Differential Equations Quiz: Linear Odes With Initial Conditions

What this quiz covers

This quiz focuses on Linear Odes With Initial Conditions, 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. $\frac{3}{4}(e^2 - e^{-2})$ (correct answer)
2. $\frac{3}{4}e^2$
3. $0$
4. $\frac{3}{4}e^2 + \frac{5}{4}e^{-2}$

Explanation: This is a linear first-order ODE. The integrating factor is $\mu(t) = e^{\int 2 dt} = e^{2t}$. Multiplying the equation by $\mu(t)$ gives $(e^{2t}y)' = e^{2t}(4te^{2t}) = 4te^{4t}$. Integrating both sides requires integration by parts for the right side: $\int 4te^{4t} dt$. Let $u=4t$ and $dv=e^{4t}dt$. Then $du=4dt$ and $v=\frac{1}{4}e^{4t}$. The integral is $te^{4t} - \int e^{4t} dt = te^{4t} - \frac{1}{4}e^{4t} + C$. So, $e^{2t}y = te^{4t} - \frac{1}{4}e^{4t} + C$. Solving for $y$ gives $y(t) = te^{2t} - \frac{1}{4}e^{2t} + Ce^{-2t}$. Using the initial condition $y(0)=-1$: $-1 = 0 \cdot e^0 - \frac{1}{4}e^0 + Ce^0$, which simplifies to $-1 = -1/4 + C$, so $C = -3/4$. The solution is $y(t) = te^{2t} - \frac{1}{4}e^{2t} - \frac{3}{4}e^{-2t}$. Evaluating at $t=1$: $y(1) = 1e^{2} - \frac{1}{4}e^{2} - \frac{3}{4}e^{-2} = \frac{3}{4}e^2 - \frac{3}{4}e^{-2} = \frac{3}{4}(e^2 - e^{-2})$.

Question 2

1. $2$
2. $5/2$ (correct answer)
3. $3$
4. $7/2$

Explanation: The integrating factor is $\mu(t) = e^{\int \cot t dt} = e^{\ln|\sin t|} = |\sin t|$. Since the initial condition is at $t=\pi/2$, we are in an interval where $\sin t > 0$, so $\mu(t) = \sin t$. The equation becomes $(\sin(t) y)' = (\sin t)(2\cos t) = \sin(2t)$. Integrating both sides gives $\sin(t) y = \int \sin(2t) dt = -\frac{1}{2}\cos(2t) + C$. Here, $C$ is the constant of integration. To find its value, we use the initial condition $y(\pi/2)=3$. Substitute $t=\pi/2$ and $y=3$ into the integrated equation: $\sin(\pi/2) \cdot 3 = -\frac{1}{2}\cos(2 \cdot \pi/2) + C$. This gives $1 \cdot 3 = -\frac{1}{2}\cos(\pi) + C$, which simplifies to $3 = -\frac{1}{2}(-1) + C$, or $3 = \frac{1}{2} + C$. Solving for $C$ yields $C = 3 - \frac{1}{2} = \frac{5}{2}$.

Question 3

1. $1/4$
2. $2$
3. $4$
4. $8$ (correct answer)

Explanation: The ODE is linear with $P(x) = -1/x$. The integrating factor is $\mu(x) = e^{\int -1/x dx} = x^{-1}$. Multiplying by $\mu(x)$ gives $(x^{-1}y)' = x^{-1}(\alpha x^2) = \alpha x$. Integrating both sides yields $x^{-1}y = \int \alpha x dx = \frac{\alpha}{2}x^2 + C$. The general solution is $y(x) = \frac{\alpha}{2}x^3 + Cx$. Using the initial condition $y(1)=2$: $2 = \frac{\alpha}{2}(1)^3 + C(1)$, which gives $C = 2 - \frac{\alpha}{2}$. The particular solution is $y(x) = \frac{\alpha}{2}x^3 + (2 - \frac{\alpha}{2})x$. Now, we evaluate the limit: $\lim_{x \to \infty} \frac{y(x)}{x^3} = \lim_{x \to \infty} \frac{\frac{\alpha}{2}x^3 + (2 - \frac{\alpha}{2})x}{x^3} = \lim_{x \to \infty} \left( \frac{\alpha}{2} + \frac{2 - \alpha/2}{x^2} \right) = \frac{\alpha}{2}$. We are given that this limit is 4. Therefore, $\frac{\alpha}{2} = 4$, which implies $\alpha = 8$.

