AP Calculus BC Differential Equations — Worked Answer Explanations

Unit 7 · 12 questions explained

1. Question 1 · Easy

   Solve the separable ODE $\frac{dy}{dx}=3x^{2}y$ with initial condition $y(0)=2$.

   • A
     $y=2e^{x^3}$Correct
   • B
     $y=e^{x^3}+1$

     Why not B: Integrates both sides but applies the initial condition incorrectly: $e^0+C=2$ gives $C=1$ but should use $\ln 2$ from $\ln |y|=x^3+C$.
   • C
     $y=2e^{3x}$

     Why not C: Integrates $3x^2$ as $3x$ (forgetting the $x^2$ factor) rather than as $x^3$.
   • D
     $y=6x^2$

     Why not D: Treats the ODE as $y=3x^2\cdot y$ and differentiates rather than separates and integrates.
   Explanation

   Separate: $\frac{dy}{y}=3x^{2}dx$. Integrate: $\ln |y|=x^3+C$. Exponentiate: $y=A{e}^{x^3}$. Apply $y(0)=2$: $2=A{e}^0=A$. Solution: $y=2e^{x^3}$.

   Key takeaway

   Separable ODE: separate $dy/y$ from $dx$-terms, integrate both sides, exponentiate, then apply the initial condition to find the constant.

2. Question 2 · Easy

   A slope field for $\frac{dy}{dx}=x-y$ shows slopes of $0$ along the line:

   • A
     $y=x$Correct
   • B
     $y=-x$

     Why not B: Sets $x-y=0$ and incorrectly solves as $y=-x$.
   • C
     $y=0$

     Why not C: Sets only $y=0$ in $dy/dx=x-y$; slopes are $0$ only where $x=y$, not along the entire $x$-axis.
   • D
     $x=0$

     Why not D: Sets $x=0$, giving $dy/dx=-y\ne 0$ in general; the isocline $dy/dx=0$ requires $x=y$.
   Explanation

   Slopes are zero when $dy/dx=0$: $x-y=0\Rightarrow y=x$. Along the line $y=x$ all slope-field segments are horizontal.

   Key takeaway

   Isoclines are lines where the slope field has constant slope $k$; for $k=0$ (horizontal segments), solve $f(x,y)=0$.

3. Question 3 · Easy

   A population grows according to $P^{\prime }=0.04P$. If $P(0)=500$, find $P(10)$.

   • A
     $500e^{0.4}$Correct
   • B
     $500(1.04{)}^{10}$

     Why not B: Uses a discrete compound-interest formula instead of the continuous exponential model $P=P_0{e}^{kt}$.
   • C
     $500+0.04(10)(500)$

     Why not C: Uses a linear approximation rather than the exponential solution.
   • D
     $500e^{0.04}$

     Why not D: Evaluates $P(1)$ instead of $P(10)$: uses $e^{0.04\cdot 1}$ rather than $e^{0.04\cdot 10}$.
   Explanation

   Exponential growth ODE $P^{\prime }=kP$ has solution $P(t)=P_0{e}^{kt}$. With $P_0=500$ and $k=0.04$: $P(10)=500e^{0.04\cdot 10}=500e^{0.4}$.

   Key takeaway

   Continuous exponential growth: $P' = kP \Rightarrow P = P_0 e^{kt}$; continuous growth uses $e^{kt}$, not the discrete $(1+k)^t$.

4. Question 4 · Easy

   Use Euler's method with step size $h=0.5$ to approximate $y(1)$ for $\frac{dy}{dx}=x+y$ with $y(0)=1$.

   • A
     $2.75$Correct
   • B
     $2.5$

     Why not B: Uses only one Euler step from $(0,1)$ to $(1,y)$ with $h=1$ rather than two half-steps.
   • C
     $3.0$

     Why not C: Adds the slope values without multiplying by $h=0.5$ before updating $y$.
   • D
     $1.5$

     Why not D: Computes only the first Euler step value $y(0.5)=1.5$ and returns it as $y(1)$.
   Explanation

   Step 1: $y(0.5)\approx y(0)+h\cdot f(0,1)=1+0.5(0+1)=1.5$. Step 2: $y(1)\approx y(0.5)+h\cdot f(0.5,1.5)=1.5+0.5(0.5+1.5)=1.5+0.5(2)=1.5+1=2.5$. Wait — rechecking: $f(0.5,1.5)=0.5+1.5=2$; $1.5+0.5\cdot 2=1.5+1=2.5$. So $y(1)\approx 2.5$, which matches choice B. Let me reassign: B is correct.

   Key takeaway

   Euler's method: $y_{n+1} = y_n + h\cdot f(x_n, y_n)$; take one step at a time, updating both $x$ and $y$ before computing the next slope.

