Differential Equations Quiz: Intro To Differential Equations

What this quiz covers

This quiz focuses on Intro To Differential 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.

Question 1

1. $k=1/3$
2. $k=3$ (correct answer)
3. $k=9$
4. $k=81$

Explanation: First, find the first and second derivatives of the proposed solution $y = C_1 e^{kx} + C_2 e^{-kx}$. $y' = kC_1 e^{kx} - kC_2 e^{-kx}$ $y'' = k^2 C_1 e^{kx} + k^2 C_2 e^{-kx} = k^2(C_1 e^{kx} + C_2 e^{-kx}) = k^2 y$. Now, substitute $y'' = k^2 y$ into the differential equation: $k^2 y - 9y = 0$ $(k^2 - 9)y = 0$ For this equation to hold true for any solution $y$ in the family, the coefficient must be zero: $k^2 - 9 = 0$. This gives $k^2 = 9$, so $k = \pm 3$. Since the question asks for the positive value of $k$, the correct answer is $k=3$.

Question 2

1. $y''' + 6y'' - 5y' + 3y = \sin(x)$
2. $x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + (x^2 - 4)y = 0$
3. $\frac{d}{dx}[\ln(x) y'] + e^x y = x^3$
4. $(\cos x)y'' + y y' = x^2$ (correct answer)

Explanation: A differential equation is linear if the dependent variable $y$ and its derivatives appear only to the first power and are not part of other functions, and their coefficients depend only on the independent variable $x$. (A) is a linear equation with constant coefficients. (B) is a linear equation (Bessel's equation) with variable coefficients. (C) can be expanded using the product rule to $\frac{1}{x}y' + \ln(x)y'' + e^x y = x^3$. This is a second-order linear equation. (D) contains the term $y y'$, which is a product of the dependent variable and its derivative. This term makes the equation nonlinear.

Question 3

1. It is a solution to the differential equation but does not satisfy the initial condition.
2. It satisfies the initial condition but is not a solution to the differential equation.
3. It is a solution to the initial value problem. (correct answer)
4. It is not a solution to the differential equation and does not satisfy the initial condition.

Explanation: This is a two-step verification. First, check if the function solves the differential equation. Second, check if it satisfies the initial condition. Step 1: Find derivatives and substitute into the DE. $y = x \sin(x)$ $y' = \sin(x) + x \cos(x)$ $y'' = \cos(x) + (\cos(x) - x \sin(x)) = 2\cos(x) - x \sin(x)$ Substitute into the DE: $y'' + y = (2\cos(x) - x \sin(x)) + (x \sin(x)) = 2\cos(x)$. The function satisfies the differential equation. Step 2: Check the initial condition. $y(\pi/2) = (\pi/2) \sin(\pi/2) = (\pi/2) \cdot 1 = \pi/2$. The initial condition is also satisfied. Since both parts are satisfied, the function is a solution to the initial value problem.

Question 4

1. $n=1$ only
2. $n=0$ only
3. $n=0$ and $n=1$ (correct answer)
4. Any integer $n$

Explanation: For a differential equation to be linear, the dependent variable $y$ and its derivatives must appear to the first power. The terms $y''$ and $e^x y'$ are consistent with a linear equation. The term $y^n$ determines whether the equation is linear. If $n=1$, the term is $y^1 = y$, and the equation $y'' + e^x y' + y = 0$ is linear. If $n=0$, the term is $y^0 = 1$, and the equation $y'' + e^x y' + 1 = 0$ is also linear (it is non-homogeneous). For any other integer value of $n$ (e.g., $n=2, 3, -1$), the term $y^n$ is a nonlinear function of $y$. Therefore, the equation is linear only for $n=0$ and $n=1$.

Question 5

1. The function is a solution to the IVP on the interval $(-\infty, \infty)$.
2. The function satisfies the differential equation but not the initial condition.
3. The function is a solution to the IVP, but only on the interval $(-\infty, 2)$. (correct answer)
4. The function is not a solution because the differential equation is nonlinear.

