(y)

Question 12

The equation $(x^2 + y^2)dx + k xy dy = 0$ is given, where $k$ is a constant. For what value of $k$ does this equation have a specific classification that allows for the most straightforward solution method?

$k = 2$ , making the equation exact and solvable by direct integration (correct answer)
$k = -2$ , making the equation homogeneous and solvable by substitution $v = \frac{y}{x}$
$k = 1$ , making the equation exact and solvable by finding a potential function
$k = 0$ , making the equation separable and solvable by direct integration of $x^2 + y^2$

Explanation: For the equation to be exact, we need

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

where

M = x^2 + y^2

and

N = kxy

. Computing:

\frac{\partial M}{\partial y} = 2y

and

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

. Setting these equal:

2y = ky

, which gives

k = 2

. Choice B is incorrect because the equation is not homogeneous for any value of

k

(it cannot be written as

f(y/x)

). Choice C is incorrect because

k = 1

gives

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

. Choice D is incorrect because

k = 0

makes

N = 0

, which doesn't lead to a separable form.

Question 13

Consider the differential equation $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$ . A student claims this can be solved using three different methods. Which combination of solution approaches is correct?

Homogeneous substitution, exact equation after rearrangement, and Bernoulli equation transformation
Homogeneous substitution, separation of variables after algebraic manipulation, and integrating factor method
Exact equation after rearrangement, Riccati equation methods, and power series expansion around $x = 0$
Homogeneous substitution and exact equation after rearrangement, but separation is not possible (correct answer)

Explanation: The equation

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

can be rewritten as

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

, which is homogeneous (depends only on

y/x

). Rearranging to

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

, we can verify exactness:

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

. However, the variables cannot be separated because of the mixed

xy

terms. Choice A is incorrect because this is not a Bernoulli equation. Choice B is incorrect because separation is not possible. Choice C is incorrect because this is not a Riccati equation.

Question 14

Consider the differential equation $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$ . After rewriting this equation in the form $M(x,y)dx + N(x,y)dy = 0$ , which classification best describes the most efficient solution approach?

Exact equation, since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ after proper rearrangement
Homogeneous equation, since the equation can be written as $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$ for some function $f$ (correct answer)
Linear first-order equation, since it can be written in the form $\frac{dy}{dx} + P(x)y = Q(x)$
Bernoulli equation, since it has the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$ with $n \neq 0, 1$

Explanation: The equation

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

can be rewritten as

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

. Letting

v = \frac{y}{x}

, we get

\frac{dy}{dx} = \frac{1}{2}\left(\frac{1}{v} + v\right)

, which is a function of

\frac{y}{x}

only, making this a homogeneous equation. Choice A is incorrect because when rearranged to

M dx + N dy = 0

, the exactness condition fails. Choice C is incorrect because the equation cannot be written in linear form due to the

y^2

term in the numerator. Choice D is incorrect because this is not a Bernoulli equation structure.

Question 15

A researcher encounters the equation $\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y = xe^{2x}$ . To solve this second-order linear ODE, which approach correctly addresses both the homogeneous and particular solution components?

Find characteristic equation $r^2 - 4r + 4 = 0$ ; use method of undetermined coefficients with trial solution $y_p = (Ax + B)e^{2x}$
Find characteristic equation $r^2 - 4r + 4 = 0$ ; use method of undetermined coefficients with trial solution $y_p = Ax^2(Cx + D)e^{2x}$ (correct answer)
Find characteristic equation $r^2 - 4r + 4 = 0$ ; use variation of parameters since $e^{2x}$ appears in both forcing term and homogeneous solution
Transform to system of first-order equations; solve using matrix methods with eigenvalue analysis

Explanation: The characteristic equation

r^2 - 4r + 4 = (r-2)^2 = 0

has repeated root

r = 2

, giving homogeneous solution

y_h = (C_1 + C_2x)e^{2x}

. Since the forcing term

xe^{2x}

has the same exponential factor as the homogeneous solution, and

x

already appears in

y_h

, we need to multiply by

x^2

(the multiplicity of the root) to avoid duplication. Thus

y_p = x^2(Ax + B)e^{2x} = (Ax^3 + Bx^2)e^{2x}

. Choice A fails because it doesn't account for the repeated root. Choice C would work but is unnecessarily complex. Choice D is inefficient for this type of problem.

