Differential Equations Quiz: Identifying Linear First Order Odes

What this quiz covers

This quiz focuses on Identifying Linear First Order 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. $P(x) = 3x$ for all $x$ in the domain of the differential equation
2. $P(x) = 3$ for $x \neq 0$, and the equation is undefined at $x = 0$ (correct answer)
3. $P(x) = \frac{3x^2}{\sin(x)}$ for all $x$ where $\sin(x) \neq 0$
4. $P(x) = \frac{3}{x}$ for $x > 0$, but the equation is nonlinear for $x < 0$

Explanation: To convert to standard form, divide the entire equation by $x$: $\frac{dy}{dx} + \frac{3xy}{x} = \frac{\sin(x)}{x}$, which simplifies to $\frac{dy}{dx} + 3y = \frac{\sin(x)}{x}$. Therefore $P(x) = 3$ for $x \neq 0$. The equation is undefined at $x = 0$ because we cannot divide by zero. Choice A incorrectly keeps the $x$ factor. Choice C incorrectly involves $\sin(x)$ in the denominator. Choice D incorrectly suggests the equation behaves differently for positive versus negative $x$.

Question 2

1. $\frac{dr}{d\theta} + r \tan(\theta) = \sec(\theta)$
2. $x dy = (x^3 + 2y) dx$
3. $y' = 1 + \frac{y}{x+1}$
4. $(1+y^2) \frac{dy}{dx} + xy = e^x$ (correct answer)

Explanation: A first-order ODE is linear if it can be written as $a_1(x)y' + a_0(x)y = b(x)$. A. This equation is already in the standard linear form for $r(\theta)$. B. Dividing by $dx$ gives $x \frac{dy}{dx} = x^3 + 2y$. Rearranging gives $x \frac{dy}{dx} - 2y = x^3$, which is a linear equation. C. Rearranging gives $y' - \frac{1}{x+1} y = 1$, which is a linear equation. D. The term $(1+y^2)$ multiplies the derivative $\frac{dy}{dx}$. Since this coefficient depends on the dependent variable $y$, the equation is nonlinear. Linear equations require the coefficients of $y$ and its derivatives to be functions of the independent variable only.

Question 3

1. k = 0
2. k = -2 or k = 2 (correct answer)
3. k > 2 or k < -2
4. k = 2 only

Explanation: A first-order differential equation is linear if it can be expressed in the form $\frac{dy}{dx} + p(x)y = q(x)$. The given equation is $\frac{dy}{dx} + \frac{1}{x}y = (k^2-4) \ln(y)$. The term $(k^2-4) \ln(y)$ on the right-hand side makes the equation nonlinear because the dependent variable $y$ appears inside a logarithmic function. For the equation to be linear, any such nonlinear term in $y$ must vanish. This requires its coefficient to be zero. Therefore, we must have $k^2 - 4 = 0$. Solving for $k$ gives $k^2 = 4$, which yields $k = 2$ and $k = -2$. For these values of $k$, the equation becomes $\frac{dy}{dx} + \frac{y}{x} = 0$, which is a linear homogeneous equation.

Question 4

1. $\frac{dy}{dx} = \frac{x-y}{x+y}$
2. $\sin(x) \frac{dy}{dx} + y \cos(x) = \tan(x)$ (correct answer)
3. $\frac{dy}{dx} + x y = \sqrt{y}$
4. $e^y \frac{dy}{dx} + e^y x = x^2$

Explanation: A first-order ODE is linear if it can be written as $a_1(x)y' + a_0(x)y = b(x)$. A. Rearranging gives $(x+y)\frac{dy}{dx} = x-y$. The coefficient of $\frac{dy}{dx}$ is $(x+y)$, which depends on $y$. Thus, it is nonlinear. B. The equation is $\sin(x) \frac{dy}{dx} + (\cos x)y = \tan(x)$. This is in the linear form with $a_1(x) = \sin(x)$, $a_0(x) = \cos(x)$, and $b(x) = \tan(x)$. Note that the left side is the derivative of $(y \sin x)$. C. The term $\sqrt{y}$ makes the equation nonlinear as the dependent variable $y$ is raised to the power of $1/2$. This is a Bernoulli equation. D. This equation is nonlinear in $y$. However, if we make the substitution $u = e^y$, then $\frac{du}{dx} = e^y \frac{dy}{dx}$. The equation becomes $\frac{du}{dx} + ux = x^2$, which is linear in $u$. But the original equation is nonlinear with respect to $y$.

