Integrating Factor Method - Differential Equations

1. Find the general solution for $x(y)$ for the differential equation $\frac{dx}{dy} + \frac{1}{y}x = y^2$, assuming $y>0$.
   Answer: $x(y) = \frac{y^3}{4} + \frac{C}{y}$ This is a first-order linear differential equation for $x$ as a function of $y$. The equation is already in standard form $dx/dy + P(y)x = Q(y)$, with $P(y) = 1/y$ and $Q(y) = y^2$. The integrating factor is $\mu(y) = e^{\int (1/y) dy} = e^{\ln y} = y$. Multiplying the equation by $\mu(y)$ gives $d/dy(y \cdot x) = y \cdot y^2 = y^3$. Integrating both sides with respect to $y$ gives $yx = \int y^3 dy = \frac{y^4}{4} + C$. Solving for $x$ yields $x(y) = \frac{y^3}{4} + \frac{C}{y}$.
2. Let $y(x)$ be the solution to the initial value problem $y' + \frac{2}{x}y = x^2$ with the initial condition $y(1)=1$. What is the value of $y(2)$?
   Answer: $9/5$ When you encounter a first-order linear differential equation like $y' + \frac{2}{x}y = x^2$, recognize this as the standard form $y' + P(x)y = Q(x)$ where $P(x) = \frac{2}{x}$ and $Q(x) = x^2$. The solution method requires finding an integrating factor. The integrating factor is $\mu(x) = e^{\int P(x)dx} = e^{\int \frac{2}{x}dx} = e^{2\ln|x|} = x^2$. Multiplying the entire equation by $x^2$ gives: $x^2y' + 2xy = x^4$. The left side is the derivative of $x^2y$, so we have $\frac{d}{dx}(x^2y) = x^4$. Integrating both sides: $x^2y = \int x^4 dx = \frac{x^5}{5} + C$, which gives us $y = \frac{x^3}{5} + \frac{C}{x^2}$. Using the initial condition $y(1) = 1$: $1 = \frac{1}{5} + C$, so $C = \frac{4}{5}$. Therefore, $y(x) = \frac{x^3}{5} + \frac{4}{5x^2}$. At $x = 2$: $y(2) = \frac{8}{5} + \frac{4}{5 \cdot 4} = \frac{8}{5} + \frac{1}{5} = \frac{9}{5}$, confirming answer B. Answer A ($8/5$) would result from forgetting the constant term. Answer C ($8$) likely comes from computational errors or misapplying the integrating factor. Answer D ($12/5$) might arise from incorrectly computing the constant or making integration mistakes. Remember: first-order linear equations always follow this integrating factor pattern. Practice identifying $P(x)$ and computing $e^{\int P(x)dx}$ accurately—it's the foundation of the entire solution process.
3. For the differential equation $\frac{dy}{dx} + y\cot x = \csc x$, a student finds the integrating factor to be $\sin x$ and obtains $y\sin x = -\cos x + C$. However, the final answer $y = -\cot x + C\csc x$ fails verification. What went wrong?
   Answer: The integration of the right-hand side was performed incorrectly The integrating factor $\mu(x) = e^{\int \cot x dx} = e^{\ln|\sin x|} = \sin x$ is correct. Multiplying: $\sin x \frac{dy}{dx} + y\sin x \cot x = \sin x \csc x = 1$. The left side is $\frac{d}{dx}[y\sin x]$. Integrating: $y\sin x = \int 1 dx = x + C$, not $-\cos x + C$. The student incorrectly integrated the right-hand side. The correct solution is $y = \frac{x + C}{\sin x} = (x + C)\csc x$. Choices A and D are incorrect about the integrating factor and integration respectively, while C is wrong because the issue isn't with verification technique.
4. The equation $x\frac{dy}{dx} + 2y = x^3\sin x$ can be solved using an integrating factor method after a preliminary step. What is the correct sequence of operations?
