AP Calculus AB 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 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.

Question 1

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

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

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

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

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

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

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

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

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

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

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

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

1. $y=Ce^{-7x}$ (correct answer)
2. $y=Ce^{7x}$
3. $y=e^{-7x}$
4. $\ln|y|=-7$
5. $y=C-7x$

Explanation: To solve $\frac{dy}{dx}=-7y$ using separation of variables, we divide by $y$ and multiply by $dx$ to get $\frac{dy}{y}=-7dx$. Integrating both sides gives $\ln|y|=-7x+C_1$. Exponentiating yields $|y|=e^{-7x+C_1}=e^{C_1}e^{-7x}$, which becomes $y=Ce^{-7x}$ where $C=\pm e^{C_1}$. Choice B incorrectly uses a positive exponent, missing the negative sign from the differential equation. The key recognition pattern is that $\frac{dy}{dx}=ky$ always yields $y=Ce^{kx}$, with the sign of $k$ preserved in the solution.

Question 8

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

Explanation: This differential equation uses separation of variables with a rational function. Starting with $\frac{dy}{dx}=\frac{3x}{1+x^2}y$, we separate to get $\frac{dy}{y}=\frac{3x}{1+x^2}dx$. To integrate the right side, note that $\frac{d}{dx}[\ln(1+x^2)]=\frac{2x}{1+x^2}$, so $\int\frac{3x}{1+x^2}dx=\frac{3}{2}\ln(1+x^2)$. This gives $\ln|y|=\frac{3}{2}\ln(1+x^2)+C_1=\ln(1+x^2)^{3/2}+C_1$. Exponentiating yields $|y|=e^{C_1}(1+x^2)^{3/2}$, which becomes $y=C(1+x^2)^{3/2}$ where $C=\pm e^{C_1}$. Choice C omits the arbitrary constant. When integrating rational functions of the form $\frac{kx}{1+x^2}$, recognize this as a logarithmic derivative pattern.

Question 9

1. $A(t)=0.08e^{t}$
2. $A(t)=Ce^{0.08t}$ (correct answer)
3. $A(t)=e^{0.08t}$
4. $\ln A=0.08$
5. $A(t)=C+0.08t$

Explanation: The investment equation $\frac{dA}{dt}=0.08A$ is solved through separation of variables. Dividing by $A$ and multiplying by $dt$ gives $\frac{dA}{A}=0.08dt$. Integration yields $\ln|A|=0.08t+C_1$. Exponentiating both sides gives $|A|=e^{0.08t+C_1}=e^{C_1}e^{0.08t}$, resulting in $A=Ce^{0.08t}$ where $C=\pm e^{C_1}$. Choice A incorrectly places the rate 0.08 as a coefficient outside the exponential rather than in the exponent. When solving $\frac{dy}{dt}=ky$, the constant $k$ always appears in the exponent of $e$ in the solution $y=Ce^{kt}$.

Question 10

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

Explanation: To solve $\frac{dy}{dx}=\frac{4}{x}y$ using separation of variables, we divide both sides by $y$ and multiply by $dx$ to get $\frac{dy}{y}=\frac{4}{x}dx$. Integrating both sides yields $\ln|y|=4\ln|x|+C_1=\ln|x|^4+C_1$. Exponentiating gives $|y|=e^{\ln|x|^4+C_1}=e^{C_1}|x|^4$, which simplifies to $y=Cx^4$ where $C=\pm e^{C_1}$. Choice A incorrectly integrates the right side as if $y$ were not present in the original equation. When the coefficient of $y$ contains only functions of the independent variable, separation of variables transforms the exponential into a power function.

