AP Calculus BC Quiz: Initial Conditions And Separation Of Variables

What this quiz covers

This quiz focuses on Initial Conditions And Separation Of Variables, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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.

Question 1

1. $y=\frac{7}{2}x$ (correct answer)
2. $y=7x$
3. $y=Cx$
4. $y=\frac{7}{x}$
5. $y=7\ln x$

Explanation: This homogeneous differential equation requires applying an initial condition at a non-zero point. The general solution to $\frac{dy}{dx}=\frac{y}{x}$ is $y=Cx$ (found by separating variables: $\frac{dy}{y}=\frac{dx}{x}$, giving $\ln|y|=\ln|x|+\ln|C|$). Using the initial condition $y(2)=7$: $7=C(2)$, so $C=\frac{7}{2}$. Therefore, the particular solution is $y=\frac{7}{2}x$. Choice B incorrectly assumes $C=7$, forgetting to solve for $C$ using the given point $(2,7)$. When the initial condition is given at a point other than $x=0$, substitute both coordinates to find the constant.

Question 2

1. $T(t)=20+60e^{-t/10}$ (correct answer)
2. $T(t)=20+60e^{t/10}$
3. $T(t)=80e^{-t/10}$
4. $T(t)=20+Ce^{-t/10}$
5. $T(t)=80-\frac{t}{10}$

Explanation: The skill demonstrated in this problem is applying an initial condition to find the particular solution of a separable differential equation. The general solution to dT/dt = - (1/10) (T - 20) is T(t) = 20 + C e^{-t/10}, obtained by rewriting the equation, separating variables, and integrating. Using the initial condition T(0) = 80 gives 80 = 20 + C, so C = 60. Therefore, the particular solution is T(t) = 20 + 60 e^{-t/10}. A tempting distractor is T(t) = 20 + 60 e^{t/10}, which assumes heating instead of cooling, but it fails because the negative rate constant indicates exponential decay toward the ambient temperature. Always substitute the initial values into the general solution to solve for the arbitrary constant and identify the unique function satisfying both the differential equation and the initial condition.

Question 3

1. $P(t)=\frac{3}{1+2e^{-3t}}$ (correct answer)
2. $P(t)=\frac{3}{1+2e^{3t}}$
3. $P(t)=\frac{3}{1-2e^{-3t}}$
4. $P(t)=\frac{3}{1+Ce^{-3t}}$
5. $P(t)=\frac{1}{1+2e^{-3t}}$

Explanation: The skill demonstrated in this problem is applying an initial condition to find the particular solution of a separable differential equation. The general solution to dP/dt = P(3-P) is P(t) = 3 / (1 + C e^{-3t}), derived from separating variables, integrating with partial fractions, and solving for P. Applying the initial condition P(0) = 1 gives 1 = 3 / (1 + C), so C = 2. Therefore, the particular solution is P(t) = 3 / (1 + 2 e^{-3t}). A tempting distractor is P(t) = 3 / (1 + 2 e^{3t}), which uses a positive exponent, but it fails because the logistic growth requires a negative exponent for convergence to the carrying capacity. Always substitute the initial values into the general solution to solve for the arbitrary constant and identify the unique function satisfying both the differential equation and the initial condition.

Question 4

1. $y=2x-1$ (correct answer)
2. $y=2x+1$
3. $y=Cx-1$
4. $y=\dfrac{2}{x}-1$
5. $y=2\ln x-1$

Explanation: This problem requires applying the initial condition to find the particular solution after solving the differential equation using separation of variables. First, separate variables: dy/(y + 1) = dx/x, then integrate both sides to get ln|y + 1| = ln|x| + C. Exponentiate to solve for y: y + 1 = e^C x, or y = K x - 1. Use the initial condition y(1) = 1: 1 = K * 1 - 1, so K = 2, yielding y = 2x - 1. A tempting distractor is y = 2x + 1, which might result from incorrect sign when solving for y. Always plug the initial condition into the general solution to determine the constant C and verify by differentiating back.

Question 5

1. $y=2+3x$ (correct answer)
2. $y=2+3\ln x$
3. $y=2+Cx$
4. $y=2+\dfrac{3}{x}$
5. $y=5x$

Explanation: This problem requires applying the initial condition to find the particular solution after solving the differential equation using separation of variables. First, separate variables: dy/(y - 2) = dx/x, then integrate both sides to get ln|y - 2| = ln|x| + C. Exponentiate to solve for y: y - 2 = e^C x, or y = K x + 2. Use the initial condition y(1) = 5: 5 = K * 1 + 2, so K = 3, yielding y = 3x + 2. A tempting distractor is y = 2 + 3 ln x, which might result from incorrect integration. Always plug the initial condition into the general solution to determine the constant C and verify by differentiating back.