   Answer: Divide by $x$, then use integrating factor $e^{\int \frac{2}{x} dx} = x^2$ First, we need the equation in standard form $\frac{dy}{dx} + P(x)y = Q(x)$. Dividing by $x$: $\frac{dy}{dx} + \frac{2}{x}y = x^2\sin x$. Now $P(x) = \frac{2}{x}$, so the integrating factor is $\mu(x) = e^{\int \frac{2}{x} dx} = e^{2\ln|x|} = x^2$. Choice A has the wrong exponent (should be $\int \frac{2}{x} dx$, not $2\int \frac{1}{x} dx$), Choice B incorrectly multiplies by $x$, and Choice D suggests an inappropriate substitution method.
5. For what value of the constant $k$ does the solution to the initial value problem $y' + 2y = e^{kt}$ with $y(0)=3$ have a finite, non-zero limit as $x \to \infty$?
   Answer: $k = 0$ When analyzing the long-term behavior of solutions to first-order linear differential equations, you need to consider how each term in the solution behaves as the independent variable approaches infinity. To solve $y' + 2y = e^{kt}$, you'll use an integrating factor. The integrating factor is $\mu(t) = e^{2t}$. Multiplying through and integrating gives you the general solution: $y = Ce^{-2t} + \frac{e^{kt}}{k+2}$ (when $k \neq -2$). Applying the initial condition $y(0) = 3$: $3 = C + \frac{1}{k+2}$, so $C = 3 - \frac{1}{k+2}$. For the solution to have a finite, non-zero limit as $t \to \infty$, the exponential terms must behave appropriately. The term $Ce^{-2t}$ always approaches zero since the exponent is negative. The term $\frac{e^{kt}}{k+2}$ determines the limit behavior. When $k = 0$, this term becomes $\frac{1}{2}$, which is constant. Therefore, $\lim_{t \to \infty} y(t) = 0 + \frac{1}{2} = \frac{1}{2}$, giving us a finite, non-zero limit. This makes (C) correct. (A) $k = -2$ creates division by zero in our solution formula, requiring special handling and ultimately leading to unbounded growth. (B) $k = 2$ makes $e^{2t}$ grow exponentially, so the limit is infinite. (D) is incorrect since we found that $k = 0$ works. Study tip: For linear differential equations with exponential forcing functions, the key to finite limits is matching the forcing function's growth rate to the natural decay rate of the homogeneous solution.
6. For the differential equation $\frac{dy}{dx} + y\tan x = \sec x$, which of the following represents the correct integrating factor and the form of the general solution?
   Answer: Integrating factor: $\sec x$; Solution: $y = \sin x + C\cos x$ The integrating factor is $\mu(x) = e^{\int \tan x dx} = e^{\ln|\sec x|} = \sec x$. Multiplying the equation: $\sec x \frac{dy}{dx} + y\sec x \tan x = \sec^2 x$. The left side is $\frac{d}{dx}[y\sec x]$. Integrating: $y\sec x = \int \sec^2 x dx = \tan x + C$. Therefore $y = (\tan x + C)\cos x = \sin x + C\cos x$. Choice A has wrong solution form, Choice B has wrong integrating factor, Choice D has both wrong integrating factor and solution.
7. A mixing tank initially contains 100 L of pure water. A salt solution with a concentration of 0.2 kg/L enters the tank at a rate of 5 L/min. The well-mixed solution is drained from the tank at the same rate. Let $A(t)$ be the amount of salt (in kg) in the tank at time $t$ (in minutes). What is the limiting amount of salt in the tank as $t \to \infty$?
   Answer: 20 kg The rate of change of salt is $dA/dt = \text{rate in} - \text{rate out}$. The rate in is $(0.2 \text{ kg/L}) \times (5 \text{ L/min}) = 1 \text{ kg/min}$. The concentration in the tank is $A(t)/100$ kg/L, so the rate out is $(A(t)/100 \text{ kg/L}) \times (5 \text{ L/min}) = A(t)/20 \text{ kg/min}$. The DE is $dA/dt = 1 - A/20$, or $A' + (1/20)A = 1$. This is a linear equation. The integrating factor is $\mu(t) = e^{\int(1/20)dt} = e^{t/20}$. The solution is $A(t) = 20 + Ce^{-t/20}$. With $A(0)=0$, we find $C=-20$. So, $A(t) = 20 - 20e^{-t/20}$. As $t \to \infty$, the term $e^{-t/20} \to 0$, so $\lim_{t\to\infty} A(t) = 20$. Alternatively, at equilibrium, $dA/dt=0$, so $1 - A/20 = 0$, which gives $A=20$.