Question 5

1. $M(x,y)$ is a function of $x$ only, and $N(x,y)$ is a function of $y$ only.
2. $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
3. $N(x,y)$ is a function of $x$ only, and $M(x,y)$ is of the form $f(x)y + g(x)$. (correct answer)
4. $M(x,y)$ and $N(x,y)$ are both linear functions of $y$, i.e. $M = a(x)y+b(x)$ and $N=c(x)y+d(x)$.

Explanation: Let's analyze the condition in each option. A. If $M(x,y)=M(x)$ and $N(x,y)=N(y)$, we have $M(x)dx + N(y)dy = 0$. This gives $\frac{dy}{dx} = -\frac{M(x)}{N(y)}$, which is a separable equation. It is linear only if $N(y)$ is a constant and $M(x)$ is a linear function of $y$, which is not guaranteed. B. This is the condition for an exact equation, which is not equivalent to linearity. For example, $(2xy)dx + (x^2 - y^2)dy=0$ is exact but not linear. C. If $N(x,y)=N(x)$ and $M(x,y)=f(x)y+g(x)$, the equation becomes $(f(x)y+g(x))dx + N(x)dy = 0$. This can be written as $N(x)\frac{dy}{dx} + f(x)y + g(x) = 0$, or $N(x)\frac{dy}{dx} + f(x)y = -g(x)$. Assuming $N(x) \neq 0$, we can divide by it to get $\frac{dy}{dx} + \frac{f(x)}{N(x)}y = -\frac{g(x)}{N(x)}$, which is the standard form of a linear equation. This condition is sufficient. D. If $M=a(x)y+b(x)$ and $N=c(x)y+d(x)$, the equation is $(a(x)y+b(x))dx + (c(x)y+d(x))dy=0$. This gives $(c(x)y+d(x))\frac{dy}{dx} = -(a(x)y+b(x))$. The coefficient of $\frac{dy}{dx}$ depends on $y$, so the equation is generally not linear unless $c(x)=0$.

Question 6

1. The equation is $\frac{dA}{dt} = 10e^{-t/10} - \frac{A}{20}$, which is a linear equation. (correct answer)
2. The equation is $\frac{dA}{dt} = 10e^{-t/10} - \frac{A^2}{100}$, which is a nonlinear equation.
3. The equation is $\frac{dA}{dt} = 2e^{-t/10} - \frac{A}{100}$, which is a linear equation.
4. The equation is $\frac{dA}{dt} = 10e^{-t/10} - \frac{A}{20}$, but it is nonlinear because the input concentration $C(t)$ is not constant.

Explanation: Let's set up the differential equation. The rate of change of salt is $\frac{dA}{dt}$. The rate of salt flowing in is (concentration in) × (flow rate in) = $(2e^{-t/10} \text{ kg/L}) \times (5 \text{ L/min}) = 10e^{-t/10}$ kg/min. The volume of the tank is constant at 100 L because the inflow and outflow rates are equal. The concentration of salt in the tank at time $t$ is $\frac{A(t)}{100}$ kg/L. The rate of salt flowing out is (concentration in tank) × (flow rate out) = $(\frac{A(t)}{100} \text{ kg/L}) \times (5 \text{ L/min}) = \frac{A(t)}{20}$ kg/min. Therefore, the differential equation is $\frac{dA}{dt} = 10e^{-t/10} - \frac{A}{20}$. To classify this equation, we rearrange it: $\frac{dA}{dt} + \frac{1}{20}A = 10e^{-t/10}$. This is in the standard form for a linear first-order ODE, $\frac{dA}{dt} + p(t)A = q(t)$, with $p(t) = 1/20$ and $q(t) = 10e^{-t/10}$. The fact that $q(t)$ is a function of $t$ does not make the equation nonlinear. Thus, the equation is linear.

Question 7

1. The equation is a non-homogeneous linear equation. (correct answer)
2. The equation is a homogeneous linear equation.
3. The equation is a homogeneous-in-degree equation.
4. The equation is an exact equation.