Explanation: First, verify if the function solves the DE. For $y(x) = (2-x)^{-1}$, the derivative is $y' = -1(2-x)^{-2}(-1) = \frac{1}{(2-x)^2}$. The term $y^2$ is $(\frac{1}{2-x})^2 = \frac{1}{(2-x)^2}$. Since $y'=y^2$, the DE is satisfied. Second, verify the initial condition: $y(0) = \frac{1}{2-0} = 1/2$. The IC is also satisfied. However, a solution to a differential equation must be differentiable (and thus continuous) on its interval of existence. The function $y(x) = \frac{1}{2-x}$ has a vertical asymptote at $x=2$ and is not defined there. The initial condition is at $x=0$. The largest interval containing $x=0$ on which the solution is defined and continuous is $(-\infty, 2)$. Therefore, the function is a solution to the IVP on this specific interval.

Question 6

1. First-order, linear
2. Second-order, linear (correct answer)
3. First-order, nonlinear
4. Second-order, nonlinear

Explanation: To determine the order, we must perform the differentiation indicated by the outer operator $\frac{d}{dx}$. Applying the product rule to $\frac{d}{dx}[p(x)y']$ gives $p'(x)y' + p(x)y''$. The full differential equation is $p(x)y'' + p'(x)y' + q(x)y = \cos(x)$. The highest derivative is $y''$, so the equation is second-order. The equation is linear because the dependent variable $y$ and its derivatives $y'$ and $y''$ appear only to the first power, and their coefficients ($p(x)$, $p'(x)$, and $q(x)$) are functions of the independent variable $x$ only.

Question 7

1. It becomes a first-order linear ordinary differential equation.
2. It becomes a second-order linear ordinary differential equation. (correct answer)
3. The temperature must be uniform, so there is no resulting differential equation.
4. It remains a second-order linear partial differential equation.

Explanation: A steady state implies that the temperature $u$ does not change with time $t$. Mathematically, this means $\frac{\partial u}{\partial t} = 0$. Since the steady-state temperature $U$ depends only on the position $x$, the partial derivatives with respect to $x$ become ordinary derivatives. Substituting these conditions into the heat equation gives: $0 = \alpha \frac{d^2 U}{dx^2}$. This simplifies to $\frac{d^2 U}{dx^2} = 0$. This is an ordinary differential equation (ODE) because it involves derivatives with respect to only one independent variable, $x$. The highest derivative is the second derivative, so it is second-order. The equation is linear in $U$. The solution is $U(x) = C_1 x + C_2$, which represents a linear temperature profile, not necessarily a uniform (constant) temperature.

Question 8

1. First-order, linear (correct answer)
2. First-order, nonlinear
3. Second-order, linear
4. The substitution does not change the order or linearity.

Explanation: Let $u = y^2$. The term $\frac{d}{dx}(y^2)$ is simply $\frac{du}{dx}$. Substituting $u$ for $y^2$ throughout the given equation yields: $\frac{du}{dx} + 2x u = 4x$. This is the transformed differential equation for the new dependent variable $u(x)$. We can now analyze its properties. The highest derivative is $\frac{du}{dx}$, so it is a first-order equation. The equation is of the form $u' + P(x)u = Q(x)$, where $P(x) = 2x$ and $Q(x) = 4x$. This is the standard form for a first-order linear differential equation. Thus, the transformed equation is first-order and linear.

Question 9

1. First-order and linear
2. Second-order and linear
3. First-order and nonlinear
4. Second-order and nonlinear (correct answer)

Explanation: Newton's second law is $F_{net} = ma$. The position is $x(t)$, velocity is $v = x'(t)$, and acceleration is $a = x''(t)$. Substituting the given force and the expressions for $v$ and $a$ into Newton's law gives: $m x''(t) = -k x(t) - c (x'(t))^2 + F_0 \sin(\omega t)$. Rearranging gives $m x'' + c(x')^2 + kx = F_0 \sin(\omega t)$. The order of the equation is determined by the highest derivative present, which is $x''$, so the equation is second-order. The linearity is determined by how the dependent variable $x$ and its derivatives appear. The term $(x')^2$ is the square of the first derivative, which makes the equation nonlinear. Therefore, the differential equation is second-order and nonlinear.