8. Consider the differential equation $t^2\frac{dy}{dt} + 3ty = \ln(t)$. If the integrating factor method is applied, the equation can be written in the form $\frac{d}{dt}(\mu(t)y) = F(t)$. What is the function $F(t)$?
   Answer: $t \ln(t)$ When you encounter a first-order linear differential equation, the integrating factor method transforms it into a form where the left side becomes the derivative of a product. This question tests whether you can correctly identify what appears on the right side after this transformation. Starting with $t^2\frac{dy}{dt} + 3ty = \ln(t)$, first divide by $t^2$ to get standard form: $\frac{dy}{dt} + \frac{3}{t}y = \frac{\ln(t)}{t^2}$ The integrating factor is $\mu(t) = e^{\int \frac{3}{t}dt} = e^{3\ln(t)} = t^3$ Multiplying the standard form by $t^3$: $t^3\frac{dy}{dt} + 3t^2y = t^3 \cdot \frac{\ln(t)}{t^2} = t\ln(t)$ The left side is indeed $\frac{d}{dt}(t^3y)$, so $F(t) = t\ln(t)$. Choice A is correct: $t\ln(t)$ is exactly what we get when multiplying the original right side by the integrating factor. Choice B ($t^3\ln(t)$) represents a common error where students forget to divide by $t^2$ before applying the integrating factor, essentially multiplying $\ln(t)$ directly by $t^3$. Choice C ($\frac{\ln(t)}{t^2}$) is the right side of the equation in standard form before multiplying by the integrating factor. Choice D ($\ln(t)$) is simply the original right side, showing no transformation was applied. Study tip: Always convert to standard form first, then find the integrating factor. The function $F(t)$ equals the integrating factor times the right side of the standard form equation.
9. What is the general solution to the differential equation $y' + y \cot(x) = 2 \cos(x)$ for $0 < x < \pi$?
   Answer: $y = \sin(x) + C\csc(x)$ When you encounter a first-order linear differential equation like this one, recognize the standard form $y' + P(x)y = Q(x)$ where $P(x) = \cot(x)$ and $Q(x) = 2\cos(x)$. The solution method requires finding an integrating factor $\mu(x) = e^{\int P(x)dx}$. First, calculate the integrating factor: $\mu(x) = e^{\int \cot(x)dx} = e^{\ln|\sin(x)|} = \sin(x)$ (for $0 < x < \pi$ where $\sin(x) > 0$). Multiply the entire equation by $\sin(x)$: $\sin(x)y' + \sin(x)y\cot(x) = 2\sin(x)\cos(x)$ This simplifies to: $\sin(x)y' + y\cos(x) = \sin(2x)$ The left side is the derivative of $\sin(x)y$, so: $\frac{d}{dx}[\sin(x)y] = \sin(2x)$ Integrating both sides: $\sin(x)y = \int \sin(2x)dx = -\frac{1}{2}\cos(2x) + C$ Therefore: $y = \frac{-\frac{1}{2}\cos(2x) + C}{\sin(x)} = -\frac{1}{2}\cos(2x)\csc(x) + C\csc(x)$ Using the identity $\cos(2x) = 1 - 2\sin^2(x)$, this becomes $y = \sin(x) + C\csc(x)$, which is answer B. Answer A incorrectly involves a logarithmic term not present in this solution. Answer C has the wrong sign and coefficient for the cosine term. Answer D incorrectly uses secant instead of cosecant and includes terms that don't arise from this differential equation. Strategy tip: For first-order linear equations, always identify $P(x)$ correctly and remember that $\int \cot(x)dx = \ln|\sin(x)|$, making $\mu(x) = \sin(x)$.
10. The differential equation $(x^2 + 1)y' + 2xy = \cos(x)$ is a first-order linear equation. What is its general solution for $y(x)$?
