Differential Equations | Free AP Calculus AB Course

<!DOCTYPE html>

<head>

<title>AP Calculus AB: Differential Equations | The School of Mathematics</title>

<style>

:root {

--blue: #4285f4;

--green: #34a853;

--red: #ea4335;

--gray-bg: #f8f9fa;

--border: #e0e0e0;

}

* { box-sizing: border-box; }

body {

font-family: -apple-system, BlinkMacSystemFont, "Segoe UI", Roboto, Helvetica, Arial, sans-serif;

background: #ffffff;

color: #202124;

line-height: 1.65;

margin: 0;

padding: 0;

}

.wrap { max-width: 820px; margin: 0 auto; padding: 32px 20px 80px; }

h1 { font-size: 2rem; line-height: 1.25; margin-bottom: 10px; }

h2 {

font-size: 1.5rem;

margin-top: 48px;

padding-top: 8px;

border-top: 2px solid var(--border);

}

h3 { font-size: 1.15rem; margin-bottom: 6px; }

.intro { font-size: 1.05rem; color: #3c4043; }

.cta-group {

display: flex;

gap: 12px;

flex-wrap: wrap;

margin: 22px 0 32px;

}

.cta-btn {

flex: 1 1 220px;

text-align: center;

text-decoration: none;

font-weight: 700;

font-size: 1rem;

padding: 15px 18px;

border-radius: 10px;

}

.cta-primary { background: var(--blue); color: #fff; }

.cta-primary:hover { background: #3367d6; }

.cta-secondary { background: #fff; color: var(--blue); border: 2px solid var(--blue); }

.cta-secondary:hover { background: #eef3fd; }

nav.toc {

background: var(--gray-bg);

border: 1px solid var(--border);

border-radius: 10px;

padding: 18px 22px;

margin-bottom: 36px;

}

nav.toc h2 { margin-top: 0; border-top: none; font-size: 1.1rem; }

nav.toc ol { margin: 0; padding-left: 22px; }

nav.toc a { color: var(--blue); text-decoration: none; }

nav.toc a:hover { text-decoration: underline; }

.note-box {

background: #fff8e1;

border-left: 4px solid #f9a825;

border-radius: 6px;

padding: 14px 18px;

margin: 18px 0;

font-size: 0.95rem;

}

.example {

background: var(--gray-bg);

border-left: 4px solid var(--blue);

border-radius: 6px;

padding: 16px 20px;

margin: 18px 0;

}

.example p { margin: 6px 0; }

.step-math { font-family: "Cambria Math", Georgia, serif; }

.frac {

display: inline-flex;

flex-direction: column;

vertical-align: middle;

text-align: center;

font-size: 0.85em;

line-height: 1;

margin: 0 2px;

position: relative;

top: 0.1em;

}

.frac .num { border-bottom: 1.3px solid #202124; padding: 0 4px 1px; }

.frac .den { padding: 1px 4px 0; }

.problem {

border: 1px solid var(--border);

border-radius: 10px;

padding: 18px 22px 20px;

margin: 18px 0;

}

.problem .prompt { font-weight: 600; margin-bottom: 12px; }

.options { display: flex; flex-direction: column; gap: 8px; margin: 0 0 4px; }

.option-btn {

text-align: left;

padding: 10px 14px;

border: 1px solid #ccc;

border-radius: 6px;

background: #fff;

cursor: pointer;

font-size: 0.98rem;

font-family: inherit;

}

.option-btn:hover:not(:disabled) { background: #f1f3f4; }

.option-btn.correct { background: #e6f4ea; border-color: var(--green); font-weight: 600; }

.option-btn.incorrect { background: #fce8e6; border-color: var(--red); }

.option-btn:disabled { cursor: default; }

.explanation {

margin-top: 0;

padding: 0 16px;

background: var(--gray-bg);

border-left: 4px solid var(--blue);

border-radius: 4px;

max-height: 0;

opacity: 0;

overflow: hidden;

transition: max-height 0.35s ease, opacity 0.35s ease, margin-top 0.35s ease, padding 0.35s ease;

}

.explanation.show {

max-height: 600px;

opacity: 1;

margin-top: 14px;

padding: 14px 16px;

}

.explanation p { margin: 6px 0; }

.explanation .label { font-weight: 700; }

.mistake-list li { margin-bottom: 10px; }

.faq-item { margin-bottom: 18px; }

.faq-item h3 { margin-bottom: 4px; }

footer.cta-final { margin-top: 56px; }

</style>

</head>

<body>

<h1>AP Calculus AB: Differential Equations</h1>

Differential equations bring the entire course together: derivatives describe how something changes, and solving the equation recovers the original function. This free, complete lesson covers general and particular solutions, slope fields, separable differential equations, exponential growth and decay, and Newton's Law of Cooling. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the full AP Calculus AB Question Bank linked below.

