AP Calculus AB Quiz: Finding General Solutions Separation Of Variables
20 questions · exam conditions
0:00
Finding General Solutions Separation Of VariablesQuestion 1 of 20

A quantity satisfies dydx=−3xy\dfrac{dy}{dx}=-\dfrac{3}{x}y for x≠0x\ne 0. What is the general solution?

y=Cx−3y=Cx^{-3}
y=−3ln⁡∣x∣+Cy=-3\ln|x|+C
ln⁡∣y∣=−3x+C\ln|y|=-\dfrac{3}{x}+C
y=x−3+Cy=x^{-3}+C
y=Ce−3xy=Ce^{-3x}
← Back to quizzes

AP Calculus AB Quiz

AP Calculus AB Quiz: Finding General Solutions Separation Of Variables

Practice Finding General Solutions Separation Of Variables in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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 AB.

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.

All questions

Question 1

A quantity satisfies dydx=−3xy\dfrac{dy}{dx}=-\dfrac{3}{x}y for x≠0x\ne 0. What is the general solution?

  1. y=Cx−3y=Cx^{-3} (correct answer)
  2. y=−3ln⁡∣x∣+Cy=-3\ln|x|+C
  3. ln⁡∣y∣=−3x+C\ln|y|=-\dfrac{3}{x}+C
  4. y=x−3+Cy=x^{-3}+C
  5. y=Ce−3xy=Ce^{-3x}
Explanation: This problem requires solving a differential equation using separation of variables. Given dy/dx = -(3/x) y for x ≠ 0, separate as dy/y = -(3/x) dx, assuming y ≠ 0. Integrate to ln|y| = -3 ln|x| + C. This simplifies to y = C x^{-3}. The choice y = -3 ln|x| + C is incorrect as it stops at the integral without solving for y exponentially. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

Question 2

A function satisfies dydx=3xy\dfrac{dy}{dx}=\dfrac{3}{x}y for x≠0x\ne 0. What is the general solution?

  1. y=Ce3/xy=Ce^{3/x}
  2. y=Cx3y=Cx^{3} (correct answer)
  3. y=x3+Cy=x^{3}+C
  4. ln⁡∣y∣=3x+C\ln|y|=\dfrac{3}{x}+C
  5. y=3ln⁡∣x∣+Cy=3\ln|x|+C
Explanation: This problem requires solving a differential equation using separation of variables. Given dy/dx = (3/x) y for x ≠ 0, separate as dy/y = (3/x) dx, assuming y ≠ 0. Integrate: ln|y| = 3 ln|x| + C. Simplify to y = C x^3. The choice y = 3 ln|x| + C is incorrect because it omits exponentiation to power form. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

Question 3

A variable satisfies dydx=2xy\dfrac{dy}{dx}=\dfrac{2}{x}y for x≠0x\ne 0. What is the general solution?

  1. y=2ln⁡∣x∣+Cy=2\ln|x|+C
  2. ln⁡∣y∣=2x+C\ln|y|=\dfrac{2}{x}+C
  3. y=Cx2y=Cx^{2} (correct answer)
  4. y=x2+Cy=x^{2}+C
  5. y=Ce2/xy=Ce^{2/x}
Explanation: This problem requires solving a differential equation using separation of variables. For dy/dx = (2/x) y with x ≠ 0, separate as dy/y = (2/x) dx, assuming y ≠ 0. Integrate: ln|y| = 2 ln|x| + C. Simplify to y = C x^2, absorbing constants. The distractor y = 2 ln|x| + C fails by not exponentiating properly, yielding a logarithmic form instead of power. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

Question 4

A function satisfies dydx=yx\dfrac{dy}{dx}=\dfrac{y}{x}. What is the general solution for x≠0x\ne 0?

