Linear ODEs with Initial Conditions - Differential Equations

1. 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 $S(t)$ represents the amount of salt at time $t$, which differential equation with initial condition correctly models this situation?
   Answer: $\frac{dS}{dt} + \frac{2S}{100+t} = 1.5$, $S(0) = 20$ 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 $100+t$.
2. Given the initial value problem $y' + 2y = 4te^{2t}$ with $y(0)=-1$, determine the value of $y(1)$.
   Answer: $\frac{3}{4}(e^2 - e^{-2})$ 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 $t=1$: $y(1) = 1e^{2} - \frac{1}{4}e^{2} - \frac{3}{4}e^{-2} = \frac{3}{4}e^2 - \frac{3}{4}e^{-2} = \frac{3}{4}(e^2 - e^{-2})$.
3. Consider the initial value problem $t y' + 2y = t^2 - t + 1$ with the initial condition $y(1) = 1/2$. What is the value of $y(2)$?
   Answer: $41/48$ 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 $t=2$: $y(2) = \frac{4}{4} - \frac{2}{3} + \frac{1}{2} + \frac{1}{12(4)} = 1 - \frac{2}{3} + \frac{1}{2} + \frac{1}{48} = \frac{1}{3} + \frac{1}{2} + \frac{1}{48} = \frac{5}{6} + \frac{1}{48} = \frac{40+1}{48} = \frac{41}{48}$.
4. For what value of the parameter $\alpha$ does the solution to the initial value problem $y' - \frac{1}{x}y = \alpha x^2$, with $y(1)=2$, satisfy the condition $\lim_{x \to \infty} \frac{y(x)}{x^3} = 4$?
   Answer: $8$ 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 $\alpha = 8$.
5. The standard solution method for the initial value problem $y' + (\cot t)y = 2\cos t$ with $y(\pi/2)=3$ involves an integrating factor $\mu(t)$ and leads to the form $\mu(t)y(t) = \int \mu(t)(2\cos t) dt$. What is the value of the constant of integration that arises from this indefinite integral?
   Answer: $5/2$ 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 $C = 3 - \frac{1}{2} = \frac{5}{2}$.
6. Let $y(x)$ be the solution to the initial value problem $y' - \frac{2}{x+1}y = (x+1)^3$ for $x > -1$, with the initial condition $y(0)=-4$. For what value of $x$ is $y(x)=0$?
   Answer: $x = 2$ 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.
7. What is the value of the solution to the initial value problem $y' + (\tan x)y = \cos^2 x$, with $y(0)=2$, at the point $x=\pi/4$?
   Answer: $\sqrt{2} + 1/2$ 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 $x=\pi/4$: $y(\pi/4) = \cos(\pi/4)(\sin(\pi/4)+2) = \frac{\sqrt{2}}{2}(\frac{\sqrt{2}}{2}+2) = \frac{2}{4} + \frac{2\sqrt{2}}{2} = \frac{1}{2} + \sqrt{2}$.
8. 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 $A(t)$ be the amount of salt in kg in the tank at time $t$ in minutes. The process is stopped at $t=50\ln(2)$ minutes. What is the concentration of salt in the tank at that time?
   Answer: $0.250$ kg/L 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.
9. In which step did the student make their first mistake?
   Answer: Step 3 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.
10. A unique solution to the initial value problem $(x-3)y' + (\ln x)y = \frac{1}{x-5}$, with $y(4)=0$, is guaranteed to exist on the largest possible interval. What is that interval?
    Answer: $(3, 5)$ 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) Since the initial condition is at $x = 4$, we need the largest interval containing 4 where both functions are continuous. The discontinuities at $x = 3$ and $x = 5$ bracket the point $x = 4$, so the largest such interval is $(3, 5)$. Choice A is wrong because it ignores the discontinuities at $x = 3$ and $x = 5$. 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 $x = 5$. 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.
11. The function $y(x)$ satisfies $xy' + 2y = x^3$ with $y(1) = \frac{1}{2}$. What is the value of $\int_1^2 y(x) dx$?
    Answer: $\frac{9}{10}$ 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 $\int_1^2 y(x)dx = \int_1^2 \left(\frac{x^3}{5} + \frac{3}{10x^2}\right)dx = \left[\frac{x^4}{20} - \frac{3}{10x}\right]_1^2 = \left(\frac{16}{20} - \frac{3}{20}\right) - \left(\frac{1}{20} - \frac{3}{10}\right) = \frac{13}{20} - \frac{1}{20} + \frac{3}{10} = \frac{12}{20} + \frac{6}{20} = \frac{18}{20} = \frac{9}{10}$.
12. Consider the initial value problem $y' + y = Q(t)$ with $y(0)=2$, where the forcing function is $Q(t) = \begin{cases} 2, & 0 \le t < 1 \\ 0, & t \ge 1 \end{cases}$. What is the value of $y(2)$?
    Answer: $2e^{-1}$ 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 $y(2) = 2e^{1-2} = 2e^{-1}$.
13. The solution to a certain initial value problem is $y(x) = x^{-3}(\int x^3 \sec^2(x) dx + 4)$. Which of the following is the initial value problem?
    Answer: $y' + \frac{3}{x}y = \sec^2(x)$, with $y(0)$ undefined 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.

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