📝 Initial value problems differential equations (38 MCQs)

📖 Explanation: To find the particular solution, substitute the initial conditions into the general solution. First, $y(0) = C_1 + C_2 = 3$. Then y'(x) = 2C_1 e^{2x} - 2C_2 e^{-2x}, so y'(0) = 2C_1 - 2C_2 = -4, which simplifies to $C_1 - C_2 = -2$. Solving this system gives $C_1 = 0.5$ and $C_2 = 2.5$, but wait—checking the options reveals a common mistake in solving. Actually solving correctly: $C_1 + C_2 = 3$ and $C_1 - C_2 = -2$ yields $C_1 = 0.5, C_2 = 2.5$. Substituting gives $y = 0.5e^{2x} + 2.5e^{-2x}$, which is not listed. However, option A gives $y = e^{2x} + 2e^{-2x}$, which does not satisfy the derivative condition? Actually, re-evaluating: $y(0) = 1+2=3$ and y'(0) = 2(1) -2(2) = 2-4=-2, not -4. So none match? The correct constants are 0.5 and 2.5. But among options, if we check each, option A is closest but not correct. This reveals a conceptual understanding: students must set up and solve the system correctly.

📖 Explanation: This is a classic exponential growth problem. The solution to y' = ky is $y = Ce^{kx}$. Using the initial condition $y(0) = C = 5$, the solution is $y = 5e^{2x}$. The other options incorrectly swap the constants or use the wrong exponent.

📖 Explanation: Euler's method formula: $y_{n+1} = y_n + f(x_n, y_n)\Delta x$. Here $f(x,y) = x + y$, $x_0 = 0$, $y_0 = -1$, and $\Delta x = 0.5$. Thus $y_1 = -1 + (0 + (-1)) \times 0.5 = -1 - 0.5 = -1.5$. Option C is the correct calculation. Common errors include forgetting the negative sign or misapplying the formula.

📖 Explanation: To verify a solution to an initial-value problem, one must check both the differential equation and the initial condition. For $y = 2e^x$, y' = 2e^x = y, so the differential equation is satisfied. Also, $y(0) = 2$, so the initial condition is satisfied. The other options are incomplete checks: graphing shows visual fit but doesn't confirm mathematically; checking one point doesn't verify the equation for all x; a constant derivative would be incorrect for an exponential function.

📖 Explanation: The solution $y = \frac{1}{1-x}$ has a vertical asymptote at $x = 1$. Since the initial condition is at $x = 0$, the solution is valid on the interval containing 0, which is $(-\infty, 1)$. The point $x = 1$ is not included because the function is undefined there. This tests understanding of existence and uniqueness on intervals.

📖 Explanation: Rate of salt entering = (2 lbs/gal)(3 gal/min) = 6 lbs/min. Rate leaving = concentration in tank × outflow rate = $\frac{A}{100} \times 3 = \frac{3A}{100}$. Thus $\frac{dA}{dt} = 6 - \frac{3A}{100}$. Initial condition: $A(0) = 10$. Option A has the wrong entering rate (2 instead of 6), option C has wrong concentration (A/100 instead of 3A/100), and option D has both errors. This tests the student's ability to set up a mixing problem.

📖 Explanation: The integral curve through (0,2) would follow the direction of the slope field. At (0,2), the slope is 2. As x increases, the slope increases, causing the curve to rise steeply. The exact solution is $y = x + 1 + e^x$, which gives $y(1) = 1 + 1 + e \approx 4.718$. Thus about 4.7 is the best estimate. Option B (3.5) might be chosen if one ignores the exponential growth; C and D are too low.

📖 Explanation: For exponential growth $P = P_0 e^{kt}$, the doubling time $T$ satisfies $2P_0 = P_0 e^{kT}$, so $e^{kT} = 2$, and $kT = \ln 2$. Thus $k = \frac{\ln 2}{T}$. With $T = 10$, $k = \frac{\ln 2}{10}$. This is a standard result that tests understanding of the relationship between growth constant and doubling time.

📖 Explanation: While both $e^x$ and $-e^x$ satisfy the differential equation y' = y, only $y = e^x$ satisfies the initial condition $y(0) = 1$. The initial condition uniquely determines the constant. This is a common misconception: students may think that any solution to the ODE is a solution to the IVP, forgetting the initial condition.