  1. ln⁡∣y∣=1x+C\ln|y|=\dfrac{1}{x}+C
  2. y=Cxy=Cx (correct answer)
  3. y=x+Cy=x+C
  4. y=Cexy=Ce^{x}
  5. y=ln⁡∣x∣+Cy=\ln|x|+C
Explanation: This problem requires solving a differential equation using separation of variables. For dy/dx = y/x with x ≠ 0, separate as dy/y = dx/x, assuming y ≠ 0. Integrate: ln|y| = ln|x| + C. Simplify to y = C x. The distractor y = x + C fails by suggesting addition instead of the proportional solution from logs. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

Question 5

A function satisfies dydx=ycos⁡x\dfrac{dy}{dx}=y\cos x. What is the general solution for y(x)y(x)?

  1. y=Cesin⁡xy=Ce^{\sin x} (correct answer)
  2. y=Csin⁡xy=C\sin x
  3. ln⁡∣y∣=cos⁡x+C\ln|y|=\cos x+C
  4. y=esin⁡x+Cy=e^{\sin x}+C
  5. y=Ce−sin⁡xy=Ce^{-\sin x}
Explanation: This problem requires solving a differential equation using separation of variables. With dy/dx = y cos x, separate as dy/y = cos x dx, assuming y ≠ 0. Integrate for ln|y| = sin x + C. Solve to y = C e^{sin x}. The option y = e^{sin x} + C is wrong because it places C additively, not multiplicatively. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

Question 6

A quantity changes by dydx=5y\dfrac{dy}{dx}=5y. What is the general solution for y(x)y(x)?

  1. y=5x+Cy=5x+C
  2. y=Ce5xy=Ce^{5x} (correct answer)
  3. ln⁡∣y∣=5x\ln|y|=5x
  4. y=e5x+Cy=e^{5x}+C
  5. y=5ex+Cy=5e^{x}+C
Explanation: This problem requires solving a differential equation using separation of variables. For dy/dx = 5y, separate as dy/y = 5 dx, assuming y ≠ 0. Integrate: ln|y| = 5x + C. Exponentiate to y = Ce^{5x}. The distractor y = e^{5x} + C fails by placing the constant additively, not multiplicatively as required. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

Question 7

A quantity satisfies dydx=−7y\frac{dy}{dx}=-7y. What is the general solution for y(x)y(x)?

  1. y=Ce−7xy=Ce^{-7x} (correct answer)
  2. y=Ce7xy=Ce^{7x}
  3. y=e−7xy=e^{-7x}
  4. ln⁡∣y∣=−7\ln|y|=-7
  5. y=C−7xy=C-7x
Explanation: To solve dydx=−7y\frac{dy}{dx}=-7y using separation of variables, we divide by yy and multiply by dxdx to get dyy=−7dx\frac{dy}{y}=-7dx. Integrating both sides gives ln⁡∣y∣=−7x+C1\ln|y|=-7x+C_1. Exponentiating yields ∣y∣=e−7x+C1=eC1e−7x|y|=e^{-7x+C_1}=e^{C_1}e^{-7x}, which becomes y=Ce−7xy=Ce^{-7x} where C=±eC1C=\pm e^{C_1}. Choice B incorrectly uses a positive exponent, missing the negative sign from the differential equation. The key recognition pattern is that dydx=ky\frac{dy}{dx}=ky always yields y=Cekxy=Ce^{kx}, with the sign of kk preserved in the solution.

Question 8

A model satisfies dydx=3x1+x2y\frac{dy}{dx}=\frac{3x}{1+x^2}y. What is the general solution for y(x)y(x)?

