AP Calculus AB Differential Equations and Slope Fields Worksheet Solutions (PDF)

AP Calculus AB math worksheet solutions: Differential Equations and Slope Fields – Maths PDFCover of the book AP Calculus AB Math Worksheets with Answers (PDF)Download as a PDFThe AP Calculus AB worksheets book with answersPrintable PDFSee the book ›


This differential equations answers for AP Calculus AB covers differential equations and slope fields (Common Core AP CALC AB 7.1, AP CALC AB 7.2, AP CALC AB 7.3, AP CALC AB 7.4, AP CALC AB 7.5, AP CALC AB 7.6, AP CALC AB 7.7), with a free PDF to print.

This answer key gives a full worked solution for every problem in the differential equations worksheet. Compare each step, not only the final answer, because AP graders award credit for a correct setup and clear justification. Final answers are in bold. Redo any problem you missed before moving on to the test.

AP Calculus AB · Common Core State Standards: AP CALC AB 7.1, AP CALC AB 7.2, AP CALC AB 7.3, AP CALC AB 7.4, AP CALC AB 7.5, AP CALC AB 7.6, AP CALC AB 7.7

Differential equations answers: exercises 1 to 10

Exercise 1 – Verify a solution

  1. \(\frac{d}{dx}(x^{2}+C)=2x\), which equals the right side.
  2. \(\frac{d}{dx}(3e^{2x})=6e^{2x}\) and \(2y=2\cdot 3e^{2x}=6e^{2x}\). They match.

Final answer: \(\text{both functions solve their equations}\).

Exercise 2 – Verify a second-order solution

\(y^{\prime}=e^{x}+xe^{x}=(1+x)e^{x}\) and \(y^{\prime\prime}=(2+x)e^{x}\).

\(y^{\prime\prime}-2y^{\prime}+y=(2+x)e^{x}-2(1+x)e^{x}+xe^{x}=e^{x}(2+x-2-2x+x)=0\).

Final answer: \(\text{verified}\).

Exercise 3 – Reading a slope field

\(x-y\) at \((1,0)\): \(1\). At \((0,2)\): \(-2\). At \((2,2)\): \(0\).

Segments are horizontal where \(x-y=0\), that is along the line \(y=x\).

Final answer: \(1,\ -2,\ 0;\ \text{horizontal along }y=x\).

Exercise 4 – Matching an equation

At \(y=0\) the slope is \(0\) (horizontal segments). For \(y > 0\) the slope is positive and increases with \(y\); for \(y < 0\) it is negative.

The slope depends only on \(y\), so segments along a horizontal line are parallel and they change as you move up or down.

Final answer: \(\text{depends only on }y;\ \text{zero at }y=0\).

Exercise 5 – Euler, two steps

Step 1: slope at \((0,1)\) is \(1\), so \(y_{1}=1+0.1(1)=1.1\) at \(x=0.1\).

Step 2: slope at \((0.1,1.1)\) is \(1.2\), so \(y_{2}=1.1+0.1(1.2)=1.22\).

Final answer: \(y(0.2)\approx 1.22\).

Exercise 6 – Euler for y prime equals y

\(y_{1}=1+0.5(1)=1.5\), \(y_{2}=1.5+0.5(1.5)=2.25\).

The exact solution is \(y=e^{x}\), so \(y(1)=e\approx 2.718\). Euler underestimates because the solution is concave up.

Final answer: \(y(1)\approx 2.25\).

0.250.50.75111.522.53xy
Euler (orange) underestimates eˣ (blue).

Exercise 7 – Euler with a decreasing solution

\(y_{1}=2+0.25(0-2)=1.5\) at \(x=0.25\).

\(y_{2}=1.5+0.25(0.25-1.5)=1.5-0.3125=1.1875\).

Final answer: \(y(0.5)\approx 1.1875\).

Exercise 8 – Separable: exponential

Separate: \(\frac{dy}{y}=2x\,dx\). Integrate: \(\ln y=x^{2}+C\).

Exponentiate: \(y=e^{C}e^{x^{2}}=Ae^{x^{2}}\).

Final answer: \(y=Ae^{x^{2}}\).

Exercise 9 – Separable with a trigonometric factor

Separate: \(\frac{dy}{y}=\cos x\,dx\). Integrate: \(\ln|y|=\sin x+C\), so \(y=Ae^{\sin x}\).

\(y\left(\frac{\pi}{2}\right)=Ae=e\) gives \(A=1\).

Final answer: \(y=e^{\sin x}\).

Exercise 10 – Separable: y squared

Separate: \(\frac{dy}{y^{2}}=dx\). Integrate: \(-\frac{1}{y}=x+C\). \(y(0)=1\) gives \(C=-1\).