Explanation: First, let's classify the equation based on linearity. We can divide by $x^2$ to put it in standard form: $y' + \frac{1}{x}y = \frac{\ln(x)}{x^2}$. This matches the form $y' + p(x)y = q(x)$ with $p(x) = 1/x$ and $q(x) = \ln(x)/x^2$. Since it fits this form, the equation is linear. Because the term $q(x)$ is not identically zero, the linear equation is non-homogeneous. So, statement A is correct. Statement B is incorrect because the equation is non-homogeneous. For statement C, we check if it is homogeneous-in-degree. The equation $y' = \frac{\ln(x) - xy}{x^2} = \frac{\ln(x)}{x^2} - \frac{y}{x}$ is not of the form $f(y/x)$, so it is not homogeneous-in-degree. For statement D, we check for exactness. In the form $M(x,y)dx + N(x,y)dy = 0$, the equation is $(xy - \ln(x))dx + x^2 dy = 0$. Here, $M = xy - \ln(x)$ and $N = x^2$. We compute the partial derivatives: $\frac{\partial M}{\partial y} = x$ and $\frac{\partial N}{\partial x} = 2x$. Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact. Thus, A is the only correct and precise classification.

Question 8

1. It is a first-order linear homogeneous equation.
2. It is a first-order linear non-homogeneous equation. (correct answer)
3. It is a first-order nonlinear equation.
4. It is a second-order linear equation.

Explanation: To find the differential equation, we differentiate the given integral equation with respect to $t$. The derivative of the constant $2$ is $0$. By the Fundamental Theorem of Calculus, Part 1, the derivative of the integral is the integrand evaluated at $t$. So, $\frac{dy}{dt} = \frac{d}{dt} \left( 2 + \int_1^t \frac{\tau - y(\tau)}{\tau^2} d\tau \right) = 0 + \frac{t - y(t)}{t^2}$. The resulting differential equation is $\frac{dy}{dt} = \frac{t}{t^2} - \frac{y(t)}{t^2}$, which simplifies to $\frac{dy}{dt} = \frac{1}{t} - \frac{y(t)}{t^2}$. To check for linearity, we rearrange it into the standard form $\frac{dy}{dt} + p(t)y = q(t)$. This gives $\frac{dy}{dt} + \frac{1}{t^2}y = \frac{1}{t}$. This is a first-order linear differential equation with $p(t) = 1/t^2$ and $q(t) = 1/t$. Since $q(t)$ is not identically zero, the equation is non-homogeneous. Therefore, it is a first-order linear non-homogeneous equation.

Question 9

1. $F(x,y) = \frac{x^2}{y}$
2. $F(x,y) = x \sin(y)$
3. $F(x,y) = y \sin(x) + x^2$ (correct answer)
4. $F(x,y) = y^2 \cos(x)$

Explanation: We apply the test $\frac{\partial^2 F}{\partial y^2} = 0$ to each function. A. For $F(x,y) = x^2 y^{-1}$, $\frac{\partial F}{\partial y} = -x^2 y^{-2}$ and $\frac{\partial^2 F}{\partial y^2} = 2x^2 y^{-3} \neq 0$. This is nonlinear. B. For $F(x,y) = x \sin(y)$, $\frac{\partial F}{\partial y} = x \cos(y)$ and $\frac{\partial^2 F}{\partial y^2} = -x \sin(y) \neq 0$. This is nonlinear. C. For $F(x,y) = y \sin(x) + x^2$, $\frac{\partial F}{\partial y} = \sin(x)$ and $\frac{\partial^2 F}{\partial y^2} = 0$. Since the second partial derivative with respect to $y$ is zero, the equation is linear. The equation can be written as $y' - (\sin x)y = x^2$. D. For $F(x,y) = y^2 \cos(x)$, $\frac{\partial F}{\partial y} = 2y \cos(x)$ and $\frac{\partial^2 F}{\partial y^2} = 2 \cos(x) \neq 0$. This is nonlinear.