  1. y=32ln⁡(1+x2)+Cy=\frac{3}{2}\ln(1+x^2)+C
  2. ln⁡y=3x1+x2+C\ln y=\frac{3x}{1+x^2}+C
  3. y=(1+x2)3/2y=(1+x^2)^{3/2}
  4. y=C(1+x2)3/2y=C(1+x^2)^{3/2} (correct answer)
  5. ln⁡(1+x2)=32y+C\ln(1+x^2)=\frac{3}{2}y+C
Explanation: This differential equation uses separation of variables with a rational function. Starting with dydx=3x1+x2y\frac{dy}{dx}=\frac{3x}{1+x^2}y, we separate to get dyy=3x1+x2dx\frac{dy}{y}=\frac{3x}{1+x^2}dx. To integrate the right side, note that ddx[ln⁡(1+x2)]=2x1+x2\frac{d}{dx}[\ln(1+x^2)]=\frac{2x}{1+x^2}, so ∫3x1+x2dx=32ln⁡(1+x2)\int\frac{3x}{1+x^2}dx=\frac{3}{2}\ln(1+x^2). This gives ln⁡∣y∣=32ln⁡(1+x2)+C1=ln⁡(1+x2)3/2+C1\ln|y|=\frac{3}{2}\ln(1+x^2)+C_1=\ln(1+x^2)^{3/2}+C_1. Exponentiating yields ∣y∣=eC1(1+x2)3/2|y|=e^{C_1}(1+x^2)^{3/2}, which becomes y=C(1+x2)3/2y=C(1+x^2)^{3/2} where C=±eC1C=\pm e^{C_1}. Choice C omits the arbitrary constant. When integrating rational functions of the form kx1+x2\frac{kx}{1+x^2}, recognize this as a logarithmic derivative pattern.

Question 9

For an investment, dAdt=0.08A\frac{dA}{dt}=0.08A. What is the general solution for A(t)A(t)?

  1. A(t)=0.08etA(t)=0.08e^{t}
  2. A(t)=Ce0.08tA(t)=Ce^{0.08t} (correct answer)
  3. A(t)=e0.08tA(t)=e^{0.08t}
  4. ln⁡A=0.08\ln A=0.08
  5. A(t)=C+0.08tA(t)=C+0.08t
Explanation: The investment equation dAdt=0.08A\frac{dA}{dt}=0.08A is solved through separation of variables. Dividing by AA and multiplying by dtdt gives dAA=0.08dt\frac{dA}{A}=0.08dt. Integration yields ln⁡∣A∣=0.08t+C1\ln|A|=0.08t+C_1. Exponentiating both sides gives ∣A∣=e0.08t+C1=eC1e0.08t|A|=e^{0.08t+C_1}=e^{C_1}e^{0.08t}, resulting in A=Ce0.08tA=Ce^{0.08t} where C=±eC1C=\pm e^{C_1}. Choice A incorrectly places the rate 0.08 as a coefficient outside the exponential rather than in the exponent. When solving dydt=ky\frac{dy}{dt}=ky, the constant kk always appears in the exponent of ee in the solution y=Cekty=Ce^{kt}.

Question 10

For a chemical reaction, dydx=4xy\frac{dy}{dx}=\frac{4}{x}y. What is the general solution for y(x)y(x)?

  1. y=4ln⁡∣x∣+Cy=4\ln|x|+C
  2. ln⁡∣y∣=4x+C\ln|y|=\frac{4}{x}+C
  3. y=Cx4y=Cx^{4} (correct answer)
  4. y=x4y=x^{4}
  5. y=Cln⁡∣x∣4y=C\ln|x|^{4}
Explanation: To solve dydx=4xy\frac{dy}{dx}=\frac{4}{x}y using separation of variables, we divide both sides by yy and multiply by dxdx to get dyy=4xdx\frac{dy}{y}=\frac{4}{x}dx. Integrating both sides yields ln⁡∣y∣=4ln⁡∣x∣+C1=ln⁡∣x∣4+C1\ln|y|=4\ln|x|+C_1=\ln|x|^4+C_1. Exponentiating gives ∣y∣=eln⁡∣x∣4+C1=eC1∣x∣4|y|=e^{\ln|x|^4+C_1}=e^{C_1}|x|^4, which simplifies to y=Cx4y=Cx^4 where C=±eC1C=\pm e^{C_1}. Choice A incorrectly integrates the right side as if yy were not present in the original equation. When the coefficient of yy contains only functions of the independent variable, separation of variables transforms the exponential into a power function.

Question 11

A tank drains so that dVdt=−2V\frac{dV}{dt}=-2V. What is the general solution for V(t)V(t)?