\(-\frac{1}{y}=x-1\), so \(y=\frac{1}{1-x}\), valid for \(x < 1\).

Final answer: \(y=\frac{1}{1-x},\ x < 1\).

-3-2-1112345xy(0, 1)
The solution blows up at x = 1.

Exercises 11 and up

Exercise 11 – Separable: x squared times y

\(\frac{dy}{y}=3x^{2}dx\), so \(\ln|y|=x^{3}+C\) and \(y=Ae^{x^{3}}\). \(y(0)=A=5\).

Final answer: \(y=5e^{x^{3}}\).

-2-1.5-1-0.50.511.551015202530xy(0, 5)
The solution y = 5e^(x³).

Exercise 12 – Separable: a clever substitution

\(e^{x-y}=\frac{e^{x}}{e^{y}}\), so \(e^{y}dy=e^{x}dx\) and \(e^{y}=e^{x}+C\).

\(y(0)=0\) gives \(1=1+C\), so \(C=0\), \(e^{y}=e^{x}\) and \(y=x\).

Final answer: \(y=x\).

Exercise 13 – Separable: y over x

\(\frac{dy}{y}=\frac{dx}{x}\), so \(\ln y=\ln x+C\) and \(y=Ax\). \(y(1)=A=3\).

Final answer: \(y=3x\).

Exercise 14 – Decay equation

\(y=Ce^{-2x}\) and \(y(0)=4\) gives \(C=4\).

\(y(1)=4e^{-2}\approx 0.541\).

Final answer: \(y=4e^{-2x};\ y(1)\approx 0.541\).

Exercise 15 – Population growth

\(P=P_{0}e^{kt}=500e^{0.03t}\).

\(P(10)=500e^{0.3}\approx 674.9\).

Final answer: \(P=500e^{0.03t};\ P(10)\approx 674.9\).

Exercise 16 – Doubling time

\(N=N_{0}e^{kt}\). Doubling at \(t=5\): \(e^{5k}=2\).

\(k=\frac{\ln 2}{5}\approx 0.1386\) per hour.

Final answer: \(k=\frac{\ln 2}{5}\approx 0.139\).

Exercise 17 – Half-life

\(e^{-10k}=\frac{1}{2}\) gives \(k=\frac{\ln 2}{10}\).

After 30 days: \(e^{-30k}=\left(e^{-10k}\right)^{3}=\frac{1}{8}\).

Final answer: \(\frac{1}{8}\text{ (12.5\%)}\).

Exercise 18 – Newton cooling

Let \(u=T-70\): \(\frac{du}{dt}=-0.1u\), so \(u=120e^{-0.1t}\) (since \(u(0)=120\)).

\(T(t)=70+120e^{-0.1t}\) and \(T(10)=70+\frac{120}{e}\approx 114.1\).

Final answer: \(T(t)=70+120e^{-0.1t};\ T(10)\approx 114.1\,^{\circ}\text{F}\).

Exercise 19 – Continuous interest

\(A=2000e^{0.05t}\), so \(A(10)=2000e^{0.5}\approx 3297.4\).

Final answer: \(\$3297.44\).

Exercise 20 – General and particular solution

Separate: \(\frac{dy}{y-1}=x\,dx\). Integrate: \(\ln|y-1|=\frac{x^{2}}{2}+C\), so \(y=1+Ae^{x^{2}/2}\).

\(y(0)=1+A=3\) gives \(A=2\).

Final answer: \(y=1+2e^{x^{2}/2}\).

Exercise 21 – General exponential model

Separate: \(\frac{dy}{y}=k\,dt\). Integrate: \(\ln|y|=kt+C\).

Exponentiate: \(y=\pm e^{C}e^{kt}=Ae^{kt}\). At \(t=0\), \(y(0)=A\).

Final answer: \(y=y_{0}e^{kt}\).

Key facts behind this differential equations answers

Key facts
  • Verify a solution by substituting y and y′ into the equation.
  • A slope field shows dy/dx at grid points; solution curves follow the segments.
  • Euler: y_new = y_old + h · (dy/dx at the old point), x_new = x_old + h.
  • dy/dt = ky has the general solution y = Ce^(kt); growth if k is positive, decay if k is negative.

How to use this differential equations answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

Download the differential equations answers as a PDF with the button below and keep it with your notes.

Download the differential equations answers as a PDF with the button below and keep it with your notes.

Download the differential equations answers as a PDF with the button below and keep it with your notes.

Download the differential equations answers as a PDF with the button below and keep it with your notes.

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