Differential Equations - IB Mathematics Analysis and Approaches

1. Find the particular solution of $\frac{dy}{dx} = \frac{y}{x\ln x}$ given that the solution passes through $(e, e)$.
   Answer: $y = e \ln x$ First, separate the variables: $$\frac{dy}{y} = \frac{dx}{x\ln x}$$ Integrate both sides. For the right side, use the substitution $u = \ln x$, so $du = \frac{1}{x} dx$. $$\int \frac{1}{y} \, dy = \int \frac{1}{x\ln x} \, dx = \int \frac{1}{u} \, du$$ $$\ln|y| = \ln|u| + C = \ln|\ln x| + C$$ Now use the initial condition $(e, e)$ to find $C$: $$\ln|e| = \ln|\ln e| + C$$ $$1 = \ln|1| + C$$ $$1 = 0 + C \Rightarrow C = 1$$ Substitute $C=1$ back into the equation: $$\ln|y| = \ln|\ln x| + 1$$ To solve for $y$, exponentiate both sides: $$|y| = e^{\ln|\ln x| + 1} = e^{\ln|\ln x|} e^1 = |\ln x| \cdot e$$ Since the point $(e,e)$ has positive coordinates, we can drop the absolute values. $$y = e \ln x$$ * A is incorrect; this would be the solution if $C=0$ (e.g., if the curve passed through $(e, 1)$). * C is incorrect; this is the solution to $\frac{dy}{dx} = \frac{y}{x}$, a common misinterpretation of the original equation. * D is incorrect due to a sign error during integration or algebraic manipulation.
2. The tangent to the curve $y=f(x)$ at the point $(x,y)$ has a gradient of $\frac{x}{y^2+1}$. The curve passes through the point (2, 0). Find the equation of the curve.
   Answer: $2y^3 + 6y = 3x^2 - 12$ The problem gives the differential equation $\frac{dy}{dx} = \frac{x}{y^2+1}$. Separate the variables: $$(y^2+1) \, dy = x \, dx$$ Integrate both sides: $$\int (y^2+1) \, dy = \int x \, dx$$ $$\frac{y^3}{3} + y = \frac{x^2}{2} + C$$ Use the point (2, 0) to find the constant $C$: $$\frac{0^3}{3} + 0 = \frac{2^2}{2} + C$$ $$0 = 2 + C \Rightarrow C = -2$$ Substitute $C=-2$ back into the equation: $$\frac{y^3}{3} + y = \frac{x^2}{2} - 2$$ To clear the denominators and match the answer choices, multiply the entire equation by 6: $$2y^3 + 6y = 3x^2 - 12$$ * B is incorrect because it fails to find and include the constant of integration $C$. * C is incorrect and results from incorrectly separating the variables as $\frac{dy}{y^2+1} = x \, dx$. * D is incorrect because it results from integrating without the proper coefficients (e.g., $\int y^2 dy = y^3$).
3. The rate of growth of a population $P$ of bacteria is directly proportional to the square root of the population size. Initially, the population is 100. After 1 hour, the population is 225. Find the population after 3 hours.
   Answer: 625 The problem states that $\frac{dP}{dt} = k\sqrt{P}$ for some constant $k$. Separate the variables: $P^{-1/2} dP = k \, dt$. Integrate both sides: $\int P^{-1/2} dP = \int k \, dt$ which gives $2P^{1/2} = kt + C$. Use the initial condition $P(0)=100$: $2\sqrt{100} = k(0) + C \Rightarrow 2(10) = C \Rightarrow C=20$. The equation becomes $2\sqrt{P} = kt + 20$. Use the condition $P(1)=225$: $2\sqrt{225} = k(1) + 20 \Rightarrow 2(15) = k+20 \Rightarrow 30 = k+20 \Rightarrow k=10$. The particular solution is $2\sqrt{P} = 10t + 20$, which simplifies to $\sqrt{P} = 5t + 10$. To find the population after 3 hours, let $t=3$: $\sqrt{P} = 5(3) + 10 = 15 + 10 = 25$. $P = 25^2 = 625$. * A is incorrect; this would be the population at $t=2$. * B is incorrect; this results from assuming a linear growth model based on the first two points. * D is incorrect and likely results from an arithmetic error in calculating $k$ or in the final step.
