Differential Equations Quiz: Classifying Odes

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.

Question 1

1. Separable
2. First-order linear
3. Homogeneous
4. 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

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

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 3

1. A linear, autonomous system because the independent variable $t$ does not appear in the coefficients of $x$ and $y$.
2. A non-linear, autonomous system because of the $y^2$ term, and the absence of $t$ as a coefficient.
3. A non-linear, non-autonomous system due to the $y^2$ and $\sin(t)$ terms, respectively. (correct answer)
4. 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 4

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

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 5

1. The initial condition has no effect on the classification of the differential equation. (correct answer)
2. The initial condition makes the equation non-autonomous, as it specifies a point other than $x=0$.
3. The initial condition renders the equation non-linear because the specific values can affect the behavior of the solution.
4. 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 6

1. First-order, linear, and autonomous.
2. First-order, separable, and non-autonomous. (correct answer)
3. Second-order, non-linear, and autonomous.
4. First-order, non-linear, and separable.

Explanation: The modified differential equation is $\frac{dC}{dt} = k_0 e^{-\alpha t}(C_0 - C)^2$.

1. Order: The highest derivative is $\frac{dC}{dt}$, so it is a first-order equation.
2. 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.
3. Autonomy: The independent variable $t$ appears explicitly in the term $e^{-\alpha t}$. Therefore, the equation is non-autonomous.
4. 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 7

1. As a first-order linear equation, its general solution is found using an integrating factor, resulting in a family of exponential curves.
2. 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)
3. As an exact equation, its solution is an implicit function $F(x,y)=C$ which describes a family of parabolas.
4. 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 8

1. First-order, linear, homogeneous, and autonomous.
2. First-order, linear, non-homogeneous, and non-autonomous. (correct answer)
3. First-order, non-linear, because of the trigonometric term.
4. 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 $\frac{dT}{dt} + kT = kA\cos(\omega t)$.

1. Order: The highest derivative is the first, so it is first-order.
2. 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.
3. Homogeneity: The term on the right-hand side, $g(t) = kA\cos(\omega t)$, is not zero. Therefore, the equation is non-homogeneous.
4. 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 9

1. It is a separable equation because the variables $x$ and $y$ can be isolated on opposite sides of the equation.
2. It is a homogeneous equation because the coefficient functions $M(x,y)$ and $N(x,y)$ are homogeneous polynomials of the same degree.
3. It is an exact equation because the partial derivatives of the coefficient functions with respect to the opposite variables are equal.
4. 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.

Question 10

1. The functions $M(x,y)$ and $N(x,y)$ are both homogeneous polynomials of degree 2, which allows for simplification.
2. The expression $\frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})$ simplifies to a function of $y$ only.
3. The expression $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$ simplifies to a function of $x$ only. (correct answer)
4. 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 11

1. A first-order, separable equation.
2. A first-order, linear equation.
3. A first-order, homogeneous equation.
4. 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 12

1. The resulting equation is a first-order linear equation in $v$, which is solvable using an integrating factor. (correct answer)
2. The resulting equation is a second-order linear equation in $v$ because the original exponent was greater than one.
3. The resulting equation remains a non-linear Bernoulli equation, as the substitution only changes the dependent variable.
4. 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 $y' + p(x)y = q(x)y^n$ is $v = y^{1-n}$. 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' = -rac{1}{2}y^3 v'. Substitute this into the original equation: (-rac{1}{2}y^3 v') + p(x)y = q(x)y^3. Now, divide by -rac{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 13

1. Homogeneous and linear; the homogeneous method is more direct.
2. Separable and linear; the separation of variables method is more direct.
3. Exact and Bernoulli; the method for exact equations is more direct.
4. 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 14

1. The equation is exact, because $\frac{\partial}{\partial y}(y \cos(x) + 2xe^y) = \frac{\partial}{\partial x}(\sin(x) + x^2e^y - 1)$. (correct answer)
2. The equation is not exact, because the partial derivatives of the coefficient functions are unequal.
3. The equation is separable, because it can be rewritten in the form $f(x)dx = g(y)dy$.
4. The equation is first-order linear, because it can be rearranged into the form $\frac{dy}{dx} + P(x)y = Q(x)$.

Explanation: To check if the equation $M(x, y)dx + N(x, y)dy = 0$ is exact, we must verify if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Here, $M(x, y) = y \cos(x) + 2xe^y$ and $N(x, y) = \sin(x) + x^2e^y - 1$. Calculating the partial derivatives: $\frac{\partial M}{\partial y} = \cos(x) + 2xe^y$ and $\frac{\partial N}{\partial x} = \cos(x) + 2xe^y$. Since the partial derivatives are equal, the equation is exact.