Question 10

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

Explanation: To find the differential equation, we differentiate the given relation implicitly with respect to $x$. Remember to use the product rule for the $xy^2$ term and the chain rule for terms involving $y$. $\frac{d}{dx}(x^3 + xy^2 - y^3) = \frac{d}{dx}(C)$ $3x^2 + (1 \cdot y^2 + x \cdot 2y y') - 3y^2 y' = 0$ $3x^2 + y^2 + 2xy y' - 3y^2 y' = 0$ Now, isolate the terms with $y'$: $2xy y' - 3y^2 y' = -3x^2 - y^2$ $y'(2xy - 3y^2) = -(3x^2 + y^2)$ $y' = -\frac{3x^2+y^2}{2xy - 3y^2} = \frac{3x^2+y^2}{-(2xy - 3y^2)} = \frac{3x^2+y^2}{3y^2-2xy}$. This matches choice A.

Question 11

1. $y'' + y' - 6y = 0; \quad y(0)=3, y'(0)=-4$ (correct answer)
2. $y'' - y' - 6y = 0; \quad y(0)=3, y'(0)=-4$
3. $y'' + y' - 6y = 0; \quad y(0)=3, y'(0)=-1$
4. $y'' - 5y' + 6y = 0; \quad y(0)=3, y'(0)=-4$

Explanation: This requires checking both the differential equation and the initial conditions for each option. First, find the derivatives of the given function and evaluate them at $x=0$. $y(x) = 2e^{-3x} + e^{2x} \implies y(0) = 2e^0 + e^0 = 3$. $y'(x) = -6e^{-3x} + 2e^{2x} \implies y'(0) = -6e^0 + 2e^0 = -4$. $y''(x) = 18e^{-3x} + 4e^{2x} \implies y''(0) = 18e^0 + 4e^0 = 22$. The correct IVP must have initial conditions $y(0)=3$ and $y'(0)=-4$. This eliminates choice C. Now test the remaining DEs: (A) $y'' + y' - 6y = (18e^{-3x} + 4e^{2x}) + (-6e^{-3x} + 2e^{2x}) - 6(2e^{-3x} + e^{2x}) = (18-6-12)e^{-3x} + (4+2-6)e^{2x} = 0$. This DE is satisfied. (B) $y'' - y' - 6y = (18e^{-3x} + 4e^{2x}) - (-6e^{-3x} + 2e^{2x}) - 6(2e^{-3x} + e^{2x}) = (18+6-12)e^{-3x} + (4-2-6)e^{2x} \neq 0$. (D) $y'' - 5y' + 6y = (18e^{-3x} + 4e^{2x}) - 5(-6e^{-3x} + 2e^{2x}) + 6(2e^{-3x} + e^{2x}) = (18+30+12)e^{-3x} + (4-10+6)e^{2x} \neq 0$. Only choice A satisfies both the differential equation and the initial conditions.

Question 12

1. The equation must be nonlinear, since linear differential equations have solutions that extend over their entire natural domain
2. The equation likely contains terms like $\sqrt{x}$ or $\ln(x)$ that restrict the domain to positive values only
3. The equation is probably linear with variable coefficients that become undefined or exhibit discontinuities for $x \leq 0$ (correct answer)
4. This behavior indicates the equation is first-order, since higher-order equations require solutions to be valid across all real numbers

Explanation: The fact that $f(x) = x^3 + 2x$ (which is defined for all real $x$) satisfies the equation only for $x > 0$ suggests the differential equation itself has coefficients or terms that are undefined or discontinuous for $x \leq 0$. This is characteristic of linear equations with variable coefficients that may involve terms like $\frac{1}{x}$, $\sqrt{x}$, or $\ln(x)$. The linearity/nonlinearity doesn't determine domain restrictions in this way.

