AP Calculus BC Quiz: Initial Conditions And Separation Of Variables
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Initial Conditions And Separation Of VariablesQuestion 1 of 20

A quantity satisfies dydx=x(1+y)\frac{dy}{dx}=x(1+y) with y(0)=1y(0)=1. Which particular solution gives y(x)y(x)?

y=ex22−1y= e^{\frac{x^2}{2}}-1
y=2ex22−1y=2e^{\frac{x^2}{2}}-1
y=2ex2−1y=2e^{x^2}-1
y=x22+1y=\frac{x^2}{2}+1
y=ln⁡(x22)y=\ln\left(\frac{x^2}{2}\right)
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AP Calculus BC Quiz

AP Calculus BC Quiz: Initial Conditions And Separation Of Variables

Practice Initial Conditions And Separation Of Variables in AP Calculus BC 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 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.

All questions

Question 1

A quantity satisfies dydx=x(1+y)\frac{dy}{dx}=x(1+y) with y(0)=1y(0)=1. Which particular solution gives y(x)y(x)?

  1. y=ex22−1y= e^{\frac{x^2}{2}}-1
  2. y=2ex22−1y=2e^{\frac{x^2}{2}}-1 (correct answer)
  3. y=2ex2−1y=2e^{x^2}-1
  4. y=x22+1y=\frac{x^2}{2}+1
  5. y=ln⁡(x22)y=\ln\left(\frac{x^2}{2}\right)
Explanation: This problem involves applying an initial condition after solving a separable differential equation. Separating variables in dy/dx = x(1+y) gives dy/(1+y) = x dx, which integrates to ln|1+y| = x²/2 + C₁. This yields 1+y = Ce^(x²/2), or y = Ce^(x²/2) - 1 for the general solution. Applying the initial condition y(0) = 1: 1 = Ce^0 - 1 = C - 1, so C = 2. Therefore, the particular solution is y = 2e^(x²/2) - 1. Choice A (y = e^(x²/2) - 1) incorrectly uses C = 1, failing to properly apply the initial condition. When solving separable equations, always substitute the initial values into the general solution to find the specific constant value.

Question 2

A quantity satisfies dydx=yx2\frac{dy}{dx}=\frac{y}{x^2} with y(1)=2y(1)=2. What is the particular solution?

  1. y=2exy=2e^{x}
  2. y=2e1xy=2e^{\frac{1}{x}}
  3. y=2e1−1xy=2e^{1-\frac{1}{x}} (correct answer)
  4. y=e1−1xy=e^{1-\frac{1}{x}}
  5. y=2(1−1x)y=2(1-\frac{1}{x})
Explanation: This problem requires applying an initial condition to a separable differential equation. Separating variables in dy/dx = y/x² gives dy/y = dx/x², which integrates to ln|y| = -1/x + C₁. This yields the general solution y = Ce^(-1/x). Applying the initial condition y(1) = 2: 2 = Ce^(-1/1) = Ce^(-1), so C = 2e^1 = 2e. Therefore, y = 2e·e^(-1/x) = 2e^(1-1/x). Choice D (y = e^(1-1/x)) incorrectly omits the factor of 2, failing to properly apply the initial condition. When exponential solutions involve nested exponents, carefully track all multiplicative constants and use initial conditions to determine their values.

Question 3

A tank's salt amount satisfies dSdt=3tS\frac{dS}{dt}=3tS with S(0)=5S(0)=5. What is the particular solution S(t)S(t)?

  1. S(t)=5e3t2S(t)=5e^{3t^2}
  2. S(t)=5e32t2S(t)=5e^{\frac{3}{2}t^2} (correct answer)
  3. S(t)=e32t2S(t)=e^{\frac{3}{2}t^2}
  4. S(t)=5+32t2S(t)=5+\frac{3}{2}t^2
  5. S(t)=51−32t2S(t)=\frac{5}{1-\frac{3}{2}t^2}
Explanation: This problem involves applying an initial condition to determine the constant in a separable differential equation solution. Separating variables in dS/dt = 3tS gives dS/S = 3t dt, which integrates to ln|S| = (3/2)t² + C₁. This yields the general solution S(t) = Ce^((3/2)t²). Applying the initial condition S(0) = 5: 5 = Ce^((3/2)·0²) = Ce^0 = C. Therefore, C = 5 and the particular solution is S(t) = 5e^((3/2)t²). Choice A (S(t) = 5e^(3t²)) has an incorrect exponent, likely from integrating 3t as 3t² instead of (3/2)t². Remember that when integrating, ∫3t dt = (3/2)t², and initial conditions determine the multiplicative constant in exponential solutions.

