AP Calculus BC Quiz: Finding General Solutions Separation Of Variables

What this quiz covers

This quiz focuses on Finding General Solutions 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. $\arctan(y)=\frac{1}{1+t^2}+C$
2. $\arctan(y)=\arctan(t)+C$ (correct answer)
3. $\ln(1+y^2)=\ln(1+t^2)+C$
4. $y=t+C$
5. $\arctan(y)=t^2+C$

Explanation: This equation $\frac{dy}{dt}=\frac{1+y^2}{1+t^2}$ separates to $\frac{dy}{1+y^2}=\frac{dt}{1+t^2}$. Both sides have the form $\frac{du}{1+u^2}$, which integrates to $\arctan(u)$. Therefore, we get $\arctan(y)=\arctan(t)+C$. Choice C incorrectly attempts to use logarithms, but $\int\frac{du}{1+u^2}=\arctan(u)$, not $\frac{1}{2}\ln(1+u^2)$ (which would be for $\int\frac{2u\,du}{1+u^2}$). When solving separable equations: (1) separate variables, (2) recognize standard integral forms, (3) apply the correct antiderivative formulas, and (4) add the constant of integration.

Question 2

1. $\ln|B|=\arctan(t)+C$ (correct answer)
2. $B(t)=\arctan(t)+C$
3. $\ln|B|=\frac{1}{1+t^2}+C$
4. $B(t)=Ce^{\arctan(t)}$
5. $\ln|B|=\ln(1+t^2)+C$

Explanation: This differential equation $\frac{dB}{dt}=\frac{B}{1+t^2}$ requires separation of variables. We separate to get $\frac{dB}{B}=\frac{dt}{1+t^2}$. The left side integrates to $\ln|B|$, while the right side is the standard integral $\arctan(t)$, giving us $\ln|B|=\arctan(t)+C$. Choice C incorrectly treats $\frac{1}{1+t^2}$ as the antiderivative rather than recognizing it as the derivative of $\arctan(t)$. When solving separable equations: (1) separate variables, (2) recognize standard integral forms, (3) integrate both sides, and (4) include the constant of integration.

Question 3

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

Explanation: Separation of variables is the method to find the general solution for this equation. Rearrange as $\frac{dy}{y} = (1 + x^2) \, dx$. Integrate to get $\ln |y| = x + \frac{x^3}{3} + K$. Exponentiating yields $y = C e^{x + \frac{x^3}{3}}$, where $C = \pm e^K$. Choice D, $y = C (x + \frac{x^3}{3})$, is a tempting distractor that could result from forgetting to exponentiate after integrating and instead treating it linearly. To solve separable equations, always isolate variables, integrate each side, solve for the dependent variable, and incorporate the constant appropriately.

Question 4

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

Explanation: This population model requires separation of variables to find the general solution. Rewrite as $\frac{dP}{P(4 - P)} = dt$, separating P and t. Use partial fractions: $\frac{1}{P(4 - P)} = \frac{1}{4} \left( \frac{1}{P} + \frac{1}{4 - P} \right)$, then integrate to $\ln|P| - \ln|4 - P| = 4 t + K$, simplifying to $\ln\left| \frac{P}{4 - P} \right| = 4 t + K$. Exponentiating and solving gives $P(t) = \frac{4}{1 + C e^{-4 t}}$. Choice B is a tempting distractor with a positive exponent, which could stem from an error in the sign during integration or exponentiation, resulting in unstable growth. To solve separable equations, always isolate variables, integrate each side, solve for the dependent variable, and incorporate the constant appropriately.

Question 5

1. $y=\arctan(x)+C$
2. $\ln|y|=\arctan(x)+C$ (correct answer)
3. $\ln|y|=\dfrac{1}{1+x^2}+C$
4. $y=C(1+x^2)$
5. $y=Ce^{1+x^2}$

Explanation: This problem requires the skill of finding general solutions to differential equations using separation of variables. To solve dy/dx = y / (1 + x^2), separate variables by writing dy / y = dx / (1 + x^2). Integrate both sides: ∫ (1/y) dy = ∫ 1/(1 + x^2) dx, yielding ln|y| = arctan x + C. This is the general solution. A tempting distractor is choice C, ln|y| = 1/(1 + x^2) + C, which mistakenly integrates to the integrand itself instead of arctan x. For any separable equation dy/dx = g(x) h(y), rewrite as (1/h(y)) dy = g(x) dx, integrate both sides, solve for y if possible, and include the constant of integration.