Question 6

1. $V(t)=10e^{0.2t}$
2. $V(t)=10e^{-0.2t}$ (correct answer)
3. $V(t)=Ce^{-0.2t}$
4. $V(t)=e^{-0.2t}+10$
5. $V(t)=-0.2e^{10t}$

Explanation: The skill here involves applying an initial condition to find the particular solution of a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution V(t) = C e^{-0.2 t}. Plugging in the initial condition V(0) = 10 yields 10 = C e^{0}, which simplifies to C = 10. Therefore, the particular solution is V(t) = 10 e^{-0.2 t}. A tempting distractor like V(t) = C e^{-0.2 t} fails because it remains the general solution without specifying the constant using the initial condition. Always substitute the initial values into the general solution to determine the constant and obtain the unique particular solution.

Question 7

1. $P(t)=50e^{0.2t}$ (correct answer)
2. $P(t)=Ce^{0.2t}$
3. $P(t)=50e^{-0.2t}$
4. $P(t)=0.2e^{50t}$
5. $P(t)=50+0.2t$

Explanation: This problem requires applying an initial condition to find the particular solution of a differential equation. The general solution to dP/dt = 0.2P is P(t) = Ce^(0.2t), where C is an arbitrary constant. To find C, we substitute the initial condition P(0) = 50: 50 = Ce^(0.2·0) = Ce^0 = C·1 = C. Therefore, C = 50 and the particular solution is P(t) = 50e^(0.2t). Choice B shows the general solution without applying the initial condition, which is a common error. When solving differential equations with initial conditions, always substitute the initial values into the general solution to determine the specific constant.

Question 8

1. $y=\sqrt{\dfrac{2}{3}x^3+16}$ (correct answer)
2. $y=\sqrt{\dfrac{1}{3}x^3+16}$
3. $y=\sqrt{\dfrac{2}{3}x^3+C}$
4. $y=\dfrac{2}{3}x^3+16$
5. $y=\sqrt{16-\dfrac{2}{3}x^3}$

Explanation: This problem requires applying the initial condition to find the particular solution after solving the differential equation using separation of variables. First, separate variables: y dy = x^2 dx, then integrate both sides to get (1/2) y^2 = (1/3) x^3 + C. Solve for y: y^2 = (2/3) x^3 + 2C, or y = sqrt((2/3) x^3 + D). Use the initial condition y(0) = 4: 16 = D, so D = 16, yielding y = sqrt((2/3) x^3 + 16). A tempting distractor is y = sqrt((1/3) x^3 + 16), which might result from incorrect coefficient in integration. Always plug the initial condition into the general solution to determine the constant C and verify by differentiating back.

Question 9

1. $V(t)=30e^{-0.1t}$ (correct answer)
2. $V(t)=Ce^{-0.1t}$
3. $V(t)=30e^{0.1t}$
4. $V(t)=30-0.1t$
5. $V(t)=-0.1e^{30t}$

Explanation: This problem requires applying an initial condition to find the particular solution for exponential decay. The general solution to dV/dt = -0.1V is V(t) = Ce^(-0.1t), representing exponential decay with rate 0.1. Using the initial condition V(0) = 30: 30 = Ce^(-0.1·0) = Ce^0 = C·1 = C. Therefore C = 30, giving the particular solution V(t) = 30e^(-0.1t). Choice C incorrectly uses positive 0.1 in the exponent, which would represent growth rather than drainage. When the differential equation has a negative coefficient, the solution must have a negative exponent to model decay correctly.

Question 10

1. $y=2e^{x+\frac{x^2}{2}}$ (correct answer)
2. $y=Ce^{x+\frac{x^2}{2}}$
3. $y=2e^{x^2}$
4. $y=2(x+1)$
5. $y=2e^{-(x+\frac{x^2}{2})}$

Explanation: This problem requires applying an initial condition to a linear differential equation with variable coefficient. Separating dy/dx = (1+x)y gives dy/y = (1+x)dx, which integrates to ln|y| = x + x²/2 + C. The general solution is y = Ae^(x+x²/2) where A = e^C. Using the initial condition y(0) = 2: 2 = Ae^(0+0²/2) = Ae^0 = A·1 = A. Thus A = 2 and the particular solution is y = 2e^(x+x²/2). Choice C incorrectly drops the linear term x from the exponent, showing only x². Always integrate each term carefully and use the initial condition to find the multiplicative constant.