Question 4

A function satisfies dydx=yx\frac{dy}{dx}=\frac{y}{x} for x>0x>0 with y(2)=7y(2)=7. Which y(x)y(x) is correct?

  1. y=72xy=\frac{7}{2}x (correct answer)
  2. y=7xy=7x
  3. y=Cxy=Cx
  4. y=7xy=\frac{7}{x}
  5. y=7ln⁡xy=7\ln x
Explanation: This homogeneous differential equation requires applying an initial condition at a non-zero point. The general solution to dydx=yx\frac{dy}{dx}=\frac{y}{x} is y=Cxy=Cx (found by separating variables: dyy=dxx\frac{dy}{y}=\frac{dx}{x}, giving ln⁡∣y∣=ln⁡∣x∣+ln⁡∣C∣\ln|y|=\ln|x|+\ln|C|). Using the initial condition y(2)=7y(2)=7: 7=C(2)7=C(2), so C=72C=\frac{7}{2}. Therefore, the particular solution is y=72xy=\frac{7}{2}x. Choice B incorrectly assumes C=7C=7, forgetting to solve for CC using the given point (2,7)(2,7). When the initial condition is given at a point other than x=0x=0, substitute both coordinates to find the constant.

Question 5

A cooling object satisfies dTdt=−110(T−20)\frac{dT}{dt}=-\frac{1}{10}(T-20) with T(0)=80T(0)=80. Find T(t)T(t).

  1. T(t)=20+60e−t/10T(t)=20+60e^{-t/10} (correct answer)
  2. T(t)=20+60et/10T(t)=20+60e^{t/10}
  3. T(t)=80e−t/10T(t)=80e^{-t/10}
  4. T(t)=20+Ce−t/10T(t)=20+Ce^{-t/10}
  5. T(t)=80−t10T(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 6

A population follows dPdt=P(3−P)\frac{dP}{dt}=P(3-P) with P(0)=1P(0)=1. Which particular solution is correct?

  1. P(t)=31+2e−3tP(t)=\frac{3}{1+2e^{-3t}} (correct answer)
  2. P(t)=31+2e3tP(t)=\frac{3}{1+2e^{3t}}
  3. P(t)=31−2e−3tP(t)=\frac{3}{1-2e^{-3t}}
  4. P(t)=31+Ce−3tP(t)=\frac{3}{1+Ce^{-3t}}
  5. P(t)=11+2e−3tP(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−3te^{-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−3te^{-3t}). A tempting distractor is P(t) = 3 / (1 + 2 e3te^{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 7

A function satisfies dydx=y+1x\dfrac{dy}{dx}=\dfrac{y+1}{x} for x>0x>0 with y(1)=1y(1)=1; find y(x)y(x).

  1. y=2x−1y=2x-1 (correct answer)
  2. y=2x+1y=2x+1
  3. y=Cx−1y=Cx-1
  4. y=2x−1y=\dfrac{2}{x}-1
  5. y=2ln⁡x−1y=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 8

A function satisfies dydx=1x(y−2)\dfrac{dy}{dx}=\dfrac{1}{x} (y-2) for x>0x>0 with y(1)=5y(1)=5; find y(x)y(x).

  1. y=2+3xy=2+3x (correct answer)
  2. y=2+3ln⁡xy=2+3\ln x
  3. y=2+Cxy=2+Cx
  4. y=2+3xy=2+\dfrac{3}{x}
  5. y=5xy=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 9

A tank drains so that dVdt=−0.2V\dfrac{dV}{dt}=-0.2V and V(0)=10V(0)=10; which V(t)V(t) is the particular solution?

  1. V(t)=10e0.2tV(t)=10e^{0.2t}
  2. V(t)=10e−0.2tV(t)=10e^{-0.2t} (correct answer)
  3. V(t)=Ce−0.2tV(t)=Ce^{-0.2t}
  4. V(t)=e−0.2t+10V(t)=e^{-0.2t}+10
  5. V(t)=−0.2e10tV(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 10

A culture grows with dPdt=0.2P\frac{dP}{dt}=0.2P and P(0)=50P(0)=50. Which function gives P(t)P(t)?