  1. V(t)=Ce−2tV(t)=Ce^{-2t} (correct answer)
  2. V(t)=−2et+CV(t)=-2e^{t}+C
  3. ln⁡V=−2t\ln V=-2t
  4. V(t)=e−2tV(t)=e^{-2t}
  5. ln⁡V=2t+C\ln V=2t+C
Explanation: This problem uses separation of variables to solve the differential equation. Starting with dVdt=−2V\frac{dV}{dt}=-2V, we separate variables to get dVV=−2dt\frac{dV}{V}=-2dt. Integrating both sides yields ln⁡∣V∣=−2t+C1\ln|V|=-2t+C_1, where C1C_1 is an integration constant. Exponentiating gives ∣V∣=e−2t+C1=eC1e−2t|V|=e^{-2t+C_1}=e^{C_1}e^{-2t}, which simplifies to V=Ce−2tV=Ce^{-2t} where C=±eC1C=\pm e^{C_1} is an arbitrary constant. Choice B incorrectly attempts direct integration without separating variables first. Remember that when the derivative equals a constant times the function itself, the solution is always exponential.

Question 12

A population satisfies dPdt=12P\frac{dP}{dt}=\frac{1}{2}P. What is the general solution for P(t)P(t)?

  1. P(t)=12etP(t)=\frac{1}{2}e^{t}
  2. ln⁡P=12+C\ln P=\frac{1}{2}+C
  3. P(t)=Cet/2P(t)=Ce^{t/2} (correct answer)
  4. P(t)=et/2P(t)=e^{t/2}
  5. P(t)=C+t2P(t)=C+\frac{t}{2}
Explanation: The differential equation dPdt=12P\frac{dP}{dt}=\frac{1}{2}P is solved by separation of variables. Dividing by PP and multiplying by dtdt gives dPP=12dt\frac{dP}{P}=\frac{1}{2}dt. Integration yields ln⁡∣P∣=t2+C1\ln|P|=\frac{t}{2}+C_1. Exponentiating both sides gives ∣P∣=et/2+C1=eC1et/2|P|=e^{t/2+C_1}=e^{C_1}e^{t/2}, resulting in P=Cet/2P=Ce^{t/2} where C=±eC1C=\pm e^{C_1}. Choice A incorrectly treats the coefficient 12\frac{1}{2} as if it were multiplying ete^t rather than appearing in the exponent. Remember that when integrating dPP=k dt\frac{dP}{P}=k\,dt, the constant kk becomes the coefficient in the exponent, not a multiplicative factor outside.

Question 13

A culture grows so that dPdt=3P\frac{dP}{dt}=3P. What is the general solution for P(t)P(t)?

  1. P(t)=3et+CP(t)=3e^{t}+C
  2. P(t)=Ce3tP(t)=Ce^{3t} (correct answer)
  3. ln⁡P=3t\ln P=3t
  4. P(t)=e3tP(t)=e^{3t}
  5. ln⁡P=t+C\ln P=t+C
Explanation: This differential equation requires separation of variables to solve. We can rewrite dPdt=3P\frac{dP}{dt}=3P as dPP=3dt\frac{dP}{P}=3dt, separating the variables P and t on opposite sides. Integrating both sides gives ln⁡∣P∣=3t+C1\ln|P|=3t+C_1, where C1C_1 is an arbitrary constant. Exponentiating both sides yields ∣P∣=e3t+C1=eC1e3t|P|=e^{3t+C_1}=e^{C_1}e^{3t}, which we can write as P=Ce3tP=Ce^{3t} where C=±eC1C=\pm e^{C_1} is an arbitrary constant. Choice A incorrectly treats this as a simple integration problem without recognizing the multiplicative relationship. When you see dydt=ky\frac{dy}{dt}=ky where k is constant, the solution will always be an exponential function y=Cekty=Ce^{kt}.

Question 14

A population satisfies dydx=xy\frac{dy}{dx}=xy. What is the general solution for y(x)y(x)?