Question 15

1. Separable, because the variables $x$ and $y$ can be isolated on opposite sides of the equation.
2. First-order linear, because it can be written in the form $\frac{dy}{dx} + P(x)y = Q(x)$. (correct answer)
3. Homogeneous, because all terms are of the same degree in $x$ and $y$.
4. Exact, because it satisfies the condition $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.

Explanation: To classify the equation, first write it in a standard form. Dividing by $x$ gives $\frac{dy}{dx} - \frac{1}{x}y = x \cos(x)$. This matches the standard form for a first-order linear differential equation, $\frac{dy}{dx} + P(x)y = Q(x)$, with $P(x) = -\frac{1}{x}$ and $Q(x) = x \cos(x)$. This classification directly leads to the method of integrating factors for a solution.

Question 16

1. The substitution $v = y/x$, because the equation is homogeneous. (correct answer)
2. The substitution $v = x+y$, because the right-hand side is a function of $x+y$.
3. An integrating factor, because the equation can be made exact but is not linear.
4. 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 17

1. The equation is a Cauchy-Euler type, best solved with the substitution $y=x^r$.
2. The equation is linear and homogeneous, so its general solution is found directly from the characteristic equation.
3. The equation is a non-homogeneous linear equation with constant coefficients, solved by finding a complementary function from a characteristic equation with a repeated root and a particular solution using undetermined coefficients. (correct answer)
4. The equation is a non-homogeneous linear equation, solved by finding a complementary function from a characteristic equation with distinct real roots and a particular solution using variation of parameters.

Explanation: The equation is a second-order, linear, non-homogeneous ODE with constant coefficients. The solution is $y = y_c + y_p$. First, solve the associated homogeneous equation $y'' - 4y' + 4y = 0$. The characteristic equation is $r^2 - 4r + 4 = 0$, or $(r-2)^2 = 0$, which has a repeated real root $r=2$. Thus, the complementary function is $y_c = c_1e^{2x} + c_2xe^{2x}$. For the particular solution $y_p$, the method of undetermined coefficients is appropriate. Since the term $xe^{2x}$ on the right-hand side duplicates terms in $y_c$, the form of the guess for $y_p$ must be modified.

Question 18

1. The original equation must have been linear in the variable $y$.
2. The expression $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$ must be a function of $x$ only. (correct answer)
3. The original equation must have been homogeneous.
4. 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 19

1. It is a Bernoulli equation, solvable by making the substitution $v = y^{-2}$. (correct answer)
2. It is a linear equation, solvable by finding an integrating factor $\mu(t) = e^{5t}$.
3. It is a separable equation, solvable by moving all $y$ terms to one side and $t$ terms to the other.
4. It is a Bernoulli equation, solvable by making the substitution $v = y^3$.

Explanation: The equation is of the form $\frac{dy}{dt} + P(t)y = Q(t)y^n$, with $n=3$, which is the definition of a Bernoulli equation. It is non-linear due to the $y^3$ term. The standard method to solve a Bernoulli equation is to use the substitution $v = y^{1-n}$. In this case, $n=3$, so the correct substitution is $v = y^{1-3} = y^{-2}$. This transforms the equation into a linear equation in $v$.

Question 20

1. $y'' + 5y' + 4y = \ln(x)$
2. $y'' + 5xy' + 4y = 0$
3. $x^2 y'' + 5y' + 4y = 0$
4. $x^2 y'' + 5xy' + 4y = 0$ (correct answer)

Explanation: When you encounter the substitution $y = x^r$ in differential equations, you're dealing with Euler equations (also called Cauchy-Euler equations). These are second-order linear equations where each term has the same degree when you consider the order of differentiation plus the power of $x$. Let's examine why option D works perfectly. For $x^2 y'' + 5xy' + 4y = 0$, when you substitute $y = x^r$, you get $y' = rx^{r-1}$ and $y'' = r(r-1)x^{r-2}$. Substituting these: $x^2 \cdot r(r-1)x^{r-2} + 5x \cdot rx^{r-1} + 4x^r = 0$ This simplifies to $x^r[r(r-1) + 5r + 4] = 0$, giving you the characteristic equation $r^2 + 4r + 4 = 0$. Option A fails because it's a constant-coefficient equation with a non-homogeneous term $\ln(x)$ - you'd use variation of parameters, not the $x^r$ substitution. Option B has constant coefficients but includes an $xy'$ term that doesn't fit the Euler equation pattern. Option C is close but incorrect - the middle term should be $5xy'$, not just $5y'$, to maintain the proper degree structure. The key pattern to remember: Euler equations have the form $ax^2y'' + bxy' + cy = 0$ where each term's total degree (power of $x$ plus derivative order) equals 2. When you see this structure, immediately think "$y = x^r$ substitution."