Differential Equations Quiz: Linear Odes With Initial Conditions
13 questions · exam conditions
0:00
Linear Odes With Initial ConditionsQuestion 1 of 13

The standard solution method for the initial value problem y′+(cot⁡t)y=2cos⁡ty' + (\cot t)y = 2\cos t with y(π/2)=3y(\pi/2)=3 involves an integrating factor μ(t)\mu(t) and leads to the form μ(t)y(t)=∫μ(t)(2cos⁡t)dt\mu(t)y(t) = \int \mu(t)(2\cos t) dt. What is the value of the constant of integration that arises from this indefinite integral?

22
5/25/2
33
7/27/2
← Back to quizzes

Differential Equations Quiz

Differential Equations Quiz: Linear Odes With Initial Conditions

Practice Linear Odes With Initial Conditions in Differential Equations with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

All questions

Question 1

The standard solution method for the initial value problem y′+(cot⁡t)y=2cos⁡ty' + (\cot t)y = 2\cos t with y(π/2)=3y(\pi/2)=3 involves an integrating factor μ(t)\mu(t) and leads to the form μ(t)y(t)=∫μ(t)(2cos⁡t)dt\mu(t)y(t) = \int \mu(t)(2\cos t) dt. What is the value of the constant of integration that arises from this indefinite integral?

  1. 22
  2. 5/25/2 (correct answer)
  3. 33
  4. 7/27/2
Explanation: The integrating factor is μ(t)=e∫cot⁡tdt=eln⁡∣sin⁡t∣=∣sin⁡t∣\mu(t) = e^{\int \cot t dt} = e^{\ln|\sin t|} = |\sin t|. Since the initial condition is at t=π/2t=\pi/2, we are in an interval where sin⁡t>0\sin t > 0, so μ(t)=sin⁡t\mu(t) = \sin t. The equation becomes (sin⁡(t)y)′=(sin⁡t)(2cos⁡t)=sin⁡(2t)(\sin(t) y)' = (\sin t)(2\cos t) = \sin(2t). Integrating both sides gives sin⁡(t)y=∫sin⁡(2t)dt=−12cos⁡(2t)+C\sin(t) y = \int \sin(2t) dt = -\frac{1}{2}\cos(2t) + C. Here, CC is the constant of integration. To find its value, we use the initial condition y(π/2)=3y(\pi/2)=3. Substitute t=π/2t=\pi/2 and y=3y=3 into the integrated equation: sin⁡(π/2)⋅3=−12cos⁡(2⋅π/2)+C\sin(\pi/2) \cdot 3 = -\frac{1}{2}\cos(2 \cdot \pi/2) + C. This gives 1⋅3=−12cos⁡(π)+C1 \cdot 3 = -\frac{1}{2}\cos(\pi) + C, which simplifies to 3=−12(−1)+C3 = -\frac{1}{2}(-1) + C, or 3=12+C3 = \frac{1}{2} + C. Solving for CC yields C=3−12=52C = 3 - \frac{1}{2} = \frac{5}{2}.

Question 2

For what value of the parameter α\alpha does the solution to the initial value problem y′−1xy=αx2y' - \frac{1}{x}y = \alpha x^2, with y(1)=2y(1)=2, satisfy the condition lim⁡x→∞y(x)x3=4\lim_{x \to \infty} \frac{y(x)}{x^3} = 4?

  1. 1/41/4
  2. 22
  3. 44
  4. 88 (correct answer)
Explanation: The ODE is linear with P(x)=−1/xP(x) = -1/x. The integrating factor is μ(x)=e∫−1/xdx=x−1\mu(x) = e^{\int -1/x dx} = x^{-1}. Multiplying by μ(x)\mu(x) gives (x−1y)′=x−1(αx2)=αx(x^{-1}y)' = x^{-1}(\alpha x^2) = \alpha x. Integrating both sides yields x−1y=∫αxdx=α2x2+Cx^{-1}y = \int \alpha x dx = \frac{\alpha}{2}x^2 + C. The general solution is y(x)=α2x3+Cxy(x) = \frac{\alpha}{2}x^3 + Cx. Using the initial condition y(1)=2y(1)=2: 2=α2(1)3+C(1)2 = \frac{\alpha}{2}(1)^3 + C(1), which gives C=2−α2C = 2 - \frac{\alpha}{2}. The particular solution is y(x)=α2x3+(2−α2)xy(x) = \frac{\alpha}{2}x^3 + (2 - \frac{\alpha}{2})x. Now, we evaluate the limit: lim⁡x→∞y(x)x3=lim⁡x→∞α2x3+(2−α2)xx3=lim⁡x→∞(α2+2−α/2x2)=α2\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, α2=4\frac{\alpha}{2} = 4, which implies α=8\alpha = 8.

