Identifying Linear First-Order ODEs - Differential Equations

1. A general first-order differential equation can be written as $M(x,y)dx + N(x,y)dy = 0$. The equation is linear in $y$ if it can be rearranged into the form $y' + p(x)y = q(x)$. Which condition on $M$ and $N$ is sufficient to guarantee that the equation is linear in $y$?
   Answer: $N(x,y)$ is a function of $x$ only, and $M(x,y)$ is of the form $f(x)y + g(x)$. 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 $c(x)=0$.
2. Which one of the following first-order differential equations is linear?
   Answer: $\sin(x) \frac{dy}{dx} + y \cos(x) = \tan(x)$ 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 $y$.
3. Which statement best describes when the equation $a(x)\frac{dy}{dx} + b(x)y = c(x)$ is considered a first-order linear ODE?
   Answer: When $a(x)$, $b(x)$, and $c(x)$ are continuous and $a(x) \neq 0$ on the given interval 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 $c(x)$.
4. Consider the differential equation $\frac{dy}{dx} + P(x)y = Q(x)y^2 + R(x)$. Under which condition is this equation linear in $y$?
   Answer: When $Q(x) = 0$ for all $x$ in the domain of the equation 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.
5. A first-order differential equation $y' = F(x,y)$ is linear if and only if the function $F(x,y)$ is a linear function of $y$ for any fixed $x$. This is equivalent to the condition that $\frac{\partial^2 F}{\partial y^2} = 0$ for all relevant $x$ and $y$. For which of the following functions $F(x,y)$ is the equation $y' = F(x,y)$ linear?
   Answer: $F(x,y) = y \sin(x) + x^2$ 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.
6. Which of the following first-order differential equations is linear?
   Answer: $\frac{dy}{dx} + y \ln(x) = y \ln(2)$ 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.
7. Consider the differential equation $\frac{dy}{dx} = \frac{1}{x + \cos(y)}$. Which of the following statements is a correct classification of this equation?
   Answer: The equation is linear in $x$ as a function of $y$. 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.
8. The equation $\frac{dy}{dx} + P(x)y = Q(x)y^n$ is called a Bernoulli equation. For which value of $n$ is this equation linear in $y$?
   Answer: $n = 0$, because the right side becomes $Q(x)$, giving the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ 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.
9. A first-order ordinary differential equation for $y(x)$ is linear if it can be expressed in the form $a_1(x) \frac{dy}{dx} + a_0(x) y = b(x)$. Which of the following equations can not be written in this form?
   Answer: $(\frac{dy}{dx})^3 = x^3 (\frac{dy}{dx} - y)$ 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.
10. A physicist writes the equation $m\frac{dv}{dt} + kv = F(t)$ to model motion with air resistance, where $v$ is velocity, $m$ is mass, $k$ is the drag coefficient, and $F(t)$ is applied force. Under what conditions is this a linear differential equation in $v$?
    Answer: When $m$, $k$ are positive constants and $F(t)$ is any continuous function of time 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.
11. Three of the following equations are linear, while one is not. Which equation is nonlinear?
    Answer: $(1+y^2) \frac{dy}{dx} + xy = e^x$ 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.
12. Consider the differential equation $x^2 y' + xy = \ln(x)$ for $x>0$. Which of the following statements provides the most precise classification of this equation?
    Answer: The equation is a non-homogeneous linear equation. 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.
13. The equation $x\frac{dy}{dx} + 3xy = \sin(x)$ can be written in standard linear form as $\frac{dy}{dx} + P(x)y = Q(x)$. What is $P(x)$?
    Answer: $P(x) = 3$ for $x \neq 0$, and the equation is undefined at $x = 0$ 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 $x$.
14. A function $y(t)$ is defined for $t > 0$ by the integral equation $y(t) = 2 + \int_1^t \frac{\tau - y(\tau)}{\tau^2} d\tau$. By differentiating the integral equation with respect to $t$, one can obtain a differential equation for $y(t)$. Which statement correctly classifies the resulting differential equation?
    Answer: It is a first-order linear non-homogeneous equation. 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.
15. Consider the equation $\frac{dy}{dx} + \frac{y}{x-1} = \frac{y^2}{(x-1)^2}$. For what values of the initial condition $y(2) = c$ does this reduce to a linear differential equation?
    Answer: Only when $c = 0$, which makes the nonlinear term vanish throughout the solution 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.
16. Consider the differential equation $\frac{dy}{dx} + \frac{y}{x} = (k^2-4) \ln(y)$. For which set of real values of the parameter $k$ is this equation linear?
    Answer: k = -2 or k = 2 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.
17. A tank initially contains 100 liters of pure water. A salt solution with a concentration of $C(t) = 2e^{-t/10}$ kg/L flows into the tank at a rate of 5 L/min. The mixture is kept uniform by stirring and flows out at the same rate. Let $A(t)$ be the amount of salt in the tank at time $t$. The differential equation for $A(t)$ is derived from the principle rate of change = rate in - rate out. Which statement correctly describes the resulting differential equation?
    Answer: The equation is $\frac{dA}{dt} = 10e^{-t/10} - \frac{A}{20}$, which is a linear equation. 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.
18. Consider the equation $(1 + x^2)\frac{dy}{dx} + 2xy = \frac{y}{\sqrt{1 + y^2}}$. Under what transformation does this become a linear equation?
    Answer: No transformation can make this equation linear because the term $\frac{y}{\sqrt{1 + y^2}}$ is inherently nonlinear 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.
19. The differential equation $\sin(x)\frac{dy}{dx} + \cos(x)y = \tan(x)$ is linear in $y$. On which interval is it guaranteed to have a unique solution through any given initial point?
    Answer: On any open interval that does not contain points where $\sin(x) = 0$, such as $(0, \pi)$ 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.

Start this quiz