Question 10

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

Explanation: A first-order differential equation is linear if it can be written in the form $\frac{dy}{dx} + p(x)y = q(x)$. Let's analyze each option. A. $\frac{dy}{dx} + \frac{x}{y} = \cos(x)$ is nonlinear because the dependent variable $y$ appears in the denominator. B. $\frac{dy}{dx} + \sqrt{y} = x^2$ is nonlinear because $y$ appears under a square root, which is equivalent to a power of $1/2$. C. $\frac{dy}{dx} + y \ln(x) = y \ln(2)$ can be rearranged by moving the term on the right to the left side: $\frac{dy}{dx} + y \ln(x) - y \ln(2) = 0$. Factoring out $y$ gives $\frac{dy}{dx} + (\ln(x) - \ln(2))y = 0$. This is in the standard linear form with $p(x) = \ln(x) - \ln(2)$ and $q(x) = 0$. D. $\frac{dy}{dx} + y\sin(x) = \sin(y)$ is nonlinear because of the $\sin(y)$ term. The dependent variable $y$ must appear linearly, not as the argument of a nonlinear function like sine.

Question 11

1. The equation is linear in $y$ as a function of $x$.
2. The equation is linear in $x$ as a function of $y$. (correct answer)
3. The equation is an exact equation.
4. The equation is nonlinear with respect to both $y(x)$ and $x(y)$.

Explanation: First, let's analyze the equation with $y$ as the dependent variable, $y(x)$. We can write it as $(x + \cos(y))\frac{dy}{dx} = 1$. The coefficient of the derivative, $(x + \cos(y))$, depends on $y$. Therefore, the equation is nonlinear in $y(x)$. Next, let's consider $x$ as the dependent variable, $x(y)$. We can invert the derivative: $\frac{dx}{dy} = x + \cos(y)$. We can rearrange this to $\frac{dx}{dy} - x = \cos(y)$. This is in the standard linear form $\frac{dx}{dy} + p(y)x = q(y)$, with $p(y) = -1$ and $q(y) = \cos(y)$. Thus, the equation is linear in $x$ as a function of $y$. For exactness, we write the equation as $1 dx - (x+\cos(y))dy = 0$. Here $M=1$ and $N=-(x+\cos(y))$. We check $\frac{\partial M}{\partial y} = 0$ and $\frac{\partial N}{\partial x} = -1$. Since they are not equal, the equation is not exact.

Question 12

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

Explanation: We need to check if each equation can be arranged into the linear form $a_1(x) y' + a_0(x) y = b(x)$. A. $\frac{dy}{dx} - y = xy$ can be rearranged to $\frac{dy}{dx} - y - xy = 0$, which is $\frac{dy}{dx} - (1+x)y = 0$. This is linear. B. $y = x(\frac{dy}{dx} - x \cos x)$ expands to $y = x \frac{dy}{dx} - x^2 \cos x$. Rearranging gives $x \frac{dy}{dx} - y = x^2 \cos x$. This is linear. C. $\frac{dy}{dx} = (y \tan x) + e^x$ can be rearranged to $\frac{dy}{dx} - (\tan x)y = e^x$. This is linear. D. $(\frac{dy}{dx})^3 = x^3 (\frac{dy}{dx} - y)$ expands to $(\frac{dy}{dx})^3 = x^3 \frac{dy}{dx} - x^3 y$. Rearranging gives $(\frac{dy}{dx})^3 - x^3 \frac{dy}{dx} + x^3 y = 0$. This equation is nonlinear because the derivative $\frac{dy}{dx}$ is raised to the third power.

Question 13

1. When $P(x)$ and $R(x)$ are both continuous functions on the interval of interest
2. When $Q(x) = 0$ for all $x$ in the domain of the equation (correct answer)
3. When $P(x)$ is a constant function and $Q(x)$ is differentiable on the given interval
4. When $R(x) = 0$ for all $x$ in the domain and $Q(x)$ is any continuous function

Explanation: A first-order linear ODE has the standard form $\frac{dy}{dx} + P(x)y = Q(x)$ where the dependent variable $y$ appears only to the first power and its coefficient is 1. In the given equation, the term $Q(x)y^2$ makes it nonlinear unless $Q(x) = 0$, which eliminates the quadratic term entirely. Choice A is incorrect because continuity conditions don't affect linearity. Choice C is wrong because even if $P(x)$ is constant, the $y^2$ term still makes it nonlinear. Choice D is incorrect because setting $R(x) = 0$ doesn't eliminate the problematic $Q(x)y^2$ term.