4. Water is leaking from a cylindrical tank of radius $R$ and height $H$. The rate of change of the volume of water, $V$, is proportional to the square root of the height, $h$, of the water (Torricelli's Law). Given that $V=\pi R^2 h$, which differential equation represents the rate of change of the height of the water, $\frac{dh}{dt}$? (Assume $k$ is a positive constant)
   Answer: $\frac{dh}{dt} = -k\sqrt{h}$ When you encounter a related rates problem involving Torricelli's Law, you need to connect two pieces of information: how volume changes with time, and how volume relates to height. Start with what you're told: the rate of volume change is proportional to the square root of height, so $\frac{dV}{dt} = -k\sqrt{h}$ (negative because water is leaking out). You're also given that $V = \pi R^2 h$ for the cylindrical tank. To find $\frac{dh}{dt}$, differentiate the volume formula with respect to time: $\frac{dV}{dt} = \pi R^2 \frac{dh}{dt}$. Since you know $\frac{dV}{dt} = -k\sqrt{h}$, you can substitute: $-k\sqrt{h} = \pi R^2 \frac{dh}{dt}$. Solving for $\frac{dh}{dt}$: $\frac{dh}{dt} = \frac{-k\sqrt{h}}{\pi R^2}$. Since $k$ is just a positive constant, you can absorb $\frac{1}{\pi R^2}$ into a new constant, giving $\frac{dh}{dt} = -k\sqrt{h}$. Option A incorrectly keeps the radius and π separate rather than combining them into the constant. Option B gives $-kh$ instead of $-k\sqrt{h}$, missing the square root relationship from Torricelli's Law. Option C has $-\frac{k}{\sqrt{h}}$, which would mean the rate increases as height decreases—the opposite of what Torricelli's Law predicts. Option D correctly captures the proportional relationship to $\sqrt{h}$ with the proper negative sign. Remember: in related rates problems, always differentiate the constraint equation and substitute known relationships to isolate your target derivative.
5. The rate of change of a variable $y$ with respect to $x$ is given by $\frac{dy}{dx} = 2y(1-\frac{y}{10})$. Given that $y(0)=1$, find $\lim_{x\to\infty} y(x)$.
   Answer: 10 The given differential equation $\frac{dy}{dx} = 2y(1-\frac{y}{10})$ is a logistic growth model of the form $\frac{dy}{dx} = ky(1-\frac{y}{M})$, where $k=2$ is the growth rate and $M=10$ is the carrying capacity. The equilibrium solutions occur when $\frac{dy}{dx} = 0$, which is at $y=0$ and $y=10$. The solution $y=10$ is a stable equilibrium (the carrying capacity), while $y=0$ is an unstable equilibrium. For any initial condition $y(0) > 0$, the solution $y(x)$ will approach the carrying capacity $M$ as $x \to \infty$. Since the initial condition is $y(0)=1$, which is between 0 and 10, the population will grow and approach the carrying capacity. Therefore, $\lim_{x\to\infty} y(x) = 10$. This question tests the conceptual understanding of the long-term behavior of a logistic model, which does not require solving the differential equation explicitly. * A is incorrect because $y=0$ is an unstable equilibrium; solutions move away from it (unless they start there). * B is incorrect because this is the initial value, not the long-term limit. * D is incorrect because growth is limited by the carrying capacity, preventing it from becoming infinite.