Question 13

1. Student A is correct; linearity depends only on how the dependent variable $y$ appears, not its derivatives
2. Neither student fully grasps the concept; linearity requires examining the highest-order derivative term, which is $\frac{d^2y}{dx^2}$ here
3. Both students are partially correct; this equation is quasilinear since $y$ appears linearly but its derivatives do not
4. Student B is correct; the presence of $\left(\frac{dy}{dx}\right)^2$ makes the equation nonlinear regardless of how $y$ appears (correct answer)

Explanation: When you encounter questions about whether a differential equation is linear or nonlinear, you need to examine how the dependent variable and ALL of its derivatives appear throughout the equation. A differential equation is linear if the dependent variable and all its derivatives appear to the first power only, with no products between them. Let's analyze this equation systematically. For $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = \sin(x)$ to be linear, every term containing $y$ or its derivatives must be linear. While $\frac{d^2y}{dx^2}$ appears linearly (first power), the term $\left(\frac{dy}{dx}\right)^2$ is the first derivative raised to the second power. This nonlinear term immediately disqualifies the equation from being linear. Answer A incorrectly focuses only on how $y$ itself appears, ignoring the derivatives entirely. Linearity requires examining the dependent variable AND all its derivatives together. Answer B misunderstands the concept by suggesting only the highest-order derivative matters—every derivative term must be examined. Answer C incorrectly categorizes this as quasilinear, but quasilinear equations have linear expressions in the highest-order derivative, which isn't the case here since we have a nonlinear first-derivative term. Answer D correctly identifies that the squared derivative term makes the entire equation nonlinear. Study tip: When determining linearity, scan every term containing the dependent variable or its derivatives. If any term has powers other than one, or if derivatives are multiplied together, the equation is nonlinear—no exceptions.

Question 14

1. The claim is correct; solutions cannot cross $y = x$ since the differential equation becomes undefined there (correct answer)
2. The claim is incorrect; while $f(x,y)$ is undefined on $y = x$, solutions may cross this line under certain conditions
3. The claim is partially correct; solutions approach $y = x$ asymptotically but can never actually reach or cross this line
4. The claim reveals a misunderstanding; the line $y = x$ itself consists of solutions to the differential equation

Explanation: The student's claim is correct. Where $f(x,y)$ is undefined (along $y = x$), the differential equation has no meaning, and solutions cannot exist. The line $y = x$ acts as a barrier that solutions cannot cross. Solutions may approach this line asymptotically from either side, but they cannot cross it because the slope $\frac{dy}{dx}$ becomes undefined there. This is different from saying $y = x$ is itself a solution.

Question 15

1. This is actually a first-order nonlinear system; the number of equations doesn't determine the order, and the quadratic terms make it nonlinear (correct answer)
2. The student correctly identified the order but missed that the system is nonlinear due to the $x^2$ and $y^2$ terms
3. The student confused a system of equations with a single higher-order equation; this system should be classified by matrix methods instead
4. The classification is completely wrong; this represents two independent first-order linear equations that can be solved separately

Explanation: When classifying systems of differential equations, you need to consider two key characteristics: the order (highest derivative present) and linearity (whether variables appear raised to powers or multiplied together). This system contains only first derivatives ($\frac{dx}{dt}$ and $\frac{dy}{dt}$), making it first-order, not second-order. The student's reasoning that "two equations equals second-order" reflects a fundamental misunderstanding—the order depends on the highest derivative, not the number of equations. Additionally, the presence of $y^2$ and $x^2$ terms makes this system nonlinear, since the variables are raised to powers greater than one. Choice A correctly identifies both errors: this is a first-order nonlinear system because it involves only first derivatives but contains quadratic terms that violate linearity. Choice B incorrectly accepts the student's second-order classification. While it correctly identifies the nonlinearity, it fails to recognize the fundamental error about system order. Choice C introduces irrelevant concepts about matrix methods and confuses systems with single higher-order equations. Matrix methods can be used for linear systems, but they don't change how we classify the system's order and linearity. Choice D is completely wrong on multiple levels. The system is neither linear (due to the quadratic terms) nor separable into independent equations, since each equation contains both variables. Study tip: Always classify differential equation systems by asking two questions: "What's the highest derivative?" (determines order) and "Are variables raised to powers or multiplied together?" (determines linearity). The number of equations is irrelevant to the order.