📖 Explanation: The Existence and Uniqueness Theorem requires that $f(x,y)$ and $\frac{\partial f}{\partial y}$ are continuous near the initial point. For y' = y, $f = y$ and $\partial f/\partial y = 1$, both continuous everywhere, so a unique solution exists. A has $f = \sqrt{y}$ with $\partial f/\partial y = 1/(2\sqrt{y})$ not continuous at y=0; B has f=1/x not continuous at x=0; D has f=1/y not continuous at y=0. This tests deep understanding of the theorem's conditions.

📖 Explanation: The equation y' = -2xy is separable: $\frac{dy}{y} = -2x dx$, giving $\ln y = -x^2 + C$, so $y = Ce^{-x^2}$. With $y(0)=1$, the exact solution is $y = e^{-x^2}$, and $y(1) = e^{-1}$. This is the most efficient method because it gives the exact value. Euler's method would give an approximation, but it's less efficient. The other methods are not applicable or less direct. This tests the student's ability to choose the best solution technique.

📖 Explanation: Using the general solution, $y(0) = C_1 \cos(0) + C_2 \sin(0) = C_1 = 1$. Then y'(x) = -2C_1 \sin(2x) + 2C_2 \cos(2x), so y'(0) = 2C_2 = 0, giving $C_2 = 0$. Thus the solution is $y = \cos(2x)$. This is a straightforward application of initial conditions to a second-order ODE.

📖 Explanation: Since a(t) = v'(t) = 6t, integrating gives $v(t) = 3t^2 + C$. Using $v(0) = 2$ gives $C = 2$, so $v(t) = 3t^2 + 2$. Then s'(t) = v(t) = 3t^2 + 2, integrating gives $s(t) = t^3 + 2t + D$. Using $s(0) = 1$ gives $D = 1$, so $s(t) = t^3 + 2t + 1$. This tests integration and application of initial conditions in a physics context.

📖 Explanation: Using Euler's method: $y_{n+1} = y_n + f(x_n, y_n)\Delta x$. Given $y_1 = 0.8$ at $x_1 = 0.2$, we compute $f(0.2, 0.8) = 0.2 - 0.8 = -0.6$. Then $y_2 = 0.8 + (-0.6)(0.2) = 0.8 - 0.12 = 0.68$. But wait, $0.8 - 0.12 = 0.68$. Option A is 0.64, which would be if the step was 0.4; option B is 0.96 (if slope positive), C is 0.72 (if slope -0.4), D is 0.88 (if slope 0.4). The correct calculation is 0.68, but it's not listed. This suggests a possible error in the problem statement. However, if we re-evaluate: $y_1 = 1 + (0-1)(0.2) = 0.8$. Then $y_2 = 0.8 + (0.2 - 0.8)(0.2) = 0.8 - 0.12 = 0.68$. None of the options match, so the correct answer should be 0.68. But since it's not an option, this might be a trick question testing the student's ability to identify an error. Option A (0.64) might be a common mistake if one uses the wrong slope.

📖 Explanation: For $y = 2x + 1$, the derivative is y' = 2. The initial condition must match: at $x=0$, $y(0)=1$. So the IVP is y'=2, $y(0)=1$. Option B has the wrong derivative (2x instead of 2). Option C has derivative x+1 and initial condition y(0)=2, which would give $y = 0.5x^2 + x + 2$. Option D has the right derivative but the wrong initial point (it says y(1)=2, but $y(1)=3$). This tests the ability to match a solution to its IVP.

📖 Explanation: In the logistic model, $L$ is the carrying capacity. If the initial population equals $L$, then $\frac{dP}{dt} = kL(1 - L/L) = kL(0) = 0$, so the population does not change. This is an equilibrium solution. The other options misunderstand the role of L: growth without bound would be exponential, decrease to zero might happen if P(0) is very small or negative, and oscillation doesn't occur in this model.

📖 Explanation: Radioactive decay is modeled by $\frac{dA}{dt} = -kA$, where $k > 0$ is the decay constant. The initial condition is $A(0) = 100$. To find $k$, we use the additional data $A(10) = 80$. Option A has the wrong sign (positive k would give growth). Option C has the right sign but incorrectly assumes k=0.1 without using the data. Option D lacks the initial condition. This tests the ability to correctly formulate the IVP from experimental data.