6. Find the general solution of the differential equation $y \frac{dy}{dx} = x^2$.
   Answer: $3y^2 - 2x^3 = C$ To solve the differential equation, we first separate the variables: $$y \, dy = x^2 \, dx$$ Next, we integrate both sides: $$\int y \, dy = \int x^2 \, dx$$ $$\frac{y^2}{2} = \frac{x^3}{3} + K$$ where $K$ is the constant of integration. To match the form of the answer choices, we can multiply the entire equation by 6 to clear the denominators: $$3y^2 = 2x^3 + 6K$$ Rearranging the terms, we get: $$3y^2 - 2x^3 = 6K$$ Since (6K) is an arbitrary constant, we can represent it by $C$. Thus, the general solution is: $$3y^2 - 2x^3 = C$$ * A is incorrect because it results from integrating without the correct coefficients (i.e., $y^2 = x^3 + C$). * B is incorrect because the coefficients for $y^2$ and $x^3$ are swapped, a common algebraic error. * D is incorrect because it comes from an incorrect integration of the left side, $\int y \, dy \neq \ln|y|$.
7. The slope field for a differential equation is such that at any point $(x,y)$, the slope is equal to the product of its coordinates. The solution curve passes through $(0, -2)$. Find the equation of this curve.
   Answer: $y = -2e^{x^2/2}$ The description of the slope field translates to the differential equation $\frac{dy}{dx} = xy$. Separate the variables: $\frac{dy}{y} = x \, dx$. Integrate both sides: $\int \frac{1}{y} \, dy = \int x \, dx$, which gives $\ln|y| = \frac{x^2}{2} + C$. Use the initial condition $(0, -2)$ to find $C$: $\ln|-2| = \frac{0^2}{2} + C \Rightarrow C = \ln(2)$. Substitute $C$ back: $\ln|y| = \frac{x^2}{2} + \ln(2)$. Exponentiate both sides to solve for $y$: $|y| = e^{\frac{x^2}{2} + \ln(2)} = e^{\frac{x^2}{2}} e^{\ln(2)} = 2e^{x^2/2}$. Since the curve must pass through a point with a negative y-coordinate $(-2)$, we must choose the negative solution for $y$. $y = -2e^{x^2/2}$. * B is incorrect because it uses the positive branch of the solution, which does not pass through $(0,-2)$. * C is incorrect and results from an error in handling the constant of integration, likely by treating $y = e^{x^2/2} + C$. * D is incorrect and arises from incorrectly separating the variables as $y \, dy = x \, dx$.
8. The value $V$ of a machine, in thousands of dollars, depreciates over time $t$ in years according to the differential equation $\frac{dV}{dt} = -k(V-5)$, where 5 is the machine's scrap value. The initial value is $65,000. After 2 years, its value is $20,000. What is the value of the machine after 4 years?
   Answer: $8,750 The differential equation is $\frac{dV}{dt} = -k(V-5)$. The general solution is $V(t) = 5 + (V_0 - 5)e^{-kt}$. The values are in thousands of dollars, so $V_0 = 65$. $$V(t) = 5 + (65 - 5)e^{-kt} = 5 + 60e^{-kt}$$ We are given that $V(2) = 20$. We use this to find the decay constant. $$20 = 5 + 60e^{-2k}$$ $$15 = 60e^{-2k}$$ $$\frac{15}{60} = e^{-2k} \implies \frac{1}{4} = e^{-2k}$$ This means $(e^{-k})^2 = (\frac{1}{2})^2$, so $e^{-k} = \frac{1}{2}$. Now we can write the particular solution as: $$V(t) = 5 + 60(e^{-k})^t = 5 + 60(\frac{1}{2})^t$$ We need to find the value after 4 years, i.e., $V(4)$. $$V(4) = 5 + 60(\frac{1}{2})^4 = 5 + 60(\frac{1}{16}) = 5 + \frac{15}{4} = 5 + 3.75 = 8.75$$ Since the value is in thousands of dollars, the value is $8,750. * A is incorrect and likely results from an algebraic error in the final calculation. * C is incorrect and might result from assuming linear decay or another incorrect model. * D is incorrect; this is the value at $t=3$, not $t=4$.