Question 3

Let y(x)y(x) be the solution to the initial value problem y′−2x+1y=(x+1)3y' - \frac{2}{x+1}y = (x+1)^3 for x>−1x > -1, with the initial condition y(0)=−4y(0)=-4. For what value of xx is y(x)=0y(x)=0?

  1. x=10−1x = \sqrt{10}-1
  2. x=53−1x = \sqrt[3]{5}-1
  3. x=2x = 2 (correct answer)
  4. No such value of xx exists.
Explanation: When you encounter a first-order linear differential equation like this one, recognize the standard form y′+P(x)y=Q(x)y' + P(x)y = Q(x) and use an integrating factor to solve it systematically. First, rewrite the equation in standard form: y′−2x+1y=(x+1)3y' - \frac{2}{x+1}y = (x+1)^3. Here, P(x)=−2x+1P(x) = -\frac{2}{x+1} and Q(x)=(x+1)3Q(x) = (x+1)^3. The integrating factor is μ(x)=e∫P(x)dx=e−2ln⁡(x+1)=1(x+1)2\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: 1(x+1)2y′−2(x+1)3y=x+1\frac{1}{(x+1)^2}y' - \frac{2}{(x+1)^3}y = x+1 The left side is the derivative of y(x+1)2\frac{y}{(x+1)^2}, so: ddx[y(x+1)2]=x+1\frac{d}{dx}\left[\frac{y}{(x+1)^2}\right] = x+1 Integrating both sides: y(x+1)2=x22+x+C\frac{y}{(x+1)^2} = \frac{x^2}{2} + x + C Therefore: y=(x+1)2(x22+x+C)y = (x+1)^2\left(\frac{x^2}{2} + x + C\right) Using the initial condition y(0)=−4y(0) = -4: −4=(1)2(0+0+C)=C-4 = (1)^2\left(0 + 0 + C\right) = C So y=(x+1)2(x22+x−4)y = (x+1)^2\left(\frac{x^2}{2} + x - 4\right) Setting y=0y = 0: Since (x+1)2>0(x+1)^2 > 0 for x>−1x > -1, we need x22+x−4=0\frac{x^2}{2} + x - 4 = 0, or x2+2x−8=0x^2 + 2x - 8 = 0. Factoring: (x+4)(x−2)=0(x+4)(x-2) = 0, giving x=−4x = -4 or x=2x = 2. Since x>−1x > -1, we have x=2x = 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 4

A student attempts to solve the initial value problem xy′+y=2xxy' + y = 2x with y(1)=3y(1)=3. The student's work is shown below: Step 1: Write in standard form: y′+1xy=2y' + \frac{1}{x}y = 2. Step 2: Find the integrating factor: μ(x)=e∫1xdx=eln⁡x=x\mu(x) = e^{\int \frac{1}{x} dx} = e^{\ln x} = x. Step 3: Multiply the standard form by μ(x)\mu(x) to get (xy)′=2(xy)' = 2. Step 4: Integrate both sides: xy=2x+Cxy = 2x + C. Step 5: Apply the initial condition: 1(3)=2(1)+C  ⟹  C=11(3) = 2(1) + C \implies C=1. Step 6: Solve for yy: y(x)=2+1xy(x) = 2 + \frac{1}{x}.

In which step did the student make their first mistake?

  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 xx gives the standard form y′+1xy=2y' + \frac{1}{x}y = 2. Step 2 is correct; the integrating factor for P(x)=1/xP(x)=1/x is indeed xx. The error occurs in Step 3. When multiplying the standard form equation by the integrating factor μ(x)=x\mu(x)=x, every term must be multiplied. The left side becomes x(y′+1xy)=xy′+y=(xy)′x(y' + \frac{1}{x}y) = xy' + y = (xy)', which is correct. The right side should be μ(x)Q(x)=x(2)=2x\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(xy)' = 2x. All subsequent steps are based on this initial error.