Question 4

1. $2e^{-1}$ (correct answer)
2. $2e^{-2}$
3. $1$
4. $e^{-1}$

Explanation: We must solve this problem in two parts. First, for $0 \le t < 1$, we solve $y' + y = 2$ with $y(0)=2$. This is a linear ODE with integrating factor $\mu(t) = e^t$. The solution is $y(t) = 2 + Ce^{-t}$. Using $y(0)=2$, we get $2 = 2 + C$, so $C=0$. Thus, $y(t) = 2$ for $0 \le t < 1$. To ensure continuity, the initial condition for the second interval is $y(1)=2$. Now, for $t \ge 1$, we solve $y' + y = 0$ with $y(1)=2$. The general solution is $y(t) = Ke^{-t}$. Using $y(1)=2$, we get $2 = Ke^{-1}$, which implies $K=2e$. So, for $t \ge 1$, the solution is $y(t) = 2e \cdot e^{-t} = 2e^{1-t}$. We need to find $y(2)$. Using the solution for $t \ge 1$, we get $y(2) = 2e^{1-2} = 2e^{-1}$.

Question 5

1. $x = \sqrt{10}-1$
2. $x = \sqrt[3]{5}-1$
3. $x = 2$ (correct answer)
4. No such value of $x$ exists.

Explanation: When you encounter a first-order linear differential equation like this one, recognize the standard form $y' + P(x)y = Q(x)$ and use an integrating factor to solve it systematically. First, rewrite the equation in standard form: $y' - \frac{2}{x+1}y = (x+1)^3$. Here, $P(x) = -\frac{2}{x+1}$ and $Q(x) = (x+1)^3$. The integrating factor is $\mu(x) = e^{\int P(x)dx} = e^{-2\ln(x+1)} = \frac{1}{(x+1)^2}$. Multiplying the entire equation by this factor gives: $\frac{1}{(x+1)^2}y' - \frac{2}{(x+1)^3}y = x+1$ The left side is the derivative of $\frac{y}{(x+1)^2}$, so: $\frac{d}{dx}\left[\frac{y}{(x+1)^2}\right] = x+1$ Integrating both sides: $\frac{y}{(x+1)^2} = \frac{x^2}{2} + x + C$ Therefore: $y = (x+1)^2\left(\frac{x^2}{2} + x + C\right)$ Using the initial condition $y(0) = -4$: $-4 = (1)^2\left(0 + 0 + C\right) = C$ So $y = (x+1)^2\left(\frac{x^2}{2} + x - 4\right)$ Setting $y = 0$: Since $(x+1)^2 > 0$ for $x > -1$, we need $\frac{x^2}{2} + x - 4 = 0$, or $x^2 + 2x - 8 = 0$. Factoring: $(x+4)(x-2) = 0$, giving $x = -4$ or $x = 2$. Since $x > -1$, we have $x = 2$. Answer choices A and B likely come from algebraic errors in solving the quadratic or misapplying the quadratic formula. Choice D is incorrect because we found a valid solution. Always verify your solution satisfies both the differential equation and initial condition, and check domain restrictions carefully.

Question 6

1. $(0, \infty)$
2. $(0, 3) \cup (3, 5) \cup (5, \infty)$
3. $(3, 5)$ (correct answer)
4. $(3, \infty)$

Explanation: When you encounter an initial value problem asking for the "largest interval" where a unique solution exists, you're dealing with existence and uniqueness theorems for first-order linear differential equations. First, rewrite the equation in standard form: $y' + \frac{\ln x}{x-3}y = \frac{1}{(x-3)(x-5)}$ For a unique solution to exist, both the coefficient function $P(x) = \frac{\ln x}{x-3}$ and the forcing function $Q(x) = \frac{1}{(x-3)(x-5)}$ must be continuous on an interval containing the initial point $x = 4$. Let's identify where these functions have discontinuities:

• $P(x)$ is discontinuous at $x = 0$ (where $\ln x$ is undefined) and $x = 3$ (where the denominator is zero)
• $Q(x)$ is discontinuous at $x = 3$ and $x = 5$ (where denominators are zero)

Question 7

1. Step 1
2. Step 3 (correct answer)
3. Step 4
4. Step 5

Explanation: Let's analyze the student's steps. Step 1 is correct; dividing the original equation by $x$ gives the standard form $y' + \frac{1}{x}y = 2$. Step 2 is correct; the integrating factor for $P(x)=1/x$ is indeed $x$. The error occurs in Step 3. When multiplying the standard form equation by the integrating factor $\mu(x)=x$, every term must be multiplied. The left side becomes $x(y' + \frac{1}{x}y) = xy' + y = (xy)'$, which is correct. The right side should be $\mu(x)Q(x) = x(2) = 2x$. The student incorrectly wrote the right side as just 2. So, the equation in Step 3 should have been $(xy)' = 2x$. All subsequent steps are based on this initial error.