9. Find the general solution to the differential equation $\frac{dy}{dx} = \frac{\cos^2 y}{\tan x}$.
   Answer: $\tan y = \ln|\sin x| + C$ First, separate the variables: $$\frac{dy}{\cos^2 y} = \frac{dx}{\tan x}$$ Recall that $\frac{1}{\cos^2 y} = \sec^2 y$ and $\frac{1}{\tan x} = \cot x$. The equation becomes: $$\sec^2 y \, dy = \cot x \, dx$$ Integrate both sides: $$\int \sec^2 y \, dy = \int \cot x \, dx$$ The integral of $\sec^2 y$ is $\tan y$. The integral of $\cot x = \frac{\cos x}{\sin x}$ is $\ln|\sin x|$. Therefore, the general solution is: $$\tan y = \ln|\sin x| + C$$ * B is incorrect because $\int \cot x \, dx \neq \ln|\cos x|$. This is a common integration error. * C is incorrect because $\int \sec^2 y \, dy \neq \sec y$. * D is incorrect due to multiple integration errors on both sides of the equation.
10. A hot object at temperature $100^\circ\text{C}$ is placed in a room with a constant temperature of $20^\circ\text{C}$. The rate of cooling of the object is proportional to the difference between its temperature and the room's temperature. After 5 minutes, the object's temperature is $60^\circ\text{C}$. Which equation models the temperature $T$ of the object after $t$ minutes?
    Answer: $T(t) = 20 + 80(2)^{-t/5}$ Newton's Law of Cooling gives the differential equation $\frac{dT}{dt} = -k(T - T_{ambient})$, where $T_{ambient} = 20$. So, $\frac{dT}{dt} = -k(T-20)$. The general solution to this DE is $T(t) = T_{ambient} + (T_0 - T_{ambient})e^{-kt}$. Given $T_0 = 100$, the equation is $T(t) = 20 + (100-20)e^{-kt} = 20 + 80e^{-kt}$. We are given that $T(5) = 60$. We use this to find $k$: $$60 = 20 + 80e^{-5k}$$ $$40 = 80e^{-5k}$$ $$\frac{1}{2} = e^{-5k}$$ Now we can express $e^{-k}$ as $(e^{-5k})^{1/5} = (1/2)^{1/5}$. So $e^{-kt} = (e^{-k})^t = ((1/2)^{1/5})^t = (1/2)^{t/5} = 2^{-t/5}$. Substituting this back into the temperature equation gives: $$T(t) = 20 + 80(2)^{-t/5}$$ * B is incorrect because it represents heating, resulting from a sign error in the exponent. * C is incorrect because it uses an incorrect initial temperature difference (60 instead of 80). * D is incorrect because it is the solution to $\frac{dT}{dt} = -kT$, which ignores the ambient room temperature.
11. The solution to the differential equation $\frac{dy}{dx} = 2xy^2$ passes through the point (0, 1/2). Find the particular solution for $y$.
    Answer: $y = \frac{1}{2-x^2}$ First, separate the variables: $$\frac{dy}{y^2} = 2x \, dx$$ Integrate both sides: $$\int y^{-2} \, dy = \int 2x \, dx$$ $$-y^{-1} = x^2 + C$$ $$-\frac{1}{y} = x^2 + C$$ Now, use the initial condition (0, 1/2) to find the constant $C$: $$-\frac{1}{1/2} = 0^2 + C$$ $$-2 = C$$ Substitute $C=-2$ back into the equation: $$-\frac{1}{y} = x^2 - 2$$ Finally, solve for $y$: $$\frac{1}{y} = -(x^2 - 2) = 2 - x^2$$ $$y = \frac{1}{2-x^2}$$ * B is incorrect due to a sign error in the final step of solving for $y$. * C is incorrect due to a sign error when calculating the constant of integration $C$. * D is incorrect because it results from incorrectly separating the variables as $\frac{dy}{y} = 2x \, dx$, which leads to a logarithmic integral.
12. Given the logistic differential equation $\frac{dP}{dt} = P(5-P)$ with initial condition $P(0)=1$, which of the following is an expression for $P(t)$?