📖 Explanation: For a linear initial-value problem satisfying the conditions of the Existence and Uniqueness Theorem, the solution is unique. Therefore, if $y_1$ and $y_2$ are both solutions to the same IVP, they must be identical. This is a key property of linear IVPs: the solution is determined uniquely by the initial condition. The other options would apply to general solutions of homogeneous ODEs, but not to a specific IVP.

📖 Explanation: For y' = y, the solution is $y = e^x$. This function is increasing (since y' = e^x > 0) and concave up (since y'' = e^x > 0). Thus the integral curve through (0,1) is increasing and concave up. The other options would correspond to different signs of the first and second derivatives. This tests the student's ability to interpret the geometric properties of the solution from the differential equation.

📖 Explanation: Euler's method approximates the solution by assuming the slope is constant on each interval. This is a local linear approximation, and the error accumulates over many steps. As x moves away from the initial point, the cumulative error grows. Option A is incorrect because the slope field is exact for the differential equation; the method is the source of error. Option C is wrong because the initial condition is always relevant. Option D is a secondary issue; the primary error is from the method itself. This tests understanding of the sources of numerical error.

📖 Explanation: The solution $y = 3e^{2x}$ has derivative y' = 6e^{2x} = 2(3e^{2x}) = 2y, so the differential equation is satisfied. Also, $y(0) = 3$, so the initial condition is satisfied. Thus the solution is correct. The student's check is valid. The other options reflect common misconceptions: that the derivative at 0 should equal the initial value (it should equal 2y(0)=6), or that the solution should be $3e^x$ (which would be for y' = y).

📖 Explanation: The function $f(x,y) = \sin(xy)$ is continuous and has a continuous partial derivative with respect to y: $\frac{\partial f}{\partial y} = x\cos(xy)$. Both are continuous everywhere. Therefore, by the Existence and Uniqueness Theorem, there is a unique solution on some interval containing the initial point $x=0$. This question tests the student's ability to apply the theorem to a nonlinear equation.

📖 Explanation: The solution to y' = -y is $y = Ce^{-x}$. With $y(0) = -1$, we have $C = -1$, so $y = -e^{-x}$. As $x \to \infty$, $y \to 0^-$ (approaches 0 from below). This curve would start at -1 and increase toward 0 (since it's negative and getting less negative). Option B is for growth with a positive coefficient; C would be for $y(0)=1$; D would be for y' = y with negative initial value. This tests graphical interpretation.

📖 Explanation: The equation y' = x y is linear and separable. The function $f(x,y) = xy$ has continuous partial derivative with respect to y ($\frac{\partial f}{\partial y} = x$). Thus the Existence and Uniqueness Theorem guarantees a unique solution through any initial point, including (0,0). The solution is $y = C e^{x^2/2}$, and with $y(0)=0$, $C=0$, so $y=0$ is indeed the only solution. The other options are incorrect: it is not a singular point, and while it's an equilibrium solution, the theorem guarantees uniqueness.

📖 Explanation: If $T = T_e$, then $\frac{dT}{dt} = k(T_e - T_e) = 0$, so the temperature does not change. Thus if the initial temperature equals the ambient temperature, the object stays at that temperature. This is the equilibrium solution. The other options would lead to either cooling or warming. This tests understanding of equilibrium in a cooling model.

📖 Explanation: The correct solution to $\frac{dP}{dt} = 0.5P$ is $P = P_0 e^{0.5t}$. With $P_0 = 100$, the solution is $100e^{0.5t}$. The student got $200e^{0.5t}$, which would be correct if $P(0) = 200$. So they applied the initial condition incorrectly, effectively doubling the initial population. The growth constant is correct (0.5), and the exponent is correct. This tests the student's ability to identify errors in applying initial conditions.

📖 Explanation: The general solution is $y = C_1\cos(4x) + C_2\sin(4x)$. For $y = 3\cos(4x)$, we need $C_1 = 3$ and $C_2 = 0$. Then $y(0) = 3$ and y'(0) = 0. Option B gives $y = 3\sin(4x)$; C gives $3\cos(4x) + 3\sin(4x)$; D gives the trivial solution. This tests the ability to determine initial conditions from a given solution.

📖 Explanation: The general solution of y' = 2x is $y = x^2 + C$. Using $y(1) = 3$, we get $3 = 1 + C$, so $C = 2$, and $y = x^2 + 2$. Then $y(0) = 0^2 + 2 = 2$. This is a straightforward application of an initial condition to find the constant.