Question 8

1. $0.500$ kg/L
2. $0.375$ kg/L
3. $0.250$ kg/L (correct answer)
4. $0.125$ kg/L

Explanation: The rate of change of salt, $dA/dt$, is the rate salt enters minus the rate salt leaves. Rate in = (concentration in) * (flow rate in) = $(0.5 \text{ kg/L}) * (2 \text{ L/min}) = 1$ kg/min. Rate out = (concentration in tank) * (flow rate out) = $(\frac{A(t)}{100} \text{ kg/L}) * (2 \text{ L/min}) = \frac{A}{50}$ kg/min. The differential equation is $\frac{dA}{dt} = 1 - \frac{A}{50}$. The initial condition is $A(0)=0$ (pure water). Rewriting in standard linear form: $A' + \frac{1}{50}A = 1$. The integrating factor is $\mu(t) = e^{t/50}$. The solution is $A(t) = 50 + Ce^{-t/50}$. Using $A(0)=0$, we find $0 = 50 + C$, so $C=-50$. The amount of salt at time $t$ is $A(t) = 50(1 - e^{-t/50})$. We need to find the amount at $t=50\ln(2)$: $A(50\ln 2) = 50(1 - e^{-50\ln(2)/50}) = 50(1 - e^{-\ln 2}) = 50(1 - 1/2) = 25$ kg. The question asks for the concentration, which is amount/volume. Concentration = $25 \text{ kg} / 100 \text{ L} = 0.25$ kg/L.

Question 9

1. $\sqrt{2} + 1/2$ (correct answer)
2. $\sqrt{2} + 1$
3. $2\sqrt{2}$
4. $1/2$

Explanation: The ODE is linear. The integrating factor is $\mu(x) = e^{\int \tan x dx} = e^{\ln|\sec x|} = |\sec x|$. Near $x=0$, $\sec x$ is positive, so $\mu(x)=\sec x$. Multiplying the ODE by $\mu(x)$ gives $(\sec(x) y)' = \sec(x) \cos^2(x) = \cos(x)$. Integrating both sides yields $\sec(x) y = \int \cos(x) dx = \sin(x) + C$. The general solution is $y(x) = \cos(x)(\sin(x)+C)$. Using the initial condition $y(0)=2$: $2 = \cos(0)(\sin(0)+C) = 1(0+C)$, which gives $C=2$. The particular solution is $y(x) = \cos(x)(\sin(x)+2)$. Evaluating at $x=\pi/4$: $y(\pi/4) = \cos(\pi/4)(\sin(\pi/4)+2) = \frac{\sqrt{2}}{2}(\frac{\sqrt{2}}{2}+2) = \frac{2}{4} + \frac{2\sqrt{2}}{2} = \frac{1}{2} + \sqrt{2}$.

Question 10

1. $y' - \frac{3}{x}y = \sec^2(x)$, with $y(0)=4$
2. $y' + \frac{3}{x}y = \sec^2(x)$, with $y(x_0)$ undetermined
3. $xy' + 3y = x \sec^2(x)$, with $y(0)=4$
4. $y' + \frac{3}{x}y = \sec^2(x)$, with $y(0)$ undefined (correct answer)

Explanation: The solution is given in the form $y(x) = \frac{1}{\mu(x)} (\int \mu(x)Q(x)dx + C)$. By comparing formats, we can identify the integrating factor as $\mu(x) = x^3$. Since $\mu(x) = e^{\int P(x)dx}$, we have $\int P(x)dx = \ln(x^3) = 3\ln(x)$. Thus, $P(x) = 3/x$. The standard form of the ODE is $y' + P(x)y = Q(x)$, which is $y' + \frac{3}{x}y = Q(x)$. We also see that $\mu(x)Q(x) = x^3 \sec^2(x)$, which implies $Q(x) = \sec^2(x)$. So the differential equation is $y' + \frac{3}{x}y = \sec^2(x)$. The constant of integration is given as $C=4$. The expression $\mu(x)y(x) = \int \mu(x)Q(x) dx + C$ is valid up to the point of applying an initial condition. The coefficients $P(x) = 3/x$ and $Q(x) = \sec^2(x)$ are not defined at $x=0$, so an initial condition cannot be specified at $x=0$. Therefore, $y(0)$ is undefined. The form of the solution implies the differential equation but does not constrain the initial condition to a specific value without more information.