Question 5

A 100-liter tank initially contains pure water. A brine solution with a salt concentration of 0.5 kg/L is pumped into the tank at a rate of 2 L/min. The well-mixed solution is pumped out at the same rate. Let A(t)A(t) be the amount of salt in kg in the tank at time tt in minutes. The process is stopped at t=50ln⁡(2)t=50\ln(2) minutes. What is the concentration of salt in the tank at that time?

  1. 0.5000.500 kg/L
  2. 0.3750.375 kg/L
  3. 0.2500.250 kg/L (correct answer)
  4. 0.1250.125 kg/L
Explanation: The rate of change of salt, dA/dtdA/dt, is the rate salt enters minus the rate salt leaves. Rate in = (concentration in) * (flow rate in) = (0.5 kg/L)∗(2 L/min)=1(0.5 \text{ kg/L}) * (2 \text{ L/min}) = 1 kg/min. Rate out = (concentration in tank) * (flow rate out) = (A(t)100 kg/L)∗(2 L/min)=A50(\frac{A(t)}{100} \text{ kg/L}) * (2 \text{ L/min}) = \frac{A}{50} kg/min. The differential equation is dAdt=1−A50\frac{dA}{dt} = 1 - \frac{A}{50}. The initial condition is A(0)=0A(0)=0 (pure water). Rewriting in standard linear form: A′+150A=1A' + \frac{1}{50}A = 1. The integrating factor is μ(t)=et/50\mu(t) = e^{t/50}. The solution is A(t)=50+Ce−t/50A(t) = 50 + Ce^{-t/50}. Using A(0)=0A(0)=0, we find 0=50+C0 = 50 + C, so C=−50C=-50. The amount of salt at time tt is A(t)=50(1−e−t/50)A(t) = 50(1 - e^{-t/50}). We need to find the amount at t=50ln⁡(2)t=50\ln(2): A(50ln⁡2)=50(1−e−50ln⁡(2)/50)=50(1−e−ln⁡2)=50(1−1/2)=25A(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 kg/100 L=0.2525 \text{ kg} / 100 \text{ L} = 0.25 kg/L.

Question 6

A tank initially contains 100 gallons of water with 20 pounds of salt dissolved in it. Water containing 0.5 pounds of salt per gallon flows in at 3 gallons per minute, and the well-mixed solution flows out at 2 gallons per minute. If S(t)S(t) represents the amount of salt at time tt, which differential equation with initial condition correctly models this situation?

  1. dSdt+2S100+t=1.5\frac{dS}{dt} + \frac{2S}{100+t} = 1.5, S(0)=20S(0) = 20 (correct answer)
  2. dSdt+2S100−t=1.5\frac{dS}{dt} + \frac{2S}{100-t} = 1.5, S(0)=20S(0) = 20
  3. dSdt+3S100+t=1.5\frac{dS}{dt} + \frac{3S}{100+t} = 1.5, S(0)=20S(0) = 20
  4. dSdt+2S100+2t=1.5\frac{dS}{dt} + \frac{2S}{100+2t} = 1.5, S(0)=20S(0) = 20
Explanation: The rate of salt change is: rate in - rate out. Salt flows in at (0.5 lb/gal)(3 gal/min)=1.5(0.5 \text{ lb/gal})(3 \text{ gal/min}) = 1.5 lb/min. The volume at time tt is 100+3t−2t=100+t100 + 3t - 2t = 100 + t gallons (net gain of 1 gal/min). The concentration at time tt is S(t)100+t\frac{S(t)}{100+t} lb/gal. Salt flows out at rate S(t)100+t⋅2=2S(t)100+t\frac{S(t)}{100+t} \cdot 2 = \frac{2S(t)}{100+t} lb/min. Therefore: dSdt=1.5−2S100+t\frac{dS}{dt} = 1.5 - \frac{2S}{100+t}, or dSdt+2S100+t=1.5\frac{dS}{dt} + \frac{2S}{100+t} = 1.5. Option B incorrectly uses 100−t100-t, C uses wrong outflow coefficient 3 instead of 2, and D incorrectly uses 100+2t100+2t instead of 100+t100+t.

Question 7

The function y(x)y(x) satisfies xy′+2y=x3xy' + 2y = x^3 with y(1)=12y(1) = \frac{1}{2}. What is the value of ∫12y(x)dx\int_1^2 y(x) dx?