Question 11

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

Explanation: This problem uses separation of variables to solve the differential equation. Starting with $\frac{dV}{dt}=-2V$, we separate variables to get $\frac{dV}{V}=-2dt$. Integrating both sides yields $\ln|V|=-2t+C_1$, where $C_1$ is an integration constant. Exponentiating gives $|V|=e^{-2t+C_1}=e^{C_1}e^{-2t}$, which simplifies to $V=Ce^{-2t}$ where $C=\pm e^{C_1}$ 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

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

Explanation: The differential equation $\frac{dP}{dt}=\frac{1}{2}P$ is solved by separation of variables. Dividing by $P$ and multiplying by $dt$ gives $\frac{dP}{P}=\frac{1}{2}dt$. Integration yields $\ln|P|=\frac{t}{2}+C_1$. Exponentiating both sides gives $|P|=e^{t/2+C_1}=e^{C_1}e^{t/2}$, resulting in $P=Ce^{t/2}$ where $C=\pm e^{C_1}$. Choice A incorrectly treats the coefficient $\frac{1}{2}$ as if it were multiplying $e^t$ rather than appearing in the exponent. Remember that when integrating $\frac{dP}{P}=k\,dt$, the constant $k$ becomes the coefficient in the exponent, not a multiplicative factor outside.

Question 13

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

Explanation: This differential equation requires separation of variables to solve. We can rewrite $\frac{dP}{dt}=3P$ as $\frac{dP}{P}=3dt$, separating the variables P and t on opposite sides. Integrating both sides gives $\ln|P|=3t+C_1$, where $C_1$ is an arbitrary constant. Exponentiating both sides yields $|P|=e^{3t+C_1}=e^{C_1}e^{3t}$, which we can write as $P=Ce^{3t}$ where $C=\pm e^{C_1}$ is an arbitrary constant. Choice A incorrectly treats this as a simple integration problem without recognizing the multiplicative relationship. When you see $\frac{dy}{dt}=ky$ where k is constant, the solution will always be an exponential function $y=Ce^{kt}$.

Question 14

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

Explanation: This differential equation requires separation of variables since the derivative involves both x and y multiplicatively. From $\frac{dy}{dx}=xy$, we separate to get $\frac{dy}{y}=x dx$. Integrating both sides gives $\ln|y|=\frac{x^2}{2}+C_1$, where $C_1$ is a constant of integration. Exponentiating yields $|y|=e^{x^2/2+C_1}=e^{C_1}e^{x^2/2}$, which becomes $y=Ce^{x^2/2}$ where $C=\pm e^{C_1}$. Choice A incorrectly treats this as direct integration of x, missing the crucial y term in the original equation. When you see $\frac{dy}{dx}=f(x)g(y)$, always separate variables before integrating.

Question 15

1. $P(t)=\ln t + C$
2. $\ln P=\ln t$
3. $P(t)=Ct$ (correct answer)
4. $P(t)=\dfrac{t}{P}+C$
5. $P(t)=\dfrac{C}{t}$

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

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

Explanation: This problem requires separation of variables with a rational coefficient. Starting with $\frac{dy}{dx}=\frac{4}{x}y$, we separate variables to get $\frac{dy}{y}=\frac{4}{x}dx$. Integrating both sides yields $\ln|y|=4\ln|x|+C_1=\ln|x|^4+C_1$, since $\int\frac{4}{x}dx=4\ln|x|$. Exponentiating gives $|y|=e^{\ln|x|^4+C_1}=e^{C_1}|x|^4$, which simplifies to $y=Cx^4$ where $C=\pm e^{C_1}$ for $x>0$. Choice A incorrectly treats this as direct integration without the y term. When the coefficient is $\frac{k}{x}$ for constant k, expect power functions rather than logarithms in the solution.

Question 17

1. $T(t)=10+Ce^{4t}$ (correct answer)
2. $T(t)=Ce^{4t}-10$
3. $T(t)=10+e^{4t}$
4. $\ln T=4t+C$
5. $T(t)=10+4t+C$

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

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

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

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

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

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

Explanation: This problem requires solving a differential equation using separation of variables. For $dy/dx = -2y / x^2$ with $x \ne 0$, separate as $dy/y = -2 \, dx / x^2$, assuming $y \ne 0$. Integrate to $\ln|y| = 2/x + C$. Exponentiate for $y = C e^{2/x}$. The distractor $y = e^{2/x} + C$ fails by adding the constant additively. 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.