    Answer: $y = \frac{\sin(x) + C}{x^2 + 1}$ First, put the equation in standard form $y' + P(x)y = Q(x)$ by dividing by $(x^2 + 1)$: $y' + \frac{2x}{x^2+1}y = \frac{\cos(x)}{x^2+1}$. The integrating factor is $\mu(x) = e^{\int \frac{2x}{x^2+1} dx} = e^{\ln(x^2+1)} = x^2+1$. Multiplying the standard form equation by $\mu(x)$ gives $d/dx((x^2+1)y) = \cos(x)$. Integrating both sides yields $(x^2+1)y = \int \cos(x) dx = \sin(x) + C$. Finally, solving for $y$ gives $y = \frac{\sin(x) + C}{x^2 + 1}$.
11. To solve $y' + P(x)y = Q(x)$, the integrating factor is typically chosen as $\mu(x) = e^{\int P(x)dx}$. Suppose instead, an integrating factor $\mu^*(x) = A \cdot e^{\int P(x)dx}$ is used, where $A$ is a positive constant other than 1. How does the resulting family of solutions $y^*(x)$ relate to the family of solutions $y(x)$ obtained using $\mu(x)$?
    Answer: The family of solutions $y^*(x)$ is identical to the family of solutions $y(x)$. Using $\mu^*(x) = A\mu(x)$, the differential equation becomes $d/dx(\mu^*(x)y) = \mu^*(x)Q(x)$. Integrating gives $\mu^*(x)y = \int \mu^*(x)Q(x)dx + C^*$, where $C^*$ is an arbitrary constant. Substituting $\mu^*(x) = A\mu(x)$ gives $A\mu(x)y = \int A\mu(x)Q(x)dx + C^* = A\int \mu(x)Q(x)dx + C^*$. Dividing by $A\mu(x)$ yields $y^*(x) = \frac{1}{\mu(x)}(\int \mu(x)Q(x)dx + \frac{C^*}{A})$. Since $C^*$ is an arbitrary constant, $C = C^*/A$ is also an arbitrary constant. Thus, $y^*(x) = \frac{1}{\mu(x)}(\int \mu(x)Q(x)dx + C)$, which represents the exact same family of solutions as $y(x)$.
12. Consider the initial value problem $y' - y = xe^x$ with the initial condition $y(0) = 2$. What is the value of $y(2)$?
    Answer: $4e^2$ The equation is in standard form with $P(x) = -1$ and $Q(x) = xe^x$. The integrating factor is $\mu(x) = e^{\int -1 dx} = e^{-x}$. Multiplying the equation by $\mu(x)$ gives $d/dx(e^{-x}y) = e^{-x}(xe^x) = x$. Integrating both sides, we get $e^{-x}y = \int x dx = \frac{x^2}{2} + C$. The general solution is $y(x) = (\frac{x^2}{2} + C)e^x$. Using the initial condition $y(0)=2$, we find $2 = (0 + C)e^0$, so $C=2$. The particular solution is $y(x) = (\frac{x^2}{2} + 2)e^x$. Evaluating at $x=2$ gives $y(2) = (\frac{2^2}{2} + 2)e^2 = (2+2)e^2 = 4e^2$.
13. A student attempts to solve the initial value problem $xy' + y = 4x^3$ with $y(1)=5$. Their work is shown below: Step 1: Rearrange to $y' + (1/x)y = 4x^2$. Step 2: The integrating factor is $\mu(x) = e^{\int(1/x)dx} = e^{\ln x} = x$. Step 3: Multiply by $\mu(x)$: $x(y' + (1/x)y) = x(4x^2)$, which simplifies to $d/dx(xy) = 4x^3$. Step 4: Integrate both sides: $xy = \int 4x^3 dx = x^4$. Step 5: Solve for $y$: $y = x^3$. Step 6: Apply initial condition: $y(1) = 1^3 = 1$. Since this does not match $y(1)=5$, the student concludes there is no solution. In which step did the first error occur?