  1. 710\frac{7}{10}
  2. 45\frac{4}{5}
  3. 910\frac{9}{10} (correct answer)
  4. 11
Explanation: First, rewrite in standard form: dydx+2xy=x2\frac{dy}{dx} + \frac{2}{x}y = x^2 for x≠0x \neq 0. The integrating factor is μ(x)=e∫2xdx=e2ln⁡∣x∣=x2\mu(x) = e^{\int \frac{2}{x}dx} = e^{2\ln|x|} = x^2 (for x>0x > 0). Multiplying by x2x^2: x2dydx+2xy=x4x^2\frac{dy}{dx} + 2xy = x^4. This gives ddx[x2y]=x4\frac{d}{dx}[x^2y] = x^4. Integrating: x2y=x55+Cx^2y = \frac{x^5}{5} + C. So y=x35+Cx2y = \frac{x^3}{5} + \frac{C}{x^2}. Using y(1)=12y(1) = \frac{1}{2}: 12=15+C\frac{1}{2} = \frac{1}{5} + C, so C=310C = \frac{3}{10}. Therefore y(x)=x35+310x2y(x) = \frac{x^3}{5} + \frac{3}{10x^2}. Now ∫12y(x)dx=∫12(x35+310x2)dx=[x420−310x]12=(1620−320)−(120−310)=1320−120+310=1220+620=1820=910\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}.

Question 8

Given the initial value problem y′+2y=4te2ty' + 2y = 4te^{2t} with y(0)=−1y(0)=-1, determine the value of y(1)y(1).

  1. 34(e2−e−2)\frac{3}{4}(e^2 - e^{-2}) (correct answer)
  2. 34e2\frac{3}{4}e^2
  3. 00
  4. 34e2+54e−2\frac{3}{4}e^2 + \frac{5}{4}e^{-2}
Explanation: This is a linear first-order ODE. The integrating factor is μ(t)=e∫2dt=e2t\mu(t) = e^{\int 2 dt} = e^{2t}. Multiplying the equation by μ(t)\mu(t) gives (e2ty)′=e2t(4te2t)=4te4t(e^{2t}y)' = e^{2t}(4te^{2t}) = 4te^{4t}. Integrating both sides requires integration by parts for the right side: ∫4te4tdt\int 4te^{4t} dt. Let u=4tu=4t and dv=e4tdtdv=e^{4t}dt. Then du=4dtdu=4dt and v=14e4tv=\frac{1}{4}e^{4t}. The integral is te4t−∫e4tdt=te4t−14e4t+Cte^{4t} - \int e^{4t} dt = te^{4t} - \frac{1}{4}e^{4t} + C. So, e2ty=te4t−14e4t+Ce^{2t}y = te^{4t} - \frac{1}{4}e^{4t} + C. Solving for yy gives y(t)=te2t−14e2t+Ce−2ty(t) = te^{2t} - \frac{1}{4}e^{2t} + Ce^{-2t}. Using the initial condition y(0)=−1y(0)=-1: −1=0⋅e0−14e0+Ce0-1 = 0 \cdot e^0 - \frac{1}{4}e^0 + Ce^0, which simplifies to −1=−1/4+C-1 = -1/4 + C, so C=−3/4C = -3/4. The solution is y(t)=te2t−14e2t−34e−2ty(t) = te^{2t} - \frac{1}{4}e^{2t} - \frac{3}{4}e^{-2t}. Evaluating at t=1t=1: y(1)=1e2−14e2−34e−2=34e2−34e−2=34(e2−e−2)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 9

Consider the initial value problem y′+y=Q(t)y' + y = Q(t) with y(0)=2y(0)=2, where the forcing function is Q(t)={2,0≤t<10,t≥1Q(t) = \begin{cases} 2, & 0 \le t < 1 \\ 0, & t \ge 1 \end{cases}. What is the value of y(2)y(2)?