Question 6

1. $S^2=3t^2+C$
2. $S(t)=3t^2+C$
3. $\ln|S|=\frac{3}{2}t^2+C$
4. $\frac{1}{2}S^2=\frac{3}{2}t^2+C$ (correct answer)
5. $\frac{1}{2}S^2=3t+C$

Explanation: This problem uses separation of variables for the equation $\frac{dS}{dt}=\frac{3t}{S}$. We separate by multiplying both sides by $S$ and $dt$ to get $S\,dS=3t\,dt$. Integrating both sides yields $\frac{1}{2}S^2=\frac{3}{2}t^2+C$, where we use the power rule on both sides. Choice A incorrectly omits the coefficient $\frac{1}{2}$ from integrating $S\,dS$, while choice E incorrectly integrates $3t$ as $3t$ instead of $\frac{3}{2}t^2$. For separable equations: (1) separate variables completely, (2) integrate each side independently, (3) include all coefficients from integration, and (4) add the constant of integration.

Question 7

1. $2\sqrt{v}=\frac{5}{2}t^2+C$ (correct answer)
2. $\sqrt{v}=\frac{5}{2}t^2+C$
3. $2v^{3/2}=\frac{5}{2}t^2+C$
4. $\ln|v|=\frac{5}{2}t^2+C$
5. $2\sqrt{v}=5t+C$

Explanation: This equation $\frac{dv}{dt}=5t\sqrt{v}$ requires separation of variables. We rewrite $\sqrt{v}$ as $v^{1/2}$ and separate to get $v^{-1/2}dv=5t\,dt$. Integrating the left side using the power rule gives $2v^{1/2}=2\sqrt{v}$, while the right side gives $\frac{5}{2}t^2$. Therefore, $2\sqrt{v}=\frac{5}{2}t^2+C$. Choice B incorrectly omits the coefficient 2 from integrating $v^{-1/2}$, while choice E incorrectly integrates $5t$ as $5t$ instead of $\frac{5}{2}t^2$. When solving: (1) convert radicals to powers, (2) separate variables, (3) apply power rule carefully, and (4) simplify the final form.

Question 8

1. $\ln|T-20|=-4t+C$ (correct answer)
2. $T(t)=20+Ce^{-4t}$
3. $T(t)=20+Ce^{4t}$
4. $\ln|T|=-4t+C$
5. $\ln|T-20|=4t+C$

Explanation: This is Newton's law of cooling with $\frac{dT}{dt}=-4(T-20)$, solved using separation of variables. We separate to get $\frac{dT}{T-20}=-4dt$. Integrating both sides yields $\ln|T-20|=-4t+C$, where we use the substitution $u=T-20$ on the left side. Choice E incorrectly has a positive $4t$ instead of negative, likely from mishandling the negative sign in the original equation. When solving separable equations: (1) separate variables preserving all signs, (2) use substitution when needed, (3) integrate both sides, and (4) check that signs match the original equation.

Question 9

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

Explanation: For $\frac{dy}{dx}=(x^2+1)e^{y}$, we separate variables to get $e^{-y}dy=(x^2+1)dx$. Integrating the left side gives $-e^{-y}$ (since $\int e^{-y}dy=-e^{-y}$), while the right side gives $\frac{1}{3}x^3+x$. Therefore, $-e^{-y}=\frac{1}{3}x^3+x+C$. Choice A incorrectly omits the negative sign when integrating $e^{-y}dy$, a common error with exponential functions. For separable equations: (1) separate variables, (2) rewrite to standard forms if needed, (3) integrate carefully with correct signs, and (4) verify your antiderivatives.

Question 10

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

Explanation: For $\frac{dy}{dx}=\frac{x}{y+2}$, we use separation of variables by writing $(y+2)dy=x\,dx$. Integrating the left side requires the substitution $u=y+2$, giving $\int(y+2)dy=\frac{1}{2}(y+2)^2$. The right side integrates to $\frac{1}{2}x^2$, so we have $\frac{1}{2}(y+2)^2=\frac{1}{2}x^2+C$. Choice E incorrectly uses $y$ instead of $(y+2)$ in the integration, missing the shift in the denominator. For separable equations: (1) separate variables, (2) identify needed substitutions, (3) integrate both sides carefully, and (4) include all terms from the original equation.