    Answer: $P(t) = \frac{5e^{5t}}{4+e^{5t}}$ Separate the variables: $\frac{dP}{P(5-P)} = dt$. Use partial fraction decomposition for the left side: $\frac{1}{P(5-P)} = \frac{A}{P} + \frac{B}{5-P}$. This yields $1 = A(5-P) + BP$. Setting $P=0$ gives $A=1/5$. Setting $P=5$ gives $B=1/5$. The integral becomes: $\int \frac{1}{5}(\frac{1}{P} + \frac{1}{5-P}) dP = \int dt$. $\frac{1}{5}(\ln|P| - \ln|5-P|) = t + C_1$. $\ln|\frac{P}{5-P}| = 5t + C_2$. Exponentiating both sides: $\frac{P}{5-P} = e^{5t+C_2} = Ae^{5t}$. Use the initial condition $P(0)=1$: $\frac{1}{5-1} = Ae^0 \Rightarrow A = \frac{1}{4}$. So, $\frac{P}{5-P} = \frac{1}{4}e^{5t}$. Now, solve for $P$: $4P = (5-P)e^{5t} = 5e^{5t} - Pe^{5t}$. $P(4+e^{5t}) = 5e^{5t}$. $P(t) = \frac{5e^{5t}}{4+e^{5t}}$. * B is incorrect due to an error in the integration step, likely dropping the factor of 5. * C is incorrect due to an error in calculating the constant $A$, finding $A=4$ instead of $1/4$. * D is incorrect due to multiple errors in the integration and algebraic manipulation.
13. Which of the following is the solution to the differential equation $\frac{dy}{dx} = (x+y)^2$, found using the substitution $u=x+y$?
    Answer: $y = \tan(x+C) - x$ When you encounter a differential equation that doesn't separate easily, look for substitution opportunities. Here, the term $(x+y)^2$ suggests using $u = x+y$, which transforms this into a separable equation. Starting with $u = x+y$, we have $\frac{du}{dx} = 1 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{du}{dx} - 1$. Substituting into the original equation: $\frac{du}{dx} - 1 = u^2$, which gives us $\frac{du}{dx} = u^2 + 1$. This separates as $\frac{du}{u^2 + 1} = dx$. Integrating both sides: $\int \frac{du}{u^2 + 1} = \int dx$. The left side integrates to $\arctan(u)$, giving us $\arctan(u) = x + C$. Taking the tangent of both sides: $u = \tan(x + C)$. Since $u = x + y$, we have $x + y = \tan(x + C)$, so $y = \tan(x + C) - x$. Option A has the wrong form entirely—it resembles integration of $u^{-2}$ rather than $\frac{1}{u^2 + 1}$. Option B incorrectly places the constant outside the tangent function, which would arise from integrating incorrectly. Option C mistakenly treats this as if $y$ alone appeared in the denominator, suggesting confusion about the substitution process. Remember: when you see expressions like $(x+y)^n$ in differential equations, try the substitution $u = x+y$. This often converts complex equations into standard separable forms, and don't forget that $\int \frac{du}{u^2+1} = \arctan(u)$.
14. The equation of a curve passing through $(\frac{\pi}{6}, 1)$ has a slope at any point $(x,y)$ given by $\frac{dy}{dx} = \sqrt{4-y^2}$. Find the equation of the curve.