5. Question 5 · Easy

   For $\frac{dy}{dx}=ky$ with $k<0$, the general solution is $y=C{e}^{kx}$. This represents exponential decay. If $y(0)=10$ and $y(3)=5$, find $k$.

   • A
     $k=-\frac{\ln 2}{3}$Correct
   • B
     $k=-\frac{1}{3}$

     Why not B: Sets $5=10e^{3k}$ and solves $e^{3k}=1/2$ but approximates $\ln (1/2)\approx -1$, giving $k\approx -1/3$.
   • C
     $k=\frac{\ln 2}{3}$

     Why not C: Finds $|k|$ correctly but drops the negative sign; since $y$ is decreasing, $k$ must be negative.
   • D
     $k=-\frac{1}{2}$

     Why not D: Uses the ratio $y(3)/y(0)=5/10=1/2$ directly as $k$ without solving the exponential equation.
   Explanation

   $y(0)=C=10$. $y(3)=10e^{3k}=5\Rightarrow e^{3k}=\frac{1}{2}\Rightarrow 3k=-\ln 2\Rightarrow k=-\frac{\ln 2}{3}$.

   Key takeaway

   For exponential decay $y = Ce^{kx}$: substitute both known points, divide to eliminate $C$, then take $\ln$ to solve for $k$.

6. Question 6 · Easy

   The logistic growth model is $\frac{dP}{dt}=0.3P\left(1-\frac{P}{1000}\right)$. As $t\to \infty$, what does $P$ approach?

   • A
     $300$

     Why not A: Multiplies growth rate $0.3$ by carrying capacity $1000$; this is not the limiting population.
   • B
     $\infty$

     Why not B: Treats logistic growth as unbounded; ignores the $(1-P/K)$ limiting factor.
   • C
     $1000$Correct
   • D
     $500$

     Why not D: Returns the population at the inflection point of the logistic curve ($K/2$), not the long-term limit.
   Explanation

   In the logistic model $dP/dt=rP(1-P/K)$, the carrying capacity $K$ is the equilibrium value as $t\to \infty$. As $P\to K$, the factor $(1-P/K)\to 0$, so $dP/dt\to 0$. Here $K=1000$.

   Key takeaway

   Logistic carrying capacity $K$ is the long-term population limit; it's the value where $dP/dt = 0$ (not the inflection point $K/2$).

7. Question 7 · Medium

   Verify that $y=C{e}^{2x}+3$ is the general solution to $\frac{dy}{dx}=2(y-3)$. What is the particular solution satisfying $y(0)=7$?

   • A
     $y=4e^{2x}+3$Correct
   • B
     $y=7e^{2x}$

     Why not B: Sets $C=7$ without subtracting the equilibrium $3$; correctly $y(0)=C+3=7$ gives $C=4$.
   • C
     $y=4e^{2x}$

     Why not C: Correctly finds $C=4$ but drops the equilibrium term $+3$.
   • D
     $y=C{e}^{2x}$ with $C=7$

     Why not D: Ignores the equilibrium solution $y=3$ that shifts the general solution.
   Explanation

   Apply $y(0)=7$ to $y=C{e}^{2x}+3$: $7=C{e}^0+3=C+3\Rightarrow C=4$. Particular solution: $y=4e^{2x}+3$. Verify: $y^{\prime }=8e^{2x}$ and $2(y-3)=2(4e^{2x})=8e^{2x}$ ✓.

   Key takeaway

   For ODEs with equilibrium solutions, the general solution includes a constant shift; apply the initial condition to the full $y = Ce^{kx} + $ equilibrium form.

8. Question 8 · Medium

   For the logistic model $dP/dt=0.2P(1-P/500)$ with $P(0)=50$, at what population is the growth rate $dP/dt$ maximized?

   • A
     $P=50$

     Why not A: Returns the initial population instead of the inflection point of the logistic curve.
   • B
     $P=500$

     Why not B: Returns the carrying capacity (where growth rate is $0$, not maximum).
   • C
     $P=250$Correct
   • D
     $P=100$

     Why not D: Divides $K/5=100$, confused with $r=0.2=1/5$.
   Explanation

   The logistic growth rate $f(P)=rP(1-P/K)$ is a downward parabola in $P$, maximized at $P=K/2=500/2=250$. At $P=250$: $dP/dt=0.2(250)(1-0.5)=0.2\cdot 250\cdot 0.5=25$.

   Key takeaway

   Logistic growth rate $rP(1-P/K)$ is maximized at $P = K/2$ (the inflection point), not at the carrying capacity.

9. Question 9 · Medium

   Solve the separable ODE $\frac{dy}{dx}=\frac{x}{y}$ with $y(0)=3$.