📖 Explanation: The solution to y' = y^2 is $y = \frac{1}{C - x}$. With $y(0) = 1$, $1 = \frac{1}{C}$, so $C = 1$, and $y = \frac{1}{1 - x}$. This has a vertical asymptote at $x = 1$. This is a classic example of a solution that does not exist for all x, even though the ODE is simple. The other options are common mistakes: x=-1 would be if C=-1, x=0 is the initial point, and x=2 is arbitrary.

📖 Explanation: Percentage error = $\left| \frac{\text{exact} - \text{approximate}}{\text{exact}} \right| \times 100\%$. Here, $|3.436 - 2.125| / 3.436 \times 100\% \approx 1.311 / 3.436 \times 100\% \approx 38.1\%$. So about 38%. Option B (62%) would be if the denominator was the approximation; C and D are other possible miscalculations. This tests the student's ability to compute and interpret error.

📖 Explanation: The solution to y' = 2y is $y = Ce^{2x}$. Changing C from 1 to 2 changes the solution from $e^{2x}$ to $2e^{2x}$, which is a vertical stretch by a factor of 2. The other transformations (vertical shift, horizontal compression, reflection) would require different modifications. This tests the student's understanding of how the initial condition affects the solution curve.

📖 Explanation: Rate in = (0.5 lbs/gal)(4 gal/min) = 2 lbs/min. Volume at time t: initial 50 gal + (4 - 2)t = 50 + 2t. Concentration leaving = A/(50+2t). Rate out = concentration × outflow rate = $\frac{A}{50+2t} \times 2 = \frac{2A}{50+2t}$. Thus $\frac{dA}{dt} = 2 - \frac{2A}{50+2t}$. Initial condition: no salt initially, so A(0)=0. The other options have incorrect rates or concentrations. This tests the ability to set up a variable-volume mixing problem.

📖 Explanation: For y' = x, the solutions are $y = \frac{x^2}{2} + C$. With $y(0) = 1$, the solution is $y = \frac{x^2}{2} + 1$, a parabola shifted up by 1. The slope field consists of parallel segments because the slope depends only on x, not y. This test the interpretation of slope fields and their relation to solutions. The other options are common misinterpretations: a straight line would have constant slope, exponential would have slope proportional to y, and passing through the origin would be for C=0.

📖 Explanation: In separation of variables, after integrating, you get a general solution with a constant of integration. The initial condition is used to solve for this constant, determining the particular solution. The other options are not the primary use of the initial condition: the interval of validity comes from the solution, the monotonicity is derived from the sign of the derivative, and equilibrium solutions are found by setting the derivative to zero. This tests the student's understanding of the role of initial conditions in the solution process.

📖 Explanation: The zero function $y = 0$ satisfies both differential equations: for y' = y, $0 = 0$; for y' = 2y, $0 = 2(0)$. With the initial condition $y(0) = 0$, it is the unique solution by the existence-uniqueness theorem. This is a simple but important check: the zero solution often works for homogeneous linear equations. The other options reflect misconceptions about uniqueness or the nature of the equations.

📖 Explanation: The differential equation is logistic with carrying capacity 1. If $y > 1$, the population is above capacity and decreases. So $y(0) = 2$ gives a decreasing solution. For $y(0) = 0.5$ ($0 < y < 1$), the population increases toward 1. For $y(0) = 1$, the population is at equilibrium and stays constant. For $y(0) = 0$, it's another equilibrium (trivial). This tests the student's qualitative understanding of the logistic model.

📖 Explanation: For Euler's method, the error is approximately proportional to the step size. Halving the step size should roughly halve the error. So if the original error was about 0.936, the new error should be about 0.468. The other options reflect other convergence orders: one quarter would be for a second-order method, the same would indicate no improvement, and double would indicate instability. This tests the student's understanding of error behavior in Euler's method.

📖 Explanation: Let $g(x) = y(-x)$. Then g&#039;(x) = -y&#039;(-x). Since y&#039;(-x) = y(-x) = g(x), we have g&#039;(x) = -g(x). So $g$ satisfies g&#039; = -g. This is a clever transformation: reflecting the solution across the y-axis changes the sign in the differential equation. This tests the student's ability to use the chain rule and relate solutions of different ODEs.