Question 14

1. For any real value of $c$, since linearity depends only on the form of the equation
2. Only when $c = \pm 1$, which are the critical values for this type of equation
3. Only when $c > 0$, since negative initial values lead to complex-valued solutions
4. Only when $c = 0$, which makes the nonlinear term vanish throughout the solution (correct answer)

Explanation: When you encounter a differential equation that appears nonlinear, look for special conditions that might eliminate the nonlinear terms. This equation has the form $\frac{dy}{dx} + P(x)y = Q(x)y^2$, which is a Bernoulli equation due to the $y^2$ term. The key insight is that if $y(x) = 0$ is a solution, then the nonlinear term $\frac{y^2}{(x-1)^2}$ vanishes identically, reducing the equation to the linear form $\frac{dy}{dx} + \frac{y}{x-1} = 0$. To have $y(x) = 0$ as a solution, you need the initial condition $y(2) = 0$, which means $c = 0$. You can verify this: if $y(x) = 0$ for all $x$, then $\frac{dy}{dx} = 0$, and substituting gives $0 + 0 = 0$, which is satisfied. For any other initial condition $c \neq 0$, the solution will be nonzero in some neighborhood of $x = 2$, making the $y^2$ term nonzero and keeping the equation nonlinear. Answer A is wrong because linearity absolutely depends on the solution behavior, not just the equation's form. Answer B incorrectly suggests $c = \pm 1$ are special—these values don't eliminate the nonlinear term. Answer C misunderstands the issue entirely; the sign of $c$ doesn't determine whether the equation becomes linear, and negative values don't inherently cause complex solutions. Remember: for Bernoulli equations, the trivial solution $y = 0$ always makes the equation linear by eliminating all nonlinear terms.

Question 15

1. When $a(x)$, $b(x)$, and $c(x)$ are all continuous functions on the interval of interest
2. When $a(x) \neq 0$ on the interval and $a(x)$, $b(x)$, $c(x)$ can be any functions
3. When $a(x)$, $b(x)$, and $c(x)$ are continuous and $a(x) \neq 0$ on the given interval (correct answer)
4. When $a(x) = 1$, $b(x)$ is continuous, and $c(x)$ is any integrable function on the interval

Explanation: A first-order linear ODE in the form $a(x)\frac{dy}{dx} + b(x)y = c(x)$ requires two conditions: (1) $a(x) \neq 0$ so we can divide through to get standard form $\frac{dy}{dx} + P(x)y = Q(x)$, and (2) the coefficients $a(x)$, $b(x)$, and $c(x)$ should be continuous to ensure existence and uniqueness of solutions. Choice A omits the crucial condition $a(x) \neq 0$. Choice B doesn't require continuity, which is needed for solution theory. Choice D unnecessarily restricts $a(x) = 1$ and has incorrect requirements for $c(x)$.

Question 16

1. $n = 0$, because the right side becomes $Q(x)$, giving the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ (correct answer)
2. $n = 1$, because the equation becomes $\frac{dy}{dx} + P(x)y = Q(x)y$, which can be rearranged to linear form
3. $n = -1$, because this allows the substitution $v = y^2$ to linearize the equation effectively
4. $n = 2$, because this creates a quadratic relationship that can be solved using linear algebraic methods

Explanation: When you encounter a Bernoulli equation, you need to identify what makes a differential equation "linear in y." A linear differential equation has the dependent variable y and its derivatives appearing only to the first power, with no products or higher powers of y. The general form $\frac{dy}{dx} + P(x)y = Q(x)y^n$ becomes linear when the right side doesn't introduce nonlinear terms in y. This happens when $n = 0$, making $y^n = y^0 = 1$. The equation then becomes $\frac{dy}{dx} + P(x)y = Q(x)$, which is the standard first-order linear differential equation form. Option A correctly identifies this: when $n = 0$, the $y^n$ term becomes just a constant multiplier, leaving you with the classic linear form where y appears only to the first power. Option B is incorrect because when $n = 1$, you get $\frac{dy}{dx} + P(x)y = Q(x)y$, which rearranges to $\frac{dy}{dx} = [Q(x) - P(x)]y$. This is separable, not linear in the standard sense. Option C misunderstands the substitution method. While $n = -1$ does allow certain substitutions, the equation $\frac{dy}{dx} + P(x)y = \frac{Q(x)}{y}$ is nonlinear because of the $\frac{1}{y}$ term. Option D is wrong because $n = 2$ gives $\frac{dy}{dx} + P(x)y = Q(x)y^2$, which is quadratic in y, not linear. Remember: linearity means the dependent variable appears only to the first power throughout the equation.