  1. y=x22+Cy=\frac{x^2}{2}+C
  2. ln⁡y=x22\ln y=\frac{x^2}{2}
  3. y=Cex2/2y=Ce^{x^2/2} (correct answer)
  4. y=12ex2+Cy=\frac{1}{2}e^{x^2}+C
  5. ln⁡x=y22+C\ln x=\frac{y^2}{2}+C
Explanation: This differential equation requires separation of variables since the derivative involves both x and y multiplicatively. From dydx=xy\frac{dy}{dx}=xy, we separate to get dyy=xdx\frac{dy}{y}=x dx. Integrating both sides gives ln⁡∣y∣=x22+C1\ln|y|=\frac{x^2}{2}+C_1, where C1C_1 is a constant of integration. Exponentiating yields ∣y∣=ex2/2+C1=eC1ex2/2|y|=e^{x^2/2+C_1}=e^{C_1}e^{x^2/2}, which becomes y=Cex2/2y=Ce^{x^2/2} where C=±eC1C=\pm e^{C_1}. Choice A incorrectly treats this as direct integration of x, missing the crucial y term in the original equation. When you see dydx=f(x)g(y)\frac{dy}{dx}=f(x)g(y), always separate variables before integrating.

Question 15

A population follows dPdt=Pt\frac{dP}{dt}=\frac{P}{t} for t>0t>0. What is the general solution?

  1. P(t)=ln⁡t+CP(t)=\ln t + C
  2. ln⁡P=ln⁡t\ln P=\ln t
  3. P(t)=CtP(t)=Ct (correct answer)
  4. P(t)=tP+CP(t)=\dfrac{t}{P}+C
  5. P(t)=CtP(t)=\dfrac{C}{t}
Explanation: This differential equation dP/dt = P/t requires separation of variables to solve. Separating gives us dP/P = dt/t, moving all P terms to one side and all t terms to the other. Integrating both sides yields ln|P| = ln|t| + C₁, where C₁ is the constant of integration. Using properties of logarithms, we can rewrite this as ln|P| = ln|t| + ln|C| = ln|Ct|, where C = e^C₁. Exponentiating both sides gives P = Ct, which is the general solution. Choice A incorrectly suggests P = ln(t) + C, but this would give dP/dt = 1/t, not P/t as required. Remember that when you have matching forms on both sides after separation (like dP/P and dt/t), the solutions often involve power functions rather than logarithms.

Question 16

A quantity satisfies dydx=4xy\frac{dy}{dx}=\frac{4}{x}y for x>0x>0. What is the general solution for y(x)y(x)?

  1. y=4ln⁡x+Cy=4\ln x+C
  2. y=Cx4y=Cx^4 (correct answer)
  3. ln⁡y=4x+C\ln y=\frac{4}{x}+C
  4. y=x4y=x^4
  5. ln⁡x=4y+C\ln x=4y+C
Explanation: This problem requires separation of variables with a rational coefficient. Starting with dydx=4xy\frac{dy}{dx}=\frac{4}{x}y, we separate variables to get dyy=4xdx\frac{dy}{y}=\frac{4}{x}dx. Integrating both sides yields ln⁡∣y∣=4ln⁡∣x∣+C1=ln⁡∣x∣4+C1\ln|y|=4\ln|x|+C_1=\ln|x|^4+C_1, since ∫4xdx=4ln⁡∣x∣\int\frac{4}{x}dx=4\ln|x|. Exponentiating gives ∣y∣=eln⁡∣x∣4+C1=eC1∣x∣4|y|=e^{\ln|x|^4+C_1}=e^{C_1}|x|^4, which simplifies to y=Cx4y=Cx^4 where C=±eC1C=\pm e^{C_1} for x>0x>0. Choice A incorrectly treats this as direct integration without the y term. When the coefficient is kx\frac{k}{x} for constant k, expect power functions rather than logarithms in the solution.

Question 17

A cooling object satisfies dTdt=4(T−10)\frac{dT}{dt}=4(T-10). What is the general solution for T(t)T(t)?