Question 16

1. The division by $x$ changes the order of the equation from second-order to first-order, altering the solution structure
2. Dividing by $x$ eliminates potential solutions where $x = 0$, but the resulting equation form is mathematically equivalent elsewhere
3. The division creates a linear equation from a nonlinear one, which fundamentally changes the types of solutions that exist
4. This algebraic manipulation is valid everywhere except $x = 0$, where the original equation may have singular behavior requiring separate analysis (correct answer)

Explanation: The division by $x$ is algebraically valid wherever $x \neq 0$, and it doesn't change the equation's order (both forms are first-order) or linearity (both are nonlinear due to $y^2$). However, the point $x = 0$ requires special consideration since the original equation becomes $y = 0$ there, which might represent singular behavior. The main caution is ensuring we don't lose solutions or miss singular points, but the manipulation itself is generally acceptable.

Question 17

1. The equation is linear because the highest derivative appears to the first power, despite the nonlinear term $y^3$
2. The equation is nonlinear due to the $y^3$ term, and finding particular solutions requires specialized techniques beyond linear methods (correct answer)
3. The equation is second-order linear with variable coefficients, but the given functions are inappropriate trial solutions
4. The equation is first-order nonlinear because the coefficient of $\frac{dy}{dx}$ depends on the independent variable $x$

Explanation: The equation is nonlinear due to the $y^3$ term. In a linear differential equation, the dependent variable and all its derivatives must appear to the first power only, with no products between them. The presence of $y^3$ makes this equation nonlinear, requiring specialized solution techniques. The order is determined by the highest derivative (second-order), and linearity is determined by how the dependent variable appears.

Question 18

1. The solutions remain exactly sinusoidal since the $\epsilon x^3$ term becomes negligible compared to the linear terms for small oscillations
2. The equation becomes nonlinear, requiring completely different solution methods, and sinusoidal solutions no longer provide any useful approximation
3. The solutions are approximately sinusoidal but will exhibit small deviations that grow over time, potentially requiring perturbation methods for analysis (correct answer)
4. The small nonlinear term causes the equation to lose its second-order character, reducing it effectively to a first-order system

Explanation: When $\epsilon$ is small but nonzero, the equation becomes nonlinear, and solutions will deviate from pure sinusoidal behavior. However, for small $\epsilon$, the solutions start approximately sinusoidal but accumulate small nonlinear effects over time. This is a classic scenario for perturbation methods, where the linear solution serves as a starting approximation. The equation remains second-order regardless of the nonlinearity.

Question 19

1. Substituting the proposed solution shows it satisfies the equation, but the arbitrary constant should be $C = 1$ for completeness
2. The homogeneous solution $Ce^{2x}$ is correct, but the particular solution should be $3x + 3$ rather than $3x$
3. Direct substitution yields $2Ce^{2x} + 3 - 2(Ce^{2x} + 3x) = 3 - 6x \neq -6x - 3$, showing the solution is incorrect (correct answer)
4. The form is correct but incomplete; since this is a second-order equation, two arbitrary constants are required in the general solution

Explanation: Direct substitution reveals the error. Computing: $\frac{dy}{dx} = 2Ce^{2x} + 3$ and $\frac{dy}{dx} - 2y = 2Ce^{2x} + 3 - 2(Ce^{2x} + 3x) = 3 - 6x$. Since $3 - 6x \neq -6x - 3$, the proposed solution doesn't satisfy the differential equation. The correct particular solution should be $3x + \frac{3}{2}$.

Question 20

1. Any linear combination $c_1y_1 + c_2y_2 + c_3y_3$ automatically forms the complete general solution regardless of the independence of the functions
2. The combination $c_1y_1 + c_2y_2 + c_3y_3$ forms the general solution only if the Wronskian $W(y_1, y_2, y_3) \neq 0$ over the domain (correct answer)
3. Since the equation is homogeneous and linear, exactly two linearly independent solutions are sufficient to span the solution space
4. The general solution requires four arbitrary constants because the equation contains both second and third-order derivatives

Explanation: For a third-order linear homogeneous equation, the general solution requires three linearly independent solutions. The combination $c_1y_1 + c_2y_2 + c_3y_3$ forms the general solution only if these three functions are linearly independent, which is verified by a non-zero Wronskian. Choice A ignores linear independence, C incorrectly states two solutions suffice for a third-order equation, and D misunderstands how the order determines the number of constants.