Question 16

A differential equation has the form $F\left(x, y, \frac{dy}{dx}\right) = 0$ where $F$ cannot be solved explicitly for $\frac{dy}{dx}$ , but can be written as $p^2 + 2xyp - y^2 = 0$ where $p = \frac{dy}{dx}$ . What classification and solution strategy should be used?

Lagrange equation; solve the quadratic for $p$ in terms of $x$ and $y$ , then solve the resulting first-order equations (correct answer)
Clairaut equation; differentiate with respect to $x$ and use the parametric solution method
Implicit differential equation; use implicit differentiation repeatedly until an explicit form is obtained
Parametric differential equation; introduce parameter $t = \frac{dy}{dx}$ and solve the system parametrically

Explanation: When you encounter a differential equation where

F(x, y, \frac{dy}{dx}) = 0

cannot be solved explicitly for

\frac{dy}{dx}

, you need to classify the equation type to choose the right solution strategy. The key is recognizing the structure and applying the appropriate method. The equation

p^2 + 2xyp - y^2 = 0

(where

p = \frac{dy}{dx}

) is quadratic in

p

. This is a Lagrange equation - a differential equation that's quadratic in the first derivative. The standard approach is to solve the quadratic equation for

p

using the quadratic formula, giving you

p = \frac{-2xy \pm \sqrt{4x^2y^2 + 4y^2}}{2} = -xy \pm y\sqrt{x^2 + 1}

. This yields two expressions for

\frac{dy}{dx}

, each giving you a separable first-order differential equation to solve. Option B is incorrect because Clairaut equations have the specific form

y = px + f(p)

, which this equation doesn't match. Option C misses the point - while the equation is implicit, the systematic approach for quadratic-in-

p

equations is to solve the quadratic first, not use repeated implicit differentiation. Option D incorrectly suggests introducing a parameter when we already have a workable quadratic structure. Study tip: When you see a differential equation that's quadratic in

p = \frac{dy}{dx}

, immediately think "Lagrange equation" and solve the quadratic for

p

. This transforms one complex equation into two simpler first-order equations you can handle with standard techniques.

Question 17

A first-order differential equation is called homogeneous if it can be written in the form $\frac{dy}{dx} = F(\frac{y}{x})$ . Which of the following equations has this property?

$\frac{dy}{dx} = \frac{x^2 + y}{x}$
$\frac{dy}{dx} = \frac{y^2 + xy}{x^2}$ (correct answer)
$\frac{dy}{dx} = \frac{y+1}{x+1}$
$\frac{dy}{dx} = \sin(\frac{y}{x}) + x$

Explanation: To check if an equation is homogeneous, we try to write the right-hand side as a function of the ratio

v = y/x

. For choice B, we can divide the numerator and denominator by

x^2

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

. This is a function of

v=y/x

, so the equation is homogeneous. The other choices cannot be written solely as a function of

y/x

. For example, choice A becomes

x + y/x

, and choice D becomes

\sin(y/x)+x

; the presence of

x

as a separate term violates the condition.

Question 18

Which of the following equations is a fourth-order, linear, homogeneous ordinary differential equation?

