AP Calculus AB Differential Equations — Worked Answer Explanations

Unit 7 · 7% of the AP exam · 12 questions explained

1. Question 1 · Easy

   Which of the following is the solution to $\frac{dy}{dx}=2x$ with initial condition $y(0)=3$?

   • A
     $y=2x$

     Why not A: Forgot to integrate $2x$ and ignored the initial condition.
   • B
     $y=x^2$

     Why not B: Integrated correctly but ignored the initial condition to find $C$.
   • C
     $y=x^2+3$Correct
   • D
     $y=2x+3$

     Why not D: Used $y=2x$ directly rather than integrating to get $x^2$.
   Explanation

   Integrate both sides: $y=\int 2xdx=x^2+C$. Apply $y(0)=3$: $3=0+C\Rightarrow C=3$. So $y=x^2+3$.

   Key takeaway

   For a separable ODE, integrate both sides and apply the initial condition to determine the constant of integration.

2. Question 2 · Easy

   A slope field for a differential equation shows short line segments with slope $x+y$ at each point $(x,y)$. Which differential equation corresponds to this slope field?

   • A
     $\frac{dy}{dx}=xy$

     Why not A: Confused addition and multiplication of $x$ and $y$.
   • B
     $\frac{dy}{dx}=x-y$

     Why not B: Sign error: subtracted instead of added.
   • C
     $\frac{dy}{dx}=x+y$Correct
   • D
     $\frac{dy}{dx}=y-x$

     Why not D: Reversed the roles of $x$ and $y$ in the sign.
   Explanation

   A slope field assigns to each point $(x,y)$ a small segment with slope $=f(x,y)$. The problem states slopes equal $x+y$, so the equation is $\frac{dy}{dx}=x+y$.

   Key takeaway

   A slope field at point $(x,y)$ draws a segment with slope equal to $\frac{dy}{dx}$ evaluated at that point.

3. Question 3 · Easy

   A population $P$ satisfies $\frac{dP}{dt}=0.05P$. If $P(0)=200$, find $P(t)$.

   • A
     $P(t)=200+0.05t$

     Why not A: Used a linear model instead of the exponential solution to $dP/dt=kP$.
   • B
     $P(t)=200e^{0.05t}$Correct
   • C
     $P(t)=e^{0.05t}$

     Why not C: Forgot to multiply by the initial condition $P(0)=200$.
   • D
     $P(t)=200(0.05{)}^t$

     Why not D: Wrote the solution as exponential with base $k$ instead of $e^{kt}$.
   Explanation

   The equation $\frac{dP}{dt}=kP$ has the general solution $P(t)=P_0{e}^{kt}$. Here $k=0.05$ and $P_0=P(0)=200$, giving $P(t)=200e^{0.05t}$.

   Key takeaway

   Exponential growth/decay model: $\dfrac{dP}{dt} = kP$ has solution $P(t) = P_0 e^{kt}$ where $P_0$ is the initial value.

4. Question 4 · Easy

   Solve the separable ODE $\frac{dy}{dx}=\frac{x}{y}$ with $y(0)=4$. (Give the solution for $y>0$.)

   • A
     $y=\sqrt{x^2+4}$

     Why not A: Did not apply the initial condition correctly; should get $y=\sqrt{x^2+16}$.
   • B
     $y=\sqrt{x^2+16}$Correct
   • C
     $y=x^2+4$

     Why not C: Did not take the square root after integrating.
   • D
     $y=\frac{x^2}{2}+4$

     Why not D: Integrated without separating variables properly — did not multiply both sides by $y$.
   Explanation

   Separate: $ydy=xdx$. Integrate: $\frac{y^2}{2}=\frac{x^2}{2}+C$. Apply $y(0)=4$: $\frac{16}{2}=0+C\Rightarrow C=8$. So $y^2=x^2+16\Rightarrow y=\sqrt{x^2+16}$ (taking positive root).

   Key takeaway

   Separable ODE: move all $y$ terms to one side, all $x$ terms to the other, then integrate and apply the initial condition.

5. Question 5 · Medium

   Newton's law of cooling gives $\frac{dT}{dt}=-k(T-T_s)$, where $T_s=20$°C is room temperature. If an object cools from $100$°C to $60$°C in $10$ minutes, what is the temperature at $t=20$ minutes?