Question 11

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

Explanation: For $\frac{dy}{dx}=\frac{x}{y+2}$, we use separation of variables by writing $(y+2)dy=x\,dx$. Integrating the left side requires the substitution $u=y+2$, giving $\int(y+2)dy=\frac{1}{2}(y+2)^2$. The right side integrates to $\frac{1}{2}x^2$, so we have $\frac{1}{2}(y+2)^2=\frac{1}{2}x^2+C$. Choice E incorrectly uses $y$ instead of $(y+2)$ in the integration, missing the shift in the denominator. For separable equations: (1) separate variables, (2) identify needed substitutions, (3) integrate both sides carefully, and (4) include all terms from the original equation.

Question 12

1. $\ln|Q|=-\cos t+C$
2. $\ln|Q|=\sin t+C$
3. $Q=-\sin t+C$
4. $\ln|Q|=-\sin t+C$ (correct answer)
5. $\ln|Q|=\cos t+C$

Explanation: This decay model requires separation of variables for $\frac{dQ}{dt}=-Q\cos t$. Separating variables gives $\frac{dQ}{Q}=-\cos t\,dt$. The left side integrates to $\ln|Q|$, while the right side integrates to $-\sin t$ (since $\int\cos t\,dt=\sin t+C$). Therefore, the general solution is $\ln|Q|=-\sin t+C$. Choice A incorrectly integrates $-\cos t$ as $-\cos t$ instead of $-\sin t$, confusing the derivative and integral of cosine. When separating variables: (1) move all terms with each variable to one side, (2) remember trigonometric integrals correctly, and (3) include the constant of integration.

Question 13

1. $\arctan(y)=\dfrac{x^2}{2}+C$
2. $y=\dfrac{x^2}{2}+C$
3. $\ln|y|=\dfrac{x^2}{2}+C$
4. $y=\tan\!\left(\dfrac{x^2}{2}+C\right)$ (correct answer)
5. $\arctan(y)=2x+C$

Explanation: This differential equation dy/dx = x(1+y²) requires separation of variables to solve. We rearrange to get dy/(1+y²) = x dx, separating y terms on the left and x terms on the right. Integrating the left side gives arctan(y) (since the derivative of arctan(y) is 1/(1+y²)), and integrating the right side gives x²/2. Thus arctan(y) = x²/2 + C, and solving for y yields y = tan(x²/2 + C). Choice A incorrectly stops at the implicit form without solving for y explicitly. When separating variables: (1) isolate dy and y terms on one side, (2) isolate dx and x terms on the other, (3) integrate both sides, and (4) solve for the dependent variable if possible.

Question 14

1. $T(t)=20+Ce^{kt}$ (correct answer)
2. $T(t)=Ce^{kt}-20$
3. $T(t)=20+kt$
4. $T(t)=20+Ce^{t}$
5. $T(t)=20+\dfrac{C}{e^{kt}}$

Explanation: This differential equation $\frac{dT}{dt}=k(T-20)$ requires separation of variables to find the general solution. We rewrite as $\frac{dT}{T-20}=k\,dt$ and integrate both sides to get $\ln|T-20|=kt+C_1$. Exponentiating yields $|T-20|=e^{kt+C_1}=e^{C_1}e^{kt}$, so $T-20=Ce^{kt}$ where $C=\pm e^{C_1}$. Solving for $T$ gives $T(t)=20+Ce^{kt}$. Choice B, $T(t)=Ce^{kt}-20$, incorrectly places the constant 20 with the wrong sign, which would give the wrong equilibrium temperature. For separation of variables with linear shifts: (1) identify the shift term, (2) separate with respect to the shifted variable, (3) integrate and exponentiate, and (4) add back the shift to isolate the original variable.

Question 15

1. $y^4=4x^3+C$ (correct answer)
2. $y^4=3x^3+C$
3. $y^2=3x^3+C$
4. $y^4=12x^2+C$
5. $y=\dfrac{3x^2}{y^3}+C$

Explanation: Separation of variables is used to solve this differential equation for y in terms of x. Rearrange as y^3 dy = 3 x^2 dx. Integrate to obtain (1/4) y^4 = x^3 + K. Multiply by 4 to get y^4 = 4 x^3 + C, where C = 4K. Choice B, y^4 = 3 x^3 + C, is a tempting distractor that might occur from forgetting to multiply the right side by 4 after integrating, thus mismatching the coefficients. To solve separable equations, always isolate variables, integrate each side, solve for the dependent variable, and incorporate the constant appropriately.