</p>

<a class="cta-btn cta-primary" href="https://theschoolofmathematics.com/quiz/differential-equations-quiz-1">Practice Differential Equations Free</a>

<a class="cta-btn cta-secondary" href="https://theschoolofmathematics.com/quiz/course/AP-Calculus-AB-QBank">Explore the Full AP Calculus AB Qbank</a>

</div>

<h2>What's covered in this lesson</h2>

<ol>

<li><a href="#solutions">General and Particular Solutions</a></li>

<li><a href="#slope-fields">Slope Fields</a></li>

<li><a href="#separable">Separable Differential Equations</a></li>

<li><a href="#growth-decay">Exponential Growth and Decay</a></li>

<li><a href="#cooling">Newton's Law of Cooling</a></li>

<li><a href="#mistakes">Common Mistakes to Avoid</a></li>

<li><a href="#faq">Frequently Asked Questions</a></li>

</ol>

</nav>

<h2 id="solutions">1. General and Particular Solutions</h2>

<p>A <strong>differential equation</strong> (d.e.) is any equation involving a derivative. A <strong>solution</strong> is any function that satisfies it. Since antidifferentiation always introduces an arbitrary constant, a d.e. has infinitely many solutions, one for each value of C, called the <strong>general solution</strong>. Applying a specific initial condition pins down a single <strong>particular solution</strong>.</p>

A particular solution must not only satisfy the differential equation and the initial condition, it must also be differentiable on an interval that contains the initial point. Vertical tangents or asymptotes can restrict the domain of the solution, even when the algebraic formula looks defined everywhere.

</div>

<p><strong>Worked Example:</strong> Given dy/dx = 6x² − 4x with general solution y = 2x³ − 2x² + C, find the particular solution satisfying y(1) = 5.</p>

</div>

<p class="prompt">1. Given dy/dx = 3x² + 2 with general solution y = x³ + 2x + C, find the particular solution satisfying y(0) = 4.</p>

</div>

<p><span class="label">Explanation:</span> y(0) = 0 + 0 + C = C = 4, so y = x³ + 2x + 4.</p>

</div>

<p class="prompt">2. A particular solution to a differential equation must satisfy which conditions?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p1-2',true)">A) The differential equation, the initial condition, and differentiability on an interval containing the initial point</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-2',false)">B) Only the differential equation</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-2',false)">C) Only the initial condition</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-2',false)">D) Only that it is a polynomial</button>

</div>

<p><span class="label">Explanation:</span> All three conditions are required for a function to genuinely qualify as the particular solution.</p>

</div>

<p class="prompt">3. Verify that x² + y² = r² is a solution of the d.e. dy/dx = −x/y, using implicit differentiation.</p>

</div>

<p><span class="label">Explanation:</span> Differentiating both sides with respect to x: 2x + 2y(dy/dx) = 0, which rearranges to dy/dx = −x/y, matching the given d.e.</p>

</div>

<p class="prompt">4. Why might a particular solution have a restricted domain even though its algebraic formula looks defined everywhere?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p1-4',true)">A) The solution must be differentiable throughout an interval containing the initial point, excluding any points of non-differentiability</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-4',false)">B) All differential equations automatically restrict the domain to positive x</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-4',false)">C) The algebraic formula is always wrong</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-4',false)">D) Particular solutions never have restricted domains</button>

</div>

<p><span class="label">Explanation:</span> Even when a formula is algebraically defined at a point, that point may involve a vertical tangent or an undefined derivative, disqualifying it from the actual domain of the particular solution.</p>

</div>

<h2 id="slope-fields">2. Slope Fields</h2>

<p>A <strong>slope field</strong> represents a d.e. graphically: at many points, a short line segment is drawn with the slope given by the d.e. at that point. Following segments tangentially from an initial point traces out a solution curve.</p>

<p><strong>Worked Example:</strong> The d.e. dy/dx = y tells us the slope at any point equals its y-coordinate. What is the slope at any point where y = 4?</p>

<p>The slope equals 4 at every such point, regardless of x.</p>

</div>

<p class="prompt">1. For the d.e. dy/dx = x − y, what is the slope of the solution curve at the point (3, 1)?</p>

</div>

<p><span class="label">Explanation:</span> Slope = x − y = 3 − 1 = 2.</p>

</div>

<p class="prompt">2. A slope field shows horizontal segments (slope 0) along the entire x-axis. This is consistent with which d.e.?</p>

</div>

<p><span class="label">Explanation:</span> For dy/dx = y, the slope equals y itself, so along the entire x-axis (where y = 0), the slope is 0 for every x, producing horizontal segments there.</p>

</div>

<p class="prompt">3. Match the d.e. dy/dx = 2x to its general solution family.</p>

</div>

<p><span class="label">Explanation:</span> y = ∫2x dx = x² + C