Question 11

1. $39/48$
2. $41/48$ (correct answer)
3. $13/16$
4. $3/2$

Explanation: First, write the linear ODE in standard form $y' + P(t)y = Q(t)$ by dividing by $t$: $y' + \frac{2}{t}y = t - 1 + \frac{1}{t}$. The integrating factor is $\mu(t) = e^{\int (2/t) dt} = e^{2\ln t} = t^2$. Multiply the standard form equation by $\mu(t)$ to get $(t^2 y)' = t^2(t - 1 + 1/t) = t^3 - t^2 + t$. Integrate both sides: $t^2 y = \int (t^3 - t^2 + t) dt = \frac{t^4}{4} - \frac{t^3}{3} + \frac{t^2}{2} + C$. The general solution is $y(t) = \frac{t^2}{4} - \frac{t}{3} + \frac{1}{2} + \frac{C}{t^2}$. Use the initial condition $y(1) = 1/2$: $\frac{1}{2} = \frac{1}{4} - \frac{1}{3} + \frac{1}{2} + C$. This simplifies to $0 = \frac{3-4}{12} + C$, so $C = 1/12$. The particular solution is $y(t) = \frac{t^2}{4} - \frac{t}{3} + \frac{1}{2} + \frac{1}{12t^2}$. Evaluate at $t=2$: $y(2) = \frac{4}{4} - \frac{2}{3} + \frac{1}{2} + \frac{1}{12(4)} = 1 - \frac{2}{3} + \frac{1}{2} + \frac{1}{48} = \frac{1}{3} + \frac{1}{2} + \frac{1}{48} = \frac{5}{6} + \frac{1}{48} = \frac{40+1}{48} = \frac{41}{48}$.

Question 12

1. $\frac{dS}{dt} + \frac{2S}{100+t} = 1.5$, $S(0) = 20$ (correct answer)
2. $\frac{dS}{dt} + \frac{2S}{100-t} = 1.5$, $S(0) = 20$
3. $\frac{dS}{dt} + \frac{3S}{100+t} = 1.5$, $S(0) = 20$
4. $\frac{dS}{dt} + \frac{2S}{100+2t} = 1.5$, $S(0) = 20$

Explanation: The rate of salt change is: rate in - rate out. Salt flows in at $(0.5 \text{ lb/gal})(3 \text{ gal/min}) = 1.5$ lb/min. The volume at time $t$ is $100 + 3t - 2t = 100 + t$ gallons (net gain of 1 gal/min). The concentration at time $t$ is $\frac{S(t)}{100+t}$ lb/gal. Salt flows out at rate $\frac{S(t)}{100+t} \cdot 2 = \frac{2S(t)}{100+t}$ lb/min. Therefore: $\frac{dS}{dt} = 1.5 - \frac{2S}{100+t}$, or $\frac{dS}{dt} + \frac{2S}{100+t} = 1.5$. Option B incorrectly uses $100-t$, C uses wrong outflow coefficient 3 instead of 2, and D incorrectly uses $100+2t$ instead of $100+t$.

Question 13

1. $\frac{7}{10}$
2. $\frac{4}{5}$
3. $\frac{9}{10}$ (correct answer)
4. $1$

Explanation: First, rewrite in standard form: $\frac{dy}{dx} + \frac{2}{x}y = x^2$ for $x \neq 0$. The integrating factor is $\mu(x) = e^{\int \frac{2}{x}dx} = e^{2\ln|x|} = x^2$ (for $x > 0$). Multiplying by $x^2$: $x^2\frac{dy}{dx} + 2xy = x^4$. This gives $\frac{d}{dx}[x^2y] = x^4$. Integrating: $x^2y = \frac{x^5}{5} + C$. So $y = \frac{x^3}{5} + \frac{C}{x^2}$. Using $y(1) = \frac{1}{2}$: $\frac{1}{2} = \frac{1}{5} + C$, so $C = \frac{3}{10}$. Therefore $y(x) = \frac{x^3}{5} + \frac{3}{10x^2}$. Now $\int_1^2 y(x)dx = \int_1^2 \left(\frac{x^3}{5} + \frac{3}{10x^2}\right)dx = \left[\frac{x^4}{20} - \frac{3}{10x}\right]_1^2 = \left(\frac{16}{20} - \frac{3}{20}\right) - \left(\frac{1}{20} - \frac{3}{10}\right) = \frac{13}{20} - \frac{1}{20} + \frac{3}{10} = \frac{12}{20} + \frac{6}{20} = \frac{18}{20} = \frac{9}{10}$.