    Answer: $y = 2\sin(x)$ This is a differential equation problem where you need to find a function given its derivative and an initial condition. When you see $\frac{dy}{dx} = \sqrt{4-y^2}$, recognize that this requires separation of variables to solve. To solve this differential equation, separate the variables: $\frac{dy}{\sqrt{4-y^2}} = dx$. The left side integrates to $\arcsin(\frac{y}{2})$, and the right side integrates to $x + C$. This gives you $\arcsin(\frac{y}{2}) = x + C$, so $y = 2\sin(x + C)$. Using the initial condition that the curve passes through $(\frac{\pi}{6}, 1)$: $1 = 2\sin(\frac{\pi}{6} + C)$, which means $\sin(\frac{\pi}{6} + C) = \frac{1}{2}$. Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$, we need $\frac{\pi}{6} + C = \frac{\pi}{6}$, giving us $C = 0$. Therefore, $y = 2\sin(x)$. Let's verify why the other options fail: Option A, $y = \sin(2x)$, gives $\frac{dy}{dx} = 2\cos(2x)$, which doesn't match our given derivative. Option B, $y = 2\sin(x + \frac{\pi}{6})$, doesn't satisfy the initial condition since $2\sin(\frac{\pi}{6} + \frac{\pi}{6}) = 2\sin(\frac{\pi}{3}) = \sqrt{3} \neq 1$. Option C, $y = 2\cos(x)$, gives the wrong derivative and fails the initial condition. When solving differential equations on the IB exam, always verify your solution by checking both the derivative relationship and the initial condition. This two-step verification catches most algebraic errors.
15. The rate of change of a quantity $N$ is modeled by the differential equation $\frac{dN}{dt} = -0.2(N-10)$. Given that $N(0)=50$, find the value of $t$ for which $N(t)=20$.
    Answer: $5\ln(4)$ First, solve the differential equation by separating variables: $$\frac{dN}{N-10} = -0.2 \, dt$$ Integrating both sides gives: $$\ln|N-10| = -0.2t + C$$ Exponentiating both sides: $$N-10 = e^{-0.2t + C} = Ae^{-0.2t} \quad (\text{where } A = e^C)$$ $$N(t) = 10 + Ae^{-0.2t}$$ Use the initial condition $N(0)=50$: $$50 = 10 + Ae^0 \Rightarrow A = 40$$ So the particular solution is $N(t) = 10 + 40e^{-0.2t}$. Now, find $t$ when $N(t)=20$: $$20 = 10 + 40e^{-0.2t}$$ $$10 = 40e^{-0.2t}$$ $$\frac{1}{4} = e^{-0.2t}$$ Take the natural logarithm of both sides: $$\ln(1/4) = -0.2t$$ $$-\ln(4) = -0.2t$$ $$t = \frac{\ln(4)}{0.2} = \frac{\ln(4)}{1/5} = 5\ln(4)$$ * A is incorrect and results from an arithmetic error, likely calculating $20-10=10$ on the left side but $50-20=30$ to set up the ratio. * C is incorrect and results from using $A=50$ instead of $A=40$. * D is incorrect due to a sign error in the final step.
16. The rate of increase of the number of people, $N$, in a town is proportional to the number of people currently in the town and also proportional to the difference between the maximum number of people the town can support, $M$, and the current number of people. Which differential equation models this situation, where $k$ is a positive constant of proportionality?
    Answer: $\frac{dN}{dt} = kN(M-N)$ When you encounter problems about rates of change that depend on multiple factors, you need to carefully translate each condition into mathematical terms and combine them appropriately. The problem states the rate of increase is proportional to both the current population N and the difference (M-N). When a quantity is proportional to multiple factors simultaneously, you multiply those factors together. Since the rate equals some constant times N times (M-N), you get $\frac{dN}{dt} = kN(M-N)$, which is answer D. This is actually the famous logistic growth model. When N is small, (M-N) is large, so growth accelerates. As N approaches M, (M-N) approaches zero, slowing growth to nearly zero as the population nears the town's capacity. Answer A, $\frac{dN}{dt} = k(M-N)$, only accounts for the capacity constraint but ignores that growth should also depend on the current population size. This would give the same growth rate whether you have 10 people or 10,000 people. Answer B, $\frac{dN}{dt} = k(N + (M-N)) = kM$, simplifies to a constant growth rate, completely missing both proportionality conditions. Answer C, $\frac{dN}{dt} = \frac{k(M-N)}{N}$, incorrectly divides by N instead of multiplying, which would make the growth rate decrease as population increases—the opposite of what's described. Remember: when a rate is proportional to multiple quantities simultaneously, multiply those quantities together. Watch for this pattern in population dynamics, chemical reactions, and other real-world modeling problems.