   • A
     $20$°C

     Why not A: Reached room temperature too fast; cooling is exponential, not linear.
   • B
     $40$°CCorrect
   • C
     $36$°C

     Why not C: Arithmetic error when computing $(e^{-10k}{)}^2$ or applying the initial excess.
   • D
     $28$°C

     Why not D: Subtracted a constant linear drop of 40°C again rather than applying exponential decay.
   Explanation

   Solution: $T(t)-20=80e^{-kt}$. At $t=10$: $60-20=40=80e^{-10k}\Rightarrow e^{-10k}=\frac{1}{2}$. At $t=20$: $T-20=80(e^{-10k}{)}^2=80\cdot \frac{1}{4}=20$. So $T(20)=20+20=40$°C.

   Key takeaway

   Newton's cooling: the excess temperature $(T - T_s)$ decays exponentially; find $k$ from one data point, then evaluate at the desired time.

6. Question 6 · Medium

   On a slope field for $\frac{dy}{dx}=y-x$, the slopes along the line $y=x$ are:

   • A
     $1$ everywhere along $y=x$.

     Why not A: Evaluated $dy/dx=x$ instead of $y-x$.
   • B
     $0$ everywhere along $y=x$.Correct
   • C
     Equal to $2x$ along $y=x$.

     Why not C: Evaluated $y+x=2x$ instead of $y-x$.
   • D
     Undefined along $y=x$.

     Why not D: $dy/dx=y-x$ is defined for all $(x,y)$.
   Explanation

   On the line $y=x$, substitute $y=x$: $\frac{dy}{dx}=y-x=x-x=0$. So all segments along $y=x$ are horizontal (slope = $0$).

   Key takeaway

   To read a slope field on a specific curve, substitute the curve's equation into $dy/dx$; zero slope means horizontal segments.

7. Question 7 · Medium

   Solve $\frac{dy}{dx}=-2x{y}^2$ with $y(0)=1$.

   • A
     $y=e^{-x^2}$

     Why not A: Solved $dy/dx=-2xy$ (first-order linear) rather than the given equation.
   • B
     $y=\frac{1}{x^2+1}$Correct
   • C
     $y=\frac{1}{x^2-1}$

     Why not C: Sign error when applying initial condition: $C=-1$ not $+1$.
   • D
     $y=-2x$

     Why not D: Did not separate variables; confused the ODE with a linear one.
   Explanation

   Separate: $y^{-2}dy=-2xdx$. Integrate: $-y^{-1}=-x^2+C$, so $\frac{1}{y}=x^2-C$. With $y(0)=1$: $1=0-C\Rightarrow C=-1$. So $\frac{1}{y}=x^2+1\Rightarrow y=\frac{1}{x^2+1}$.

   Key takeaway

   Separate variables, integrate both sides including any power of $y$, then apply the initial condition carefully.

8. Question 8 · Medium

   A bacteria culture doubles every 3 hours. If the initial count is $N_0$, what is the count after $t$ hours?

   • A
     $N(t)=N_0{e}^{3t}$

     Why not A: Used $k=3$ (the doubling time) rather than $k=\ln 2/3$.
   • B
     $N(t)=N_0\cdot 2^{t/3}$Correct
   • C
     $N(t)=N_0\cdot 2^{3t}$

     Why not C: Inverted the exponent: used $3t$ instead of $t/3$.
   • D
     $N(t)=N_0(1+t/3)$

     Why not D: Used a linear growth model instead of exponential.
   Explanation

   Doubling every 3 hours means $N(3)=2N_0$. Using $N(t)=N_0{e}^{kt}$: $2N_0=N_0{e}^{3k}\Rightarrow k=\frac{\ln 2}{3}$. So $N(t)=N_0{e}^{(\ln 2/3)t}=N_0\cdot 2^{t/3}$.

   Key takeaway

   Doubling time $T_d$ gives $k = \ln 2 / T_d$; equivalently write $N(t) = N_0 \cdot 2^{t/T_d}$.