  1. T(t)=10+Ce4tT(t)=10+Ce^{4t} (correct answer)
  2. T(t)=Ce4t−10T(t)=Ce^{4t}-10
  3. T(t)=10+e4tT(t)=10+e^{4t}
  4. ln⁡T=4t+C\ln T=4t+C
  5. T(t)=10+4t+CT(t)=10+4t+C
Explanation: This Newton's Law of Cooling equation dT/dt = 4(T-10) requires separation of variables after recognizing the shifted form. We can rewrite this as dT/dt = 4T - 40, and to separate variables, we need dT/(T-10) = 4dt. Integrating both sides gives ln|T-10| = 4t + C₁. Exponentiating yields |T-10| = e^(4t+C₁) = Ce^(4t), where C = ±e^C₁. Solving for T gives T = 10 + Ce^(4t), which is the general solution. Choice B incorrectly places the 10 as -10 outside the exponential term, which would give dT/dt = 4Ce^(4t) ≠ 4(T-10). When dealing with shifted exponential models like dT/dt = k(T-a), the solution is always T = a + Ce^(kt), not Ce^(kt) - a.

Question 18

A function satisfies dydx=−4xy\dfrac{dy}{dx}=-4xy. What is the general solution for y(x)y(x)?

  1. y=Ce−2x2y=Ce^{-2x^{2}} (correct answer)
  2. y=Ce−4x2y=Ce^{-4x^{2}}
  3. ln⁡∣y∣=−2x2\ln|y|=-2x^{2}
  4. y=e−2x2+Cy=e^{-2x^{2}}+C
  5. y=−2x2+Cy=-2x^{2}+C
Explanation: This problem requires solving a differential equation using separation of variables. With dy/dx = -4 x y, separate as dy/y = -4 x dx, assuming y ≠ 0. Integrate for ln|y| = -2 x^2 + C. Solve to y = C e^{-2 x^2}. The option y = e^{-2 x^2} + C is wrong as it adds C outside. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

Question 19

A substance cools according to dTdt=−2T\dfrac{dT}{dt}=-2T. What is the general solution for T(t)T(t)?

  1. T=Ce−2tT=Ce^{-2t} (correct answer)
  2. T=−2et+CT=-2e^{t}+C
  3. ln⁡∣T∣=−2t\ln|T|=-2t
  4. T=e−2t+CT=e^{-2t}+C
  5. T=−2t+CT=-2t+C
Explanation: This problem requires solving a differential equation using separation of variables. For dT/dt = -2T, separate by writing dT/T = -2 dt, assuming T ≠ 0. Integrate: ln|T| on the left and -2t + C on the right. Solve for T by exponentiating, getting T = Ce^{-2t}. The distractor T = -2t + C fails as it implies linear decay, not exponential, which would fit dT/dt = -2. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

Question 20

A function satisfies dydx=−2yx2\dfrac{dy}{dx}=-\dfrac{2y}{x^{2}} for x≠0x\ne 0. What is the general solution?

  1. y=Ce2/xy=Ce^{2/x} (correct answer)
  2. y=Ce−2/xy=Ce^{-2/x}
  3. ln⁡∣y∣=−2x2+C\ln|y|=-\dfrac{2}{x^{2}}+C
  4. y=e2/x+Cy=e^{2/x}+C
  5. y=2x+Cy=\dfrac{2}{x}+C
Explanation: This problem requires solving a differential equation using separation of variables. For dy/dx=−2y/x2dy/dx = -2y / x^2 with x≠0x \ne 0, separate as dy/y=−2 dx/x2dy/y = -2 \, dx / x^2, assuming y≠0y \ne 0. Integrate to ln⁡∣y∣=2/x+C\ln|y| = 2/x + C. Exponentiate for y=Ce2/xy = C e^{2/x}. The distractor y=e2/x+Cy = e^{2/x} + C fails by adding the constant additively. To recognize separable equations, look for the form dy/dx=f(x)g(y)dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy=f(x) dx(1/g(y)) \, dy = f(x) \, dx for integration.