17. A curve is defined by the differential equation $\frac{dy}{dx} = \frac{ky}{x}$. Given the curve passes through the points $(1, 2)$ and $(4, 1)$, find the value of $k$.
    Answer: -0.5 First, find the general solution of the differential equation. Separate variables: $$\frac{dy}{y} = k \frac{dx}{x}$$ Integrate both sides: $$\int \frac{1}{y} dy = k \int \frac{1}{x} dx$$ $$\ln|y| = k \ln|x| + C$$ Using logarithm properties, this can be written as $\ln|y| = \ln|x^k| + C$. Exponentiating both sides gives $|y| = e^C |x^k|$, which simplifies to the general form $y = Ax^k$. Now, use the two given points to find $A$ and $k$. Using $(1, 2)$: $$2 = A(1)^k \Rightarrow A = 2$$ So the particular solution has the form $y = 2x^k$. Now use the second point $(4, 1)$: $$1 = 2(4^k)$$ $$\frac{1}{2} = 4^k$$ To solve for $k$, take the natural logarithm of both sides: $$\ln(\frac{1}{2}) = \ln(4^k)$$ $$-\ln(2) = k \ln(4)$$ $$-\ln(2) = k \ln(2^2)$$ $$-\ln(2) = k (2\ln(2))$$ Divide by $2\ln(2)$: $$k = -\frac{1}{2} = -0.5$$ * A is incorrect due to an error in solving the logarithmic equation. * C is incorrect due to a sign error. * D is incorrect due to both a sign error and an inversion error.
18. Solve the differential equation $\frac{dy}{dx} = e^{x-y}$ given the initial condition $y(0)=\ln(2)$.
    Answer: $y = \ln(e^x + 1)$ First, rewrite the differential equation using exponent properties: $\frac{dy}{dx} = e^x e^{-y}$. Separate the variables: $$\frac{dy}{e^{-y}} = e^x \, dx$$ $$e^y \, dy = e^x \, dx$$ Integrate both sides: $$\int e^y \, dy = \int e^x \, dx$$ $$e^y = e^x + C$$ Now use the initial condition $y(0)=\ln(2)$ to find $C$: $$e^{\ln(2)} = e^0 + C$$ $$2 = 1 + C \Rightarrow C = 1$$ Substitute $C=1$ back into the equation: $$e^y = e^x + 1$$ Solve for $y$ by taking the natural logarithm of both sides: $$y = \ln(e^x + 1)$$ * B is incorrect due to an arithmetic error when calculating $C$, likely $2-1=2$. * C is incorrect; this would be the solution to the much simpler equation $\frac{dy}{dx}=1$. * D is incorrect and arises from a sign error when calculating $C$, finding $C=-1$.
19. Find the general solution to the differential equation $x\frac{dy}{dx} - 2y = 0$ for $x > 0$.
    Answer: $y = Ax^2$ First, rearrange the equation to separate the variables: $$x\frac{dy}{dx} = 2y$$ $$\frac{dy}{y} = \frac{2}{x} dx$$ Integrate both sides: $$\int \frac{1}{y} \, dy = \int \frac{2}{x} \, dx$$ $$\ln|y| = 2\ln|x| + C$$ Since $x>0$, we can write this as $\ln|y| = 2\ln(x) + C$. Using logarithm properties, $2\ln(x) = \ln(x^2)$: $$\ln|y| = \ln(x^2) + C$$ To solve for $y$, exponentiate both sides: $$|y| = e^{\ln(x^2) + C} = e^{\ln(x^2)} e^C$$ $$|y| = x^2 e^C$$ Let $A = \pm e^C$, which is an arbitrary non-zero constant. We can also include the case $y=0$ (the trivial solution), so A can be any real number. $$y = Ax^2$$ * B is incorrect because it results from incorrectly handling the constant of integration after exponentiating. * C is incorrect due to an error in solving for $y$ from $\ln|y|$. * D is incorrect from incorrectly integrating $\frac{dy}{y}$ as y or some other error.

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