Question 17

1. When $m$, $k$ are positive constants and $F(t)$ is any continuous function of time (correct answer)
2. Only when $F(t)$ is a linear function of time, such as $F(t) = at + b$ for constants $a$ and $b$
3. When the drag force model $kv$ is valid, which requires $v$ to be small compared to terminal velocity
4. Only when $m$ and $k$ are both functions of $v$ that maintain the linear structure of the equation

Explanation: The equation $m\frac{dv}{dt} + kv = F(t)$ is linear in $v$ when it can be written in the standard form $\frac{dv}{dt} + P(t)v = Q(t)$. If $m$ and $k$ are constants with $m \neq 0$, we can divide by $m$ to get $\frac{dv}{dt} + \frac{k}{m}v = \frac{F(t)}{m}$, which is linear regardless of the specific form of $F(t)$. Choice B incorrectly suggests that $F(t)$ must be linear in $t$ for the ODE to be linear in $v$—these are different concepts. Choice C confuses the physical validity of the drag model with mathematical linearity. Choice D is incorrect because if $m$ and $k$ depend on $v$, the equation generally becomes nonlinear.

Question 18

1. Let $v = y^2$, then the equation becomes linear in $v$ after substituting $\frac{dv}{dx} = 2y\frac{dy}{dx}$
2. Let $u = \sqrt{1 + y^2}$, then the equation becomes linear in $u$ with appropriate substitutions
3. Let $w = \frac{1}{y}$, then the equation becomes linear in $w$ through the standard reciprocal transformation
4. No transformation can make this equation linear because the term $\frac{y}{\sqrt{1 + y^2}}$ is inherently nonlinear (correct answer)

Explanation: This equation is a Bernoulli-type equation, but the right-hand side $\frac{y}{\sqrt{1 + y^2}}$ cannot be handled by standard linearizing transformations. Let's check each option: Choice A with $v = y^2$ gives $\frac{dv}{dx} = 2y\frac{dy}{dx}$, but substituting this doesn't eliminate the $\sqrt{1 + y^2}$ term. Choice B with $u = \sqrt{1 + y^2}$ gives $\frac{du}{dx} = \frac{y}{\sqrt{1 + y^2}}\frac{dy}{dx}$, but this creates an even more complex relationship. Choice C with $w = \frac{1}{y}$ is the standard transformation for Bernoulli equations, but it doesn't address the square root term. The presence of $\sqrt{1 + y^2}$ makes this equation genuinely nonlinear with no elementary linearizing transformation.

Question 19

1. On any interval where $\sin(x) \neq 0$ and $\cos(x) \neq 0$, ensuring all coefficients are well-defined
2. On any interval where $\sin(x) > 0$ and the equation can be written in standard form
3. On any open interval that does not contain points where $\sin(x) = 0$, such as $(0, \pi)$ (correct answer)
4. On any interval where $\tan(x)$ is continuous and $\sin(x)$ maintains the same sign

Explanation: To apply existence and uniqueness theorems for linear ODEs, we need the coefficient of $\frac{dy}{dx}$ to be nonzero so we can divide by it to get standard form $\frac{dy}{dx} + P(x)y = Q(x)$. Dividing by $\sin(x)$ gives $\frac{dy}{dx} + \frac{\cos(x)}{\sin(x)}y = \frac{\tan(x)}{\sin(x)} = \frac{1}{\cos(x)}$. We need $\sin(x) \neq 0$ for this division. The resulting $P(x) = \cot(x)$ and $Q(x) = \sec(x)$ must be continuous on the interval. $\cot(x)$ is continuous wherever $\sin(x) \neq 0$, and $\sec(x)$ is continuous wherever $\cos(x) \neq 0$. However, the primary constraint comes from the leading coefficient $\sin(x)$. Choice A incorrectly requires both $\sin(x) \neq 0$ and $\cos(x) \neq 0$. Choice B incorrectly restricts to positive values. Choice D focuses on the wrong conditions.