  1. 2e−12e^{-1} (correct answer)
  2. 2e−22e^{-2}
  3. 11
  4. e−1e^{-1}
Explanation: We must solve this problem in two parts. First, for 0≤t<10 \le t < 1, we solve y′+y=2y' + y = 2 with y(0)=2y(0)=2. This is a linear ODE with integrating factor μ(t)=et\mu(t) = e^t. The solution is y(t)=2+Ce−ty(t) = 2 + Ce^{-t}. Using y(0)=2y(0)=2, we get 2=2+C2 = 2 + C, so C=0C=0. Thus, y(t)=2y(t) = 2 for 0≤t<10 \le t < 1. To ensure continuity, the initial condition for the second interval is y(1)=2y(1)=2. Now, for t≥1t \ge 1, we solve y′+y=0y' + y = 0 with y(1)=2y(1)=2. The general solution is y(t)=Ke−ty(t) = Ke^{-t}. Using y(1)=2y(1)=2, we get 2=Ke−12 = Ke^{-1}, which implies K=2eK=2e. So, for t≥1t \ge 1, the solution is y(t)=2e⋅e−t=2e1−ty(t) = 2e \cdot e^{-t} = 2e^{1-t}. We need to find y(2)y(2). Using the solution for t≥1t \ge 1, we get y(2)=2e1−2=2e−1y(2) = 2e^{1-2} = 2e^{-1}.

Question 10

A unique solution to the initial value problem (x−3)y′+(ln⁡x)y=1x−5(x-3)y' + (\ln x)y = \frac{1}{x-5}, with y(4)=0y(4)=0, is guaranteed to exist on the largest possible interval. What is that interval?

  1. (0,∞)(0, \infty)
  2. (0,3)∪(3,5)∪(5,∞)(0, 3) \cup (3, 5) \cup (5, \infty)
  3. (3,5)(3, 5) (correct answer)
  4. (3,∞)(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′+ln⁡xx−3y=1(x−3)(x−5)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)=ln⁡xx−3P(x) = \frac{\ln x}{x-3} and the forcing function Q(x)=1(x−3)(x−5)Q(x) = \frac{1}{(x-3)(x-5)} must be continuous on an interval containing the initial point x=4x = 4. Let's identify where these functions have discontinuities:
  • P(x)P(x) is discontinuous at x=0x = 0 (where ln⁡x\ln x is undefined) and x=3x = 3 (where the denominator is zero)
  • Q(x)Q(x) is discontinuous at x=3x = 3 and x=5x = 5 (where denominators are zero)
Since the initial condition is at x=4x = 4, we need the largest interval containing 4 where both functions are continuous. The discontinuities at x=3x = 3 and x=5x = 5 bracket the point x=4x = 4, so the largest such interval is (3,5)(3, 5). Choice A is wrong because it ignores the discontinuities at x=3x = 3 and x=5x = 5. Choice B incorrectly suggests the solution exists on multiple disjoint intervals, but uniqueness theorems guarantee existence on a single interval containing the initial point. Choice D ignores the critical discontinuity at x=5x = 5. Study tip: Always convert to standard form first, then find where coefficient and forcing functions are discontinuous. The solution interval is bounded by the nearest discontinuities on either side of your initial condition.

Question 11

What is the value of the solution to the initial value problem y′+(tan⁡x)y=cos⁡2xy' + (\tan x)y = \cos^2 x, with y(0)=2y(0)=2, at the point x=π/4x=\pi/4?

  1. 2+1/2\sqrt{2} + 1/2 (correct answer)
  2. 2+1\sqrt{2} + 1
  3. 222\sqrt{2}
  4. 1/21/2
Explanation: The ODE is linear. The integrating factor is μ(x)=e∫tan⁡xdx=eln⁡∣sec⁡x∣=∣sec⁡x∣\mu(x) = e^{\int \tan x dx} = e^{\ln|\sec x|} = |\sec x|. Near x=0x=0, sec⁡x\sec x is positive, so μ(x)=sec⁡x\mu(x)=\sec x. Multiplying the ODE by μ(x)\mu(x) gives (sec⁡(x)y)′=sec⁡(x)cos⁡2(x)=cos⁡(x)(\sec(x) y)' = \sec(x) \cos^2(x) = \cos(x). Integrating both sides yields sec⁡(x)y=∫cos⁡(x)dx=sin⁡(x)+C\sec(x) y = \int \cos(x) dx = \sin(x) + C. The general solution is y(x)=cos⁡(x)(sin⁡(x)+C)y(x) = \cos(x)(\sin(x)+C). Using the initial condition y(0)=2y(0)=2: 2=cos⁡(0)(sin⁡(0)+C)=1(0+C)2 = \cos(0)(\sin(0)+C) = 1(0+C), which gives C=2C=2. The particular solution is y(x)=cos⁡(x)(sin⁡(x)+2)y(x) = \cos(x)(\sin(x)+2). Evaluating at x=π/4x=\pi/4: y(π/4)=cos⁡(π/4)(sin⁡(π/4)+2)=22(22+2)=24+222=12+2y(\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 12

The solution to a certain initial value problem is y(x)=x−3(∫x3sec⁡2(x)dx+4)y(x) = x^{-3}(\int x^3 \sec^2(x) dx + 4). Which of the following is the initial value problem?