Question 11

1. $S(t)=10e^{-0.5(t-2)}$ (correct answer)
2. $S(t)=10e^{-0.5t}$
3. $S(t)=Ce^{-0.5t}$
4. $S(t)=10e^{0.5(t-2)}$
5. $S(t)=10-0.5(t-2)$

Explanation: This problem requires finding a particular solution using an initial condition given at t = 2 rather than t = 0. The general solution to dS/dt = -0.5S is S(t) = Ce^(-0.5t). To find C, we use S(2) = 10: 10 = Ce^(-0.5·2) = Ce^(-1). Solving for C gives C = 10e^1 = 10e. Therefore, S(t) = 10e·e^(-0.5t) = 10e^(1-0.5t) = 10e^(-0.5(t-2)). Choice B incorrectly assumes the initial condition is at t = 0, which would give S(t) = 10e^(-0.5t). When the initial condition is given at t = t₀ ≠ 0, the solution often takes the form involving (t - t₀) in the exponent.

Question 12

1. $A(t)=4e^{\cos t}$
2. $A(t)=e^{\sin t}$
3. $A(t)=4e^{\sin t}$ (correct answer)
4. $A(t)=4\sin t$
5. $A(t)=4+\sin t$

Explanation: This problem requires applying an initial condition to find the particular solution of a differential equation with a trigonometric coefficient. Separating variables in dA/dt = A cos t gives dA/A = cos t dt, which integrates to ln|A| = sin t + C₁. This yields the general solution A(t) = Ce^(sin t). Applying the initial condition A(0) = 4: 4 = Ce^(sin 0) = Ce^0 = C·1 = C. Therefore, C = 4 and the particular solution is A(t) = 4e^(sin t). Choice A (A(t) = 4e^(cos t)) incorrectly uses cos t in the exponent, likely confusing the integral of cos t. Remember that ∫cos t dt = sin t, and initial conditions determine the multiplicative constant in exponential solutions.

Question 13

1. $P(t)=\frac{100}{1+4e^{-t}}$ (correct answer)
2. $P(t)=\frac{100}{1+4e^{t}}$
3. $P(t)=\frac{100}{1+Ce^{-t}}$
4. $P(t)=\frac{100}{1+e^{-4t}}$
5. $P(t)=\frac{20}{1+4e^{-t}}$

Explanation: This logistic differential equation requires applying an initial condition to determine the particular solution. The general solution to $\frac{dP}{dt}=\frac{1}{100}P(100-P)$ is $P(t)=\frac{100}{1+Ce^{-t}}$ for some constant $C$. Substituting the initial condition $P(0)=20$: $20=\frac{100}{1+Ce^0}=\frac{100}{1+C}$. Solving: $20(1+C)=100$, so $20+20C=100$, giving $C=4$. Therefore, $P(t)=\frac{100}{1+4e^{-t}}$. Choice C shows the general form without determining $C$, which is incomplete. For logistic equations, always use the initial population to find the specific constant in the denominator.

Question 14

1. $y=3e^{x^{2}/2}$ (correct answer)
2. $y=Ce^{x^{2}/2}$
3. $y=3e^{x^{2}}$
4. $y=3e^{-x^{2}/2}$
5. $y=\frac{3}{2}x^{2}$

Explanation: This problem requires applying an initial condition after solving a separable differential equation. Separating variables in $\frac{dy}{dx}=xy$ gives $\frac{dy}{y}=x\,dx$. Integrating both sides yields $\ln|y|=\frac{x^2}{2}+C$, so $y=Ae^{x^2/2}$ where $A=e^C$. Using the initial condition $y(0)=3$: $3=Ae^{0^2/2}=Ae^0=A$, so $A=3$. Therefore, the particular solution is $y=3e^{x^2/2}$. Choice C incorrectly integrates to get $e^{x^2}$ instead of $e^{x^2/2}$, forgetting the factor of $\frac{1}{2}$ from integration. When solving separable equations, carefully integrate both sides and then apply the initial condition to find the constant.

Question 15

1. $P(t)=50e^{0.2t}$ (correct answer)
2. $P(t)=Ce^{0.2t}$
3. $P(t)=50e^{-0.2t}$
4. $P(t)=0.2e^{50t}$
5. $P(t)=50+0.2t$

Explanation: This problem requires applying an initial condition to find the particular solution of a differential equation. The general solution to $\frac{dP}{dt}=0.2P$ is $P(t)=Ce^{0.2t}$, where $C$ is an arbitrary constant. To find $C$, we substitute the initial condition $P(0)=50$: $50=Ce^{0.2(0)}=Ce^0=C$, so $C=50$. Therefore, the particular solution is $P(t)=50e^{0.2t}$. Choice B shows the general solution without applying the initial condition, which is a common error. When solving initial value problems, always substitute the initial condition into the general solution to determine the specific constant value.

