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AP Calculus BC

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Differential Equations practice questions — AP Calculus BC

7 free multiple-choice problems on differential equations, ordered to match AP Calculus BC difficulty, each with a full worked solution. No account needed — and there's a fresh problem set every day.

Problems & worked solutions

Problem #3564 International

Problem 1 — Differential Equations

The salt content S(t)S(t) of a tank, measured in kilograms tt minutes after the start, satisfies S′=−15S+4,S(0)=100.S' = -\frac{1}{5}S + 4, \qquad S(0) = 100. The time, in minutes, at which the tank holds 4040 kilograms of salt is:

Show answer & worked solution
  1. A. 10ln⁡210\ln 2✓ correct
  2. B. 5ln⁡25\ln 2
  3. C. 5ln⁡55\ln 5
  4. D. 10ln⁡410\ln 4
  5. E. 2ln⁡25\dfrac{2\ln 2}{5}

Factor the right-hand side: S′=−15(S−20)S' = -\dfrac{1}{5}\left(S - 20\right).

Separating the variables and integrating, ∫dSS−20=∫−15 dt⟹ln⁡∣S−20∣=−t5+k,\int \frac{dS}{S - 20} = \int -\frac{1}{5}\,dt \quad \Longrightarrow \quad \ln|S - 20| = -\frac{t}{5} + k, so S−20=Ce−t/5S - 20 = Ce^{-t/5}.

The initial condition gives 100−20=C100 - 20 = C, hence C=80C = 80 and S(t)=20+80e−t/5.S(t) = 20 + 80e^{-t/5}.

Now impose S(t)=40S(t) = 40: 80e−t/5=20⟹e−t/5=14⟹t5=ln⁡4.80e^{-t/5} = 20 \quad \Longrightarrow \quad e^{-t/5} = \frac{1}{4} \quad \Longrightarrow \quad \frac{t}{5} = \ln 4.

So t=5ln⁡4t = 5\ln 4, and since ln⁡4=2ln⁡2\ln 4 = 2\ln 2, t=10ln⁡2.t = 10\ln 2.

Problem #3563 International

Problem 2 — Differential Equations

The solution of the initial value problem y′=−2y+10y' = -2y + 10, y(0)=1y(0) = 1 is:

Show answer & worked solution
  1. A. y=5−4e−2xy = 5 - 4e^{-2x}✓ correct
  2. B. y=5+4e−2xy = 5 + 4e^{-2x}
  3. C. y=5−4e2xy = 5 - 4e^{2x}
  4. D. y=−5+6e−2xy = -5 + 6e^{-2x}
  5. E. y=10−9e−2xy = 10 - 9e^{-2x}

The constant solution comes from setting y′=0y' = 0: 0=−2y+100 = -2y + 10, so ye=5y_e = 5.

Put u=y−5u = y - 5, so that u′=y′u' = y' and y=u+5y = u + 5: u′=−2(u+5)+10=−2u−10+10=−2u.u' = -2(u + 5) + 10 = -2u - 10 + 10 = -2u.

A function whose derivative is −2-2 times itself is a multiple of e−2xe^{-2x}, so u(x)=Ce−2xu(x) = Ce^{-2x} and therefore y(x)=5+Ce−2x.y(x) = 5 + Ce^{-2x}.

The initial condition y(0)=1y(0) = 1 gives 5+C=15 + C = 1, hence C=−4C = -4.

Check: y′=8e−2xy' = 8e^{-2x}, while −2y+10=−2(5−4e−2x)+10=8e−2x-2y + 10 = -2\left(5 - 4e^{-2x}\right) + 10 = 8e^{-2x}, and y(0)=5−4=1y(0) = 5 - 4 = 1. y(x)=5−4e−2xy(x) = 5 - 4e^{-2x}

Problem #3562 International

Problem 3 — Differential Equations

A quantity y(t)y(t) satisfies y′=kyy' = ky for some constant kk, with y(0)=12y(0) = 12 and y(2)=3y(2) = 3. The value of y(4)y(4) is:

Show answer & worked solution
  1. A. 34\dfrac{3}{4}✓ correct
  2. B. 32\dfrac{3}{2}
  3. C. 14\dfrac{1}{4}
  4. D. −6-6
  5. E. 1212

Separate the variables in y′=kyy' = ky: dyy=k dt  ⟹  ln⁡∣y∣=kt+c  ⟹  y(t)=Cekt.\frac{dy}{y} = k\,dt \implies \ln|y| = kt + c \implies y(t) = Ce^{kt}.