  1. y′−3xy=sec⁡2(x)y' - \frac{3}{x}y = \sec^2(x), with y(0)=4y(0)=4
  2. y′+3xy=sec⁡2(x)y' + \frac{3}{x}y = \sec^2(x), with y(x0)y(x_0) undetermined
  3. xy′+3y=xsec⁡2(x)xy' + 3y = x \sec^2(x), with y(0)=4y(0)=4
  4. y′+3xy=sec⁡2(x)y' + \frac{3}{x}y = \sec^2(x), with y(0)y(0) undefined (correct answer)
Explanation: The solution is given in the form y(x)=1μ(x)(∫μ(x)Q(x)dx+C)y(x) = \frac{1}{\mu(x)} (\int \mu(x)Q(x)dx + C). By comparing formats, we can identify the integrating factor as μ(x)=x3\mu(x) = x^3. Since μ(x)=e∫P(x)dx\mu(x) = e^{\int P(x)dx}, we have ∫P(x)dx=ln⁡(x3)=3ln⁡(x)\int P(x)dx = \ln(x^3) = 3\ln(x). Thus, P(x)=3/xP(x) = 3/x. The standard form of the ODE is y′+P(x)y=Q(x)y' + P(x)y = Q(x), which is y′+3xy=Q(x)y' + \frac{3}{x}y = Q(x). We also see that μ(x)Q(x)=x3sec⁡2(x)\mu(x)Q(x) = x^3 \sec^2(x), which implies Q(x)=sec⁡2(x)Q(x) = \sec^2(x). So the differential equation is y′+3xy=sec⁡2(x)y' + \frac{3}{x}y = \sec^2(x). The constant of integration is given as C=4C=4. The expression μ(x)y(x)=∫μ(x)Q(x)dx+C\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/xP(x) = 3/x and Q(x)=sec⁡2(x)Q(x) = \sec^2(x) are not defined at x=0x=0, so an initial condition cannot be specified at x=0x=0. Therefore, y(0)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 13

Consider the initial value problem ty′+2y=t2−t+1t y' + 2y = t^2 - t + 1 with the initial condition y(1)=1/2y(1) = 1/2. What is the value of y(2)y(2)?

  1. 39/4839/48
  2. 41/4841/48 (correct answer)
  3. 13/1613/16
  4. 3/23/2
Explanation: First, write the linear ODE in standard form y′+P(t)y=Q(t)y' + P(t)y = Q(t) by dividing by tt: y′+2ty=t−1+1ty' + \frac{2}{t}y = t - 1 + \frac{1}{t}. The integrating factor is μ(t)=e∫(2/t)dt=e2ln⁡t=t2\mu(t) = e^{\int (2/t) dt} = e^{2\ln t} = t^2. Multiply the standard form equation by μ(t)\mu(t) to get (t2y)′=t2(t−1+1/t)=t3−t2+t(t^2 y)' = t^2(t - 1 + 1/t) = t^3 - t^2 + t. Integrate both sides: t2y=∫(t3−t2+t)dt=t44−t33+t22+Ct^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)=t24−t3+12+Ct2y(t) = \frac{t^2}{4} - \frac{t}{3} + \frac{1}{2} + \frac{C}{t^2}. Use the initial condition y(1)=1/2y(1) = 1/2: 12=14−13+12+C\frac{1}{2} = \frac{1}{4} - \frac{1}{3} + \frac{1}{2} + C. This simplifies to 0=3−412+C0 = \frac{3-4}{12} + C, so C=1/12C = 1/12. The particular solution is y(t)=t24−t3+12+112t2y(t) = \frac{t^2}{4} - \frac{t}{3} + \frac{1}{2} + \frac{1}{12t^2}. Evaluate at t=2t=2: y(2)=44−23+12+112(4)=1−23+12+148=13+12+148=56+148=40+148=4148y(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}.