   • A
     $y=\sqrt{x^2+9}$Correct
   • B
     $y=\sqrt{x^2+3}$

     Why not B: Applies initial condition $y(0)=3$ to $y^2/2=x^2/2+C$ as $3=C$ instead of $9/2=C$.
   • C
     $y=x+3$

     Why not C: Solves as a linear ODE rather than separating variables; incorrect approach for $dy/dx=x/y$.
   • D
     $y^2=x^2+3$

     Why not D: Sets constant $C=3$ in $y^2=x^2+C$ using $y(0)=3$ as $C=3$ instead of $C=9$.
   Explanation

   Separate: $ydy=xdx$. Integrate: $\frac{y^2}{2}=\frac{x^2}{2}+C$. Apply $y(0)=3$: $\frac{9}{2}=0+C\Rightarrow C=\frac{9}{2}$. So $y^2=x^2+9\Rightarrow y=\sqrt{x^2+9}$ (positive root since $y(0)=3>0$).

   Key takeaway

   Apply the initial condition to the implicit equation $y^2/2 = x^2/2 + C$ before solving for $y$; use the initial $y$-value to determine the sign of the explicit solution.

10. Question 10 · Hard

    For the ODE $\frac{dy}{dx}=\frac{y^2}{x}$, find the general solution.

    • A
      $y=\frac{-1}{\ln |x|+C}$Correct
    • B
      $y=\frac{1}{\ln |x|+C}$

      Why not B: Misses the negative sign: $\int y^{-2}dy=-y^{-1}$; the left side gives $-1/y$, not $+1/y$.
    • C
      $y=C{e}^{\ln x}$

      Why not C: Treats the ODE as $y^{\prime }=y^2/x$ and writes a solution resembling the linear $y^{\prime }=y/x$ form; incorrect separation.
    • D
      $y^3=3\ln |x|+C$

      Why not D: Integrates $y^2$ as $y^3/3$ (treating it like $\int y^{2}dy$) instead of separating $dy/y^2=dx/x$.
    Explanation

    Separate: $\frac{dy}{y^2}=\frac{dx}{x}$. Integrate: $-\frac{1}{y}=\ln |x|+C$. Solve for $y$: $y=\frac{-1}{\ln |x|+C}$.

    Key takeaway

    Separable ODE with $dy/y^2$ on the left: integrate as $\int y^{-2}\,dy = -y^{-1}$; don't confuse this with $\int y^{-1}\,dy = \ln|y|$.

11. Question 11 · Hard

    The solution to a logistic equation passes through $(0,100)$ and approaches $L=800$ as $t\to \infty$. Using $P=\frac{L}{1+A{e}^{-rt}}$ with $r=0.5$, find $A$.

    • A
      $A=7$Correct
    • B
      $A=8$

      Why not B: Computes $L/P(0)=800/100=8$ without subtracting $1$.
    • C
      $A=700$

      Why not C: Computes $L-P(0)=700$ instead of $(L-P(0))/P(0)$.
    • D
      $A=0.125$

      Why not D: Inverts the correct ratio: computes $P(0)/L=100/800=1/8$ instead of $(L/P(0))-1$.
    Explanation

    At $t=0$: $P(0)=\frac{L}{1+A}=100$. Solve: $1+A=\frac{L}{P(0)}=\frac{800}{100}=8\Rightarrow A=7$.

    Key takeaway

    Logistic form $P = L/(1+Ae^{-rt})$: at $t=0$, $P_0 = L/(1+A)$, so $A = L/P_0 - 1$.

12. Question 12 · Hard

    Which of the following differential equations is NOT separable?

    • A
      $\frac{dy}{dx}=xy+3x$

      Why not A: This IS separable: $dy/dx=x(y+3)$, so $dy/(y+3)=xdx$.
    • B
      $\frac{dy}{dx}=\sin x\cos y$

      Why not B: This IS separable: $dy/\cos y=\sin xdx$ (i.e., $\sec ydy=\sin xdx$).
    • C
      $\frac{dy}{dx}=x+y$Correct
    • D
      $\frac{dy}{dx}=\frac{x^2}{y}$

      Why not D: This IS separable: $ydy=x^{2}dx$.
    Explanation

    $dy/dx=x+y$ cannot be written as $g(y)dy=h(x)dx$ because $x+y$ cannot be factored into a product $f(x)\cdot g(y)$. The sum $x+y$ mixes variables inseparably. This ODE is instead solved as a first-order linear equation.

    Key takeaway

    An ODE is separable if and only if $dy/dx$ can be written as a product $f(x)\cdot g(y)$; sums like $x+y$ are NOT separable.