  1. P(t)=50e0.2tP(t)=50e^{0.2t} (correct answer)
  2. P(t)=Ce0.2tP(t)=Ce^{0.2t}
  3. P(t)=50e−0.2tP(t)=50e^{-0.2t}
  4. P(t)=0.2e50tP(t)=0.2e^{50t}
  5. P(t)=50+0.2tP(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 11

A function satisfies dydx=x2y\dfrac{dy}{dx}=\dfrac{x^2}{y} with y(0)=4y(0)=4; which is the particular solution?

  1. y=23x3+16y=\sqrt{\dfrac{2}{3}x^3+16} (correct answer)
  2. y=13x3+16y=\sqrt{\dfrac{1}{3}x^3+16}
  3. y=23x3+Cy=\sqrt{\dfrac{2}{3}x^3+C}
  4. y=23x3+16y=\dfrac{2}{3}x^3+16
  5. y=16−23x3y=\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 12

A tank drains so that dVdt=−0.1V\frac{dV}{dt}=-0.1V with V(0)=30V(0)=30. Which is V(t)V(t)?

  1. V(t)=30e−0.1tV(t)=30e^{-0.1t} (correct answer)
  2. V(t)=Ce−0.1tV(t)=Ce^{-0.1t}
  3. V(t)=30e0.1tV(t)=30e^{0.1t}
  4. V(t)=30−0.1tV(t)=30-0.1t
  5. V(t)=−0.1e30tV(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 13

If dydx=(1+x)y\frac{dy}{dx}=(1+x)y and y(0)=2y(0)=2, what is the particular solution y(x)y(x)?

  1. y=2ex+x22y=2e^{x+\frac{x^2}{2}} (correct answer)
  2. y=Cex+x22y=Ce^{x+\frac{x^2}{2}}
  3. y=2ex2y=2e^{x^2}
  4. y=2(x+1)y=2(x+1)
  5. y=2e−(x+x22)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 14

A tank's salt mass SS satisfies dSdt=−0.5S \frac{dS}{dt}=-0.5S. If S(2)=10S(2)=10, find S(t)S(t).

  1. S(t)=10e−0.5(t−2)S(t)=10e^{-0.5(t-2)} (correct answer)
  2. S(t)=10e−0.5tS(t)=10e^{-0.5t}
  3. S(t)=Ce−0.5tS(t)=Ce^{-0.5t}
  4. S(t)=10e0.5(t−2)S(t)=10e^{0.5(t-2)}
  5. S(t)=10−0.5(t−2)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 15

A chemical amount satisfies dAdt=Acos⁡t\frac{dA}{dt}=A\cos t with A(0)=4A(0)=4. What is the particular solution A(t)A(t)?

  1. A(t)=4ecos⁡tA(t)=4e^{\cos t}
  2. A(t)=esin⁡tA(t)=e^{\sin t}
  3. A(t)=4esin⁡tA(t)=4e^{\sin t} (correct answer)
  4. A(t)=4sin⁡tA(t)=4\sin t
  5. A(t)=4+sin⁡tA(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 16

A population satisfies dPdt=1100P(100−P)\frac{dP}{dt}=\frac{1}{100}P(100-P) with P(0)=20P(0)=20. Which P(t)P(t) is correct?

  1. P(t)=1001+4e−tP(t)=\frac{100}{1+4e^{-t}} (correct answer)
  2. P(t)=1001+4etP(t)=\frac{100}{1+4e^{t}}
  3. P(t)=1001+Ce−tP(t)=\frac{100}{1+Ce^{-t}}
  4. P(t)=1001+e−4tP(t)=\frac{100}{1+e^{-4t}}
  5. P(t)=201+4e−tP(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 dPdt=1100P(100−P)\frac{dP}{dt}=\frac{1}{100}P(100-P) is P(t)=1001+Ce−tP(t)=\frac{100}{1+Ce^{-t}} for some constant CC. Substituting the initial condition P(0)=20P(0)=20: 20=1001+Ce0=1001+C20=\frac{100}{1+Ce^0}=\frac{100}{1+C}. Solving: 20(1+C)=10020(1+C)=100, so 20+20C=10020+20C=100, giving C=4C=4. Therefore, P(t)=1001+4e−tP(t)=\frac{100}{1+4e^{-t}}. Choice C shows the general form without determining CC, which is incomplete. For logistic equations, always use the initial population to find the specific constant in the denominator.

Question 17

A solution yy satisfies dydx=xy\frac{dy}{dx}=xy and y(0)=3y(0)=3. What is the particular solution?