The condition y(0)=12y(0) = 12 gives Ce0=C=12Ce^{0} = C = 12, so y(t)=12ekty(t) = 12e^{kt}.

The condition y(2)=3y(2) = 3 pins the rate: 12e2k=3  ⟹  e2k=14.12e^{2k} = 3 \implies e^{2k} = \frac{1}{4}.

There is no need to extract kk itself, because e4ke^{4k} is the square of e2ke^{2k}: y(4)=12e4k=12(e2k)2=12⋅116.y(4) = 12e^{4k} = 12\left(e^{2k}\right)^{2} = 12\cdot\frac{1}{16}.

Hence y(4)=34.y(4) = \frac{3}{4}.

Problem #0886 US AP

Problem 4 — Differential Equations

For the ODE y′=y+xy' = y + x, the function g(x)=−x−1g(x) = -x - 1 is a particular solution. The general solution is:

Show answer & worked solution
  1. A. y=−x−1y = -x - 1
  2. B. y=Cex⋅(−x−1)y = C e^{x} \cdot (-x - 1)
  3. C. y=−x−1+Cexy = -x - 1 + C e^{x}✓ correct
  4. D. y=−x−1+Cy = -x - 1 + C

Verify gg: g′(x)=−1g'(x) = -1 and g+x=−x−1+x=−1g + x = -x - 1 + x = -1, so g′=g+xg' = g + x. ✓

The homogeneous equation y′=yy' = y has general solution CexCe^{x}.

So the general solution of y′=y+xy' = y + x is y(x)=−x−1+Cexy(x) = -x - 1 + C e^{x}.

Problem #0885 US AP

Problem 5 — Differential Equations

The constant (equilibrium) solution of y′=2y+6y' = 2y + 6 is:

Show answer & worked solution
  1. A. y=−3y = -3✓ correct
  2. B. y=3y = 3
  3. C. y=6y = 6
  4. D. y=−6y = -6

Setting y′=0y' = 0: 0=2y+6  ⟹  y=−30 = 2y + 6 \implies y = -3.

Equivalently, yp=−ba=−62=−3y_p = -\dfrac{b}{a} = -\dfrac{6}{2} = -3.

Problem #0884 US AP

Problem 6 — Differential Equations

The solution of y′=3yy' = 3y with y(0)=5y(0) = 5 is:

Show answer & worked solution
  1. A. y=5+3xy = 5 + 3x
  2. B. y=5e3xy = 5 e^{3x}✓ correct
  3. C. y=3e5xy = 3 e^{5x}
  4. D. y=5e−3xy = 5 e^{-3x}

General solution of y′=3yy' = 3y: y(x)=Ce3xy(x) = C e^{3x}.

Apply y(0)=5y(0) = 5: Ce0=C=5C e^{0} = C = 5.

So y(x)=5e3xy(x) = 5 e^{3x}.

Problem #3561 International

Problem 7 — Differential Equations

The function y(x)=4e−3xy(x) = 4e^{-3x} is a solution of the differential equation:

Show answer & worked solution
  1. A. y′=−3yy' = -3y✓ correct
  2. B. y′=3yy' = 3y
  3. C. y′=4yy' = 4y
  4. D. y′=−12yy' = -12y

Differentiate the given function: y′(x)=4⋅(−3)e−3x=−12e−3x.y'(x) = 4\cdot(-3)e^{-3x} = -12e^{-3x}.

To compare y′y' with yy, factor the given function back out of this expression: −12e−3x=−3⋅(4e−3x)=−3 y(x).-12e^{-3x} = -3\cdot\left(4e^{-3x}\right) = -3\,y(x).

The amplitude 44 appears in both yy and y′y', so it cancels in the ratio y′/yy'/y and never enters the rate; only the coefficient of xx in the exponent survives: y′=−3y.y' = -3y.

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