Question 16

1. $P(t)=\dfrac{1}{1+Ce^{-0.5t}}$ (correct answer)
2. $P(t)=\dfrac{1}{1+Ce^{0.5t}}$
3. $P(t)=\dfrac{1}{1+Ce^{-t}}$
4. $P(t)=1+Ce^{-0.5t}$
5. $P(t)=\dfrac{Ce^{0.5t}}{1+Ce^{0.5t}}$

Explanation: This problem requires using separation of variables to solve the differential equation for the bacteria population. Start by rewriting the equation as dP / [P(1 - P)] = 0.5 dt, separating the variables P and t. Use partial fractions to express 1/[P(1 - P)] as 1/P + 1/(1 - P), then integrate both sides to get ln|P| - ln|1 - P| = 0.5 t + K, which simplifies to ln|P / (1 - P)| = 0.5 t + K. Exponentiating yields |P / (1 - P)| = e^K e^{0.5 t}, and solving for P gives the logistic form P(t) = 1 / (1 + C e^{-0.5 t}), where C accounts for the constant and sign. A tempting distractor like choice E presents an equivalent form but with a different parameterization of the constant, which might confuse students but represents the same family of solutions. To solve separable equations, always isolate variables, integrate each side, solve for the dependent variable, and incorporate the constant appropriately.

Question 17

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

Explanation: This problem requires the skill of finding general solutions to differential equations using separation of variables. To solve dy/dx = x / (1 + y^2), separate variables by writing (1 + y^2) dy = x dx. Integrate both sides: ∫ (1 + y^2) dy = ∫ x dx, yielding y + (1/3) y^3 = (1/2) x^2 + C. This is the implicit general solution. A tempting distractor is choice C, ln|1 + y^2| = (1/2) x^2 + C, which incorrectly integrates the left side as a log instead of a polynomial. For any separable equation dy/dx = g(x) h(y), rewrite as (1/h(y)) dy = g(x) dx, integrate both sides, solve for y if possible, and include the constant of integration.

Question 18

1. $y=\left(\dfrac{x^2}{4}+C\right)^2$ (correct answer)
2. $\ln|y|=\dfrac{x^2}{2}+C$
3. $\sqrt{y}=\dfrac{x^2}{2}+C$
4. $y=\dfrac{x^2}{2}+C$
5. $2\sqrt{y}=\dfrac{x^2}{4}+C$

Explanation: This problem requires the skill of finding general solutions to differential equations using separation of variables. To solve dy/dx = x sqrt(y), separate variables by writing dy / sqrt(y) = x dx. Integrate both sides: ∫ y^{-1/2} dy = ∫ x dx, yielding 2 sqrt(y) = (1/2) x^2 + C. Solve for y: sqrt(y) = (1/4) x^2 + K, so y = ((1/4) x^2 + K)^2. A tempting distractor is choice C, sqrt(y) = (1/2) x^2 + C, which forgets to divide by 2 after integrating the left side. For any separable equation dy/dx = g(x) h(y), rewrite as (1/h(y)) dy = g(x) dx, integrate both sides, solve for y if possible, and include the constant of integration.

Question 19

1. $T(t)=20+Ce^{-kt}$ (correct answer)
2. $T(t)=20+Ce^{kt}$
3. $T(t)=Ce^{-kt}-20$
4. $\ln|T-20|=-kt+C$
5. $T(t)=\ln|t-20|+C$

Explanation: This problem requires the skill of finding general solutions to differential equations using separation of variables. To solve dT/dt = -k (T - 20), separate variables by writing dT / (T - 20) = -k dt. Integrate both sides: ∫ 1/(T - 20) dT = ∫ -k dt, yielding ln|T - 20| = -k t + C. Exponentiate to get T - 20 = A e^{-k t}, or T = 20 + A e^{-k t}. A tempting distractor is choice B, T(t) = 20 + C e^{k t}, which has a positive exponent and would model heating away from equilibrium instead of approaching it. For any separable equation dy/dx = g(x) h(y), rewrite as (1/h(y)) dy = g(x) dx, integrate both sides, solve for y if possible, and include the constant of integration.

Question 20

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

Explanation: This problem requires the skill of finding general solutions to differential equations using separation of variables. To solve dy/dx = (1 + y) sin x, separate variables by writing dy / (1 + y) = sin x dx. Integrate both sides: ∫ 1/(1 + y) dy = ∫ sin x dx, yielding ln|1 + y| = - cos x + C. This is the general solution in implicit form. A tempting distractor is choice B, ln|1 + y| = sin x + C, which incorrectly integrates sin x to sin x instead of -cos x. For any separable equation dy/dx = g(x) h(y), rewrite as (1/h(y)) dy = g(x) dx, integrate both sides, solve for y if possible, and include the constant of integration.