Question 16

1. $S(t)=2e^{3t}$ (correct answer)
2. $S(t)=Ce^{3t}$
3. $S(t)=2e^{-3t}$
4. $S(t)=3e^{2t}$
5. $S(t)=2+3t$

Explanation: This exponential growth problem requires applying an initial condition to determine the particular solution. The general solution to $\frac{dS}{dt}=3S$ is $S(t)=Ce^{3t}$, representing exponential growth with rate 3. Applying the initial condition $S(0)=2$: $2=Ce^{3(0)}=Ce^0=C$, so $C=2$. Thus, the particular solution is $S(t)=2e^{3t}$. Choice D reverses the roles of the initial value and growth rate, giving $S(t)=3e^{2t}$, which satisfies neither the differential equation nor the initial condition. For exponential growth/decay, the coefficient in the differential equation becomes the exponent's coefficient, while the initial value becomes the multiplicative constant.

Question 17

1. $T(t)=20+60e^{-t/5}$ (correct answer)
2. $T(t)=20+60e^{t/5}$
3. $T(t)=20+Ce^{-t/5}$
4. $T(t)=80e^{-t/5}$
5. $T(t)=20-60e^{-t/5}$

Explanation: This problem involves applying an initial condition to Newton's law of cooling. The differential equation $\frac{dT}{dt}=-\frac{1}{5}(T-20)$ has general solution $T(t)=20+Ce^{-t/5}$, where 20 is the ambient temperature. Using the initial condition $T(0)=80$, we get $80=20+Ce^0=20+C$, which gives $C=60$. Thus, the particular solution is $T(t)=20+60e^{-t/5}$. Choice C represents the general solution without determining $C$, missing the crucial step of applying the initial condition. For cooling problems, always identify the equilibrium temperature and use the initial condition to find the specific constant.

Question 18

1. $v(t)=\frac{2}{\sqrt{1-8t}}$
2. $v(t)=\frac{2}{\sqrt{1+8t}}$ (correct answer)
3. $v(t)=\frac{1}{\sqrt{t+\frac{1}{4}}}$
4. $v(t)=\frac{2}{1+8t}$
5. $v(t)=\frac{1}{\sqrt{2t+C}}$

Explanation: The skill demonstrated in this problem is applying an initial condition to find the particular solution of a separable differential equation. The general solution to dv/dt = -v^3 is v(t) = 1 / sqrt(2 t + C), derived by separating variables, integrating with power rules, and solving for v (taking the positive root since v(0) > 0). Substituting v(0) = 2 gives 2 = 1 / sqrt(C), so C = 1/4. Thus, the particular solution is v(t) = 2 / sqrt(1 + 8 t) after simplifying. A tempting distractor is v(t) = 2 / sqrt(1 - 8 t), which uses a negative sign, but it fails because the differential equation's negative rate leads to a positive term inside the square root for decreasing v. Always substitute the initial values into the general solution to solve for the arbitrary constant and identify the unique function satisfying both the differential equation and the initial condition.

Question 19

1. $y(x)=3(1+x^2)$ (correct answer)
2. $y(x)=3\sqrt{1+x^2}$
3. $y(x)=\frac{3}{1+x^2}$
4. $y(x)=C(1+x^2)$
5. $y(x)=3e^{\frac{2x}{1+x^2}}$

Explanation: The skill demonstrated in this problem is applying an initial condition to find the particular solution of a separable differential equation. The general solution to $\frac{dy}{dx} = \frac{2x}{1 + x^2} y$ is $y(x) = C (1 + x^2)$, derived by separating variables, recognizing the integral as $\ln(1 + x^2)$, and exponentiating. Applying the initial condition $y(0) = 3$ yields $3 = C (1)$, so $C = 3$. Thus, the particular solution is $y(x) = 3 (1 + x^2)$. A tempting distractor is $y(x) = 3 \sqrt{1 + x^2}$, which might arise from mishandling the integral, but it fails because the correct integration leads to a linear power, not a square root. Always substitute the initial values into the general solution to solve for the arbitrary constant and identify the unique function satisfying both the differential equation and the initial condition.

Question 20

1. $V(t)=Ce^{-0.2t}$
2. $V(t)=30e^{-0.2t}$ (correct answer)
3. $V(t)=30e^{0.2t}$
4. $V(t)=30-0.2t$
5. $V(t)=\frac{30}{1+0.2t}$

Explanation: The skill demonstrated in this problem is applying an initial condition to find the particular solution of a separable differential equation. The general solution to dV/dt = -0.2V is V(t) = C e^{-0.2t}, obtained by separating variables and integrating. Using the initial condition V(0) = 30 yields 30 = C e^{0}, so C = 30. Thus, the particular solution is V(t) = 30 e^{-0.2t}. A tempting distractor is V(t) = 30 e^{0.2t}, which assumes growth instead of decay, but it fails because the differential equation indicates a negative rate, leading to exponential decrease. Always substitute the initial values into the general solution to solve for the arbitrary constant and identify the unique function satisfying both the differential equation and the initial condition.