9. Question 9 · Medium

   Radioactive carbon-14 decays according to $\frac{dN}{dt}=-0.000121N$. If $N_0=1000$ atoms initially, how many atoms remain after $5730$ years (one half-life)?

   • A
     $0$

     Why not A: A half-life reduces quantity by half, not to zero.
   • B
     $500$Correct
   • C
     $250$

     Why not C: Applied two half-lives instead of one.
   • D
     $993$

     Why not D: Computed $1000(1-0.000121\times 5730)$ using a linear model.
   Explanation

   By definition, one half-life reduces the quantity to half. After $t=5730$ years: $N=1000\cdot e^{-0.000121\times 5730}=1000\cdot e^{-0.693}\approx 1000\cdot 0.5=500$ atoms.

   Key takeaway

   The half-life is the time at which exactly half the original amount remains; $e^{-k T_{1/2}} = 1/2$.

10. Question 10 · Hard

    A slope field for $\frac{dy}{dx}=\frac{x}{y}$ is sketched. Which statement best describes a solution curve passing through $(0,3)$?

    • A
      A line with slope $0$ through $(0,3)$.

      Why not A: The slope at $(0,3)$ is $0$, but the curve is not a straight line overall.
    • B
      The upper semicircle of radius $3$ centered at the origin.

      Why not B: Solving $ydy=xdx$ gives $y^2-x^2=C$ — a hyperbola, not a circle.
    • C
      The upper branch of the hyperbola $y^2-x^2=9$.Correct
    • D
      The parabola $y=x^2+3$.

      Why not D: Confused the ODE $dy/dx=x/y$ with $dy/dx=x$.
    Explanation

    Separate: $ydy=xdx$. Integrate: $\frac{y^2}{2}=\frac{x^2}{2}+C$, so $y^2-x^2=2C$. Applying $(0,3)$: $9-0=2C\Rightarrow 2C=9$. The solution is $y^2-x^2=9$ — the upper branch of a hyperbola (since $y(0)=3>0$).

    Key takeaway

    Identifying the shape of a solution curve requires solving the ODE, not just reading the slope at one point.

11. Question 11 · Hard

    The general solution to $\frac{dy}{dx}=ky$ is $y=C{e}^{kx}$. If $y(0)=5$ and $y(2)=20$, find $k$.

    • A
      $k=\frac{\ln 2}{2}$

      Why not A: Used $y(2)/y(0)=4$ but divided by $4$ instead of taking $\ln 4/2$.
    • B
      $k=\ln 2$Correct
    • C
      $k=2$

      Why not C: Used $k=y(2)/y(0)$ without applying the logarithm.
    • D
      $k=\frac{\ln 4}{4}$

      Why not D: Divided $\ln 4$ by $4$ instead of by $2$.
    Explanation

    With $y(0)=5$: $5=C{e}^0=C$. With $y(2)=20$: $20=5e^{2k}\Rightarrow e^{2k}=4\Rightarrow 2k=\ln 4=2\ln 2\Rightarrow k=\ln 2$.

    Key takeaway

    To find the growth constant, use two data points: apply the initial condition for $C$, then solve for $k$ using the logarithm.

12. Question 12 · Hard

    For the initial value problem $\frac{dy}{dx}=x-y$, $y(0)=1$, use Euler's method with step size $h=1$ to estimate $y(2)$.

    • A
      $0$

      Why not A: Only applied one Euler step, stopping at $y_1=0$ rather than continuing to $y_2$.
    • B
      $1$Correct
    • C
      $-1$

      Why not C: Sign error in the second Euler step: used $f(1,0)=-(1-0)$ instead of $1-0$.
    • D
      $2$

      Why not D: Added $h$ to $y$ directly instead of $h\cdot f(x_n,y_n)$.
    Explanation

    Euler step 1 at $(x_0,y_0)=(0,1)$: $f(0,1)=0-1=-1$. $y_1=1+1(-1)=0$, $x_1=1$. Euler step 2 at $(1,0)$: $f(1,0)=1-0=1$. $y_2=0+1(1)=1$. So $y(2)\approx 1$.

    Key takeaway

    Euler's method: $y_{n+1} = y_n + h \cdot f(x_n, y_n)$; apply iteratively, updating both $x$ and $y$ at each step.