  1. y=3ex2/2y=3e^{x^{2}/2} (correct answer)
  2. y=Cex2/2y=Ce^{x^{2}/2}
  3. y=3ex2y=3e^{x^{2}}
  4. y=3e−x2/2y=3e^{-x^{2}/2}
  5. y=32x2y=\frac{3}{2}x^{2}
Explanation: This problem requires applying an initial condition after solving a separable differential equation. Separating variables in dydx=xy\frac{dy}{dx}=xy gives dyy=x dx\frac{dy}{y}=x\,dx. Integrating both sides yields ln⁡∣y∣=x22+C\ln|y|=\frac{x^2}{2}+C, so y=Aex2/2y=Ae^{x^2/2} where A=eCA=e^C. Using the initial condition y(0)=3y(0)=3: 3=Ae02/2=Ae0=A3=Ae^{0^2/2}=Ae^0=A, so A=3A=3. Therefore, the particular solution is y=3ex2/2y=3e^{x^2/2}. Choice C incorrectly integrates to get ex2e^{x^2} instead of ex2/2e^{x^2/2}, forgetting the factor of 12\frac{1}{2} from integration. When solving separable equations, carefully integrate both sides and then apply the initial condition to find the constant.

Question 18

A bacteria culture satisfies dPdt=0.2P\frac{dP}{dt}=0.2P with P(0)=50P(0)=50. Which particular solution P(t)P(t) fits?

  1. P(t)=50e0.2tP(t)=50e^{0.2t} (correct answer)
  2. P(t)=Ce0.2tP(t)=Ce^{0.2t}
  3. P(t)=50e−0.2tP(t)=50e^{-0.2t}
  4. P(t)=0.2e50tP(t)=0.2e^{50t}
  5. P(t)=50+0.2tP(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 dPdt=0.2P\frac{dP}{dt}=0.2P is P(t)=Ce0.2tP(t)=Ce^{0.2t}, where CC is an arbitrary constant. To find CC, we substitute the initial condition P(0)=50P(0)=50: 50=Ce0.2(0)=Ce0=C50=Ce^{0.2(0)}=Ce^0=C, so C=50C=50. Therefore, the particular solution is P(t)=50e0.2tP(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 19

A tank's salt amount satisfies dSdt=3S\frac{dS}{dt}=3S with S(0)=2S(0)=2. What is the particular solution S(t)S(t)?

  1. S(t)=2e3tS(t)=2e^{3t} (correct answer)
  2. S(t)=Ce3tS(t)=Ce^{3t}
  3. S(t)=2e−3tS(t)=2e^{-3t}
  4. S(t)=3e2tS(t)=3e^{2t}
  5. S(t)=2+3tS(t)=2+3t
Explanation: This exponential growth problem requires applying an initial condition to determine the particular solution. The general solution to dSdt=3S\frac{dS}{dt}=3S is S(t)=Ce3tS(t)=Ce^{3t}, representing exponential growth with rate 3. Applying the initial condition S(0)=2S(0)=2: 2=Ce3(0)=Ce0=C2=Ce^{3(0)}=Ce^0=C, so C=2C=2. Thus, the particular solution is S(t)=2e3tS(t)=2e^{3t}. Choice D reverses the roles of the initial value and growth rate, giving S(t)=3e2tS(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 20

A cooling object follows dTdt=−15(T−20)\frac{dT}{dt}=-\frac{1}{5}(T-20) with T(0)=80T(0)=80. Find T(t)T(t).

  1. T(t)=20+60e−t/5T(t)=20+60e^{-t/5} (correct answer)
  2. T(t)=20+60et/5T(t)=20+60e^{t/5}
  3. T(t)=20+Ce−t/5T(t)=20+Ce^{-t/5}
  4. T(t)=80e−t/5T(t)=80e^{-t/5}
  5. T(t)=20−60e−t/5T(t)=20-60e^{-t/5}
Explanation: This problem involves applying an initial condition to Newton's law of cooling. The differential equation dTdt=−15(T−20)\frac{dT}{dt}=-\frac{1}{5}(T-20) has general solution T(t)=20+Ce−t/5T(t)=20+Ce^{-t/5}, where 20 is the ambient temperature. Using the initial condition T(0)=80T(0)=80, we get 80=20+Ce0=20+C80=20+Ce^0=20+C, which gives C=60C=60. Thus, the particular solution is T(t)=20+60e−t/5T(t)=20+60e^{-t/5}. Choice C represents the general solution without determining CC, 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.