$(\frac{d^2y}{dx^2})^2 + x^2\frac{d^2y}{dx^2} + y = 0$
$e^x \frac{d^4y}{dx^4} + \cos(x) \frac{d^3y}{dx^3} = y\frac{dy}{dx}$
$x\frac{d^4y}{dx^4} - \sin(x)\frac{d^2y}{dx^2} + (\ln x)y = 0$ (correct answer)
$\frac{d^4y}{dx^4} + 5\frac{d^2y}{dx^2} + 9y = \sin(x)$

Explanation: When classifying differential equations, you need to check three key properties: order (highest derivative), linearity (how the dependent variable and its derivatives appear), and homogeneity (whether there's a non-zero term without the dependent variable). The correct answer is C because it satisfies all three conditions for a fourth-order, linear, homogeneous ODE. The highest derivative is

\frac{d^4y}{dx^4}

, making it fourth-order. Each term contains the dependent variable

y

or its derivatives raised only to the first power with no products between them, making it linear. Finally, every term contains

y

or its derivatives (no standalone functions), making it homogeneous. Option A fails the linearity test because

(\frac{d^2y}{dx^2})^2

squares the second derivative, creating a nonlinear term. Additionally, it's only second-order, not fourth-order. Option B is fourth-order but nonlinear due to the product term

y\frac{dy}{dx}

, which multiplies the dependent variable by its first derivative. This violates the linearity requirement. Option D is fourth-order and linear, but it's non-homogeneous because of the

\sin(x)

term on the right side, which doesn't contain

y

or any of its derivatives. Remember this systematic approach: first identify the highest derivative for order, then check if all terms involving

y

and its derivatives are linear (first power only, no products), and finally verify homogeneity by confirming every term contains the dependent variable or its derivatives.

Question 19

The initial value problem $y' = \sqrt{y^2 + x^2}$ with $y(1) = 2$ is given. How does the initial condition $y(1)=2$ influence the classification of the differential equation itself?

The initial condition has no effect on the classification of the differential equation. (correct answer)
The initial condition makes the equation non-autonomous, as it specifies a point other than $x=0$ .
The initial condition renders the equation non-linear because the specific values can affect the behavior of the solution.
The initial condition helps determine if the equation is homogeneous by testing the values $(1,2)$ .

Explanation: The classification of a differential equation depends only on its algebraic form, involving the variables and their derivatives. Classifications like order, linearity, autonomy, and type (e.g., separable, exact, homogeneous) are properties of the equation

y' = f(x,y)

. An initial condition, such as

y(1)=2

, provides a specific point through which a particular solution must pass. It is used to find the value of the constant of integration after the general solution has been found. It does not alter the fundamental structure of the differential equation itself. Therefore, the initial condition has no effect on the classification.

Question 20

Consider the differential equation $(x^2+y^2)dx + 2xy \ln(x) dy = 0$ . Which classification is most appropriate for determining a solution method for this equation?

It is a separable equation because the variables $x$ and $y$ can be isolated on opposite sides of the equation.
It is a homogeneous equation because the coefficient functions $M(x,y)$ and $N(x,y)$ are homogeneous polynomials of the same degree.
It is an exact equation because the partial derivatives of the coefficient functions with respect to the opposite variables are equal.
It is not exact in its current form, but an integrating factor can be found to make it exact. (correct answer)

Explanation: Let

M(x,y) = x^2+y^2

and

N(x,y) = 2xy \ln(x)

. First, we test for exactness:

\frac{\partial M}{\partial y} = 2y

and

\frac{\partial N}{\partial x} = 2y \ln(x) + 2xy(1/x) = 2y \ln(x) + 2y

. Since

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

, the equation is not exact. It is not separable. The term

\ln(x)

prevents

N(x,y)

from being a homogeneous function, so it is not a homogeneous ODE. We must check if an integrating factor exists. Consider the test expression

\frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) = \frac{1}{x^2+y^2}(2y \ln(x) + 2y - 2y) = \frac{2y \ln(x)}{x^2+y^2}

, which depends on both

x

and

y

. Now consider

\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{1}{2xy \ln(x)}(2y - (2y \ln(x) + 2y)) = \frac{-2y \ln(x)}{2xy \ln(x)} = -\frac{1}{x}

. Since this expression depends only on

x

, an integrating factor

\mu(x)

exists. Thus, classifying it as an equation that can be made exact is the most appropriate path forward.