    Answer: Step 4 When solving first-order linear differential equations using integrating factors, you must carefully track each algebraic step, especially during integration where constants of integration are crucial. Let's trace through the student's work to find where things went wrong. Steps 1-3 are executed correctly: the equation is properly rearranged into standard form $y' + \frac{1}{x}y = 4x^2$, the integrating factor $\mu(x) = x$ is computed correctly, and multiplying through gives $\frac{d}{dx}(xy) = 4x^3$. The error occurs in Step 4 during integration. When integrating $4x^3$, the student gets $x^4$, but this is missing the constant of integration. The correct integration should yield $xy = \int 4x^3 dx = x^4 + C$, giving $y = x^3 + \frac{C}{x}$. Applying the initial condition $y(1) = 5$ allows us to solve for $C$: $5 = 1 + C$, so $C = 4$. The complete solution is $y = x^3 + \frac{4}{x}$. Choice A is wrong because Step 1 correctly rearranges to standard form. Choice B is wrong because Step 3 properly applies the integrating factor. Choice D is wrong because Step 6 correctly applies the initial condition to the (incomplete) solution from Step 5. Study tip: Always include the constant of integration when solving differential equations, even if it seems obvious. The initial conditions exist specifically to determine this constant, and forgetting it will lead to incorrect solutions that don't satisfy the given conditions.
14. A tank contains 100 gallons of brine with 10 pounds of salt. Pure water flows in at 5 gal/min and the mixture flows out at the same rate. The differential equation $\frac{dS}{dt} + \frac{S}{20} = 0$ models the salt content $S(t)$. If someone claims the integrating factor approach gives $S(t) = 10e^{-t/20}$, what verification step would confirm this is correct?
    Answer: Substitute into the original equation and verify both sides are equal When you're given a proposed solution to a differential equation, the most reliable way to verify its correctness is to substitute it back into the original equation and check that both sides are equal. This is the fundamental verification method in differential equations. Let's see why option A is correct. If $S(t) = 10e^{-t/20}$, then $\frac{dS}{dt} = 10 \cdot (-\frac{1}{20})e^{-t/20} = -\frac{1}{2}e^{-t/20}$. Substituting into the original equation: $-\frac{1}{2}e^{-t/20} + \frac{10e^{-t/20}}{20} = -\frac{1}{2}e^{-t/20} + \frac{1}{2}e^{-t/20} = 0$. Since both sides equal zero, the solution is verified. Option B checks boundary conditions and limiting behavior, which are important for physical reasonableness but don't prove the solution satisfies the differential equation itself. A function could have the right initial value and limit but still not solve the equation. Option C focuses on verifying the integrating factor computation, but even if the integrating factor is correct, errors could occur in subsequent steps. This doesn't verify the final answer. Option D examines physical interpretation, which is valuable for understanding but insufficient for mathematical verification. The solution could make physical sense yet still be mathematically incorrect. Remember: when verifying any proposed solution to a differential equation, always substitute back into the original equation first. This direct substitution is your most powerful and definitive verification tool.
15. The differential equation $\frac{dy}{dx} - \frac{y}{x} = \frac{e^x}{x^2}$ has solution $y = \frac{e^x + C}{x}$ for $x > 0$. If we extend this solution to $x < 0$, which statement about the general solution is correct?
    Answer: The solution becomes $y = \frac{e^x + C_1}{x}$ for $x > 0$ and $y = \frac{e^x + C_2}{x}$ for $x < 0$ When solving linear first-order differential equations across domains that include singular points (like $x = 0$ here), you need to consider each connected domain separately. The point $x = 0$ creates a discontinuity that prevents solutions from being extended across it. To understand why, let's examine the integrating factor method. For this equation $\frac{dy}{dx} - \frac{y}{x} = \frac{e^x}{x^2}$, the integrating factor is $\mu(x) = e^{-\int \frac{1}{x}dx} = e^{-\ln|x|} = \frac{1}{|x|}$. When you multiply through and integrate, you get different constants of integration for $x > 0$ and $x < 0$ because these are separate integration processes over disconnected intervals. The correct answer is (D) because the general solution must allow for different constants $C_1$ and $C_2$ on each side of the singular point. Each domain $(x > 0$ and $x < 0)$ represents a separate initial value problem with its own arbitrary constant. (A) is wrong because it assumes continuity across $x = 0$, which isn't possible given the singular behavior of the coefficients. (B) incorrectly modifies the solution form rather than recognizing the need for separate constants. (C) is incorrect because the integrating factor $\frac{1}{|x|}$ works perfectly well for $x < 0$; the issue isn't undefined behavior but rather separate domains. Study tip: When differential equations have singular points, always solve on each connected domain separately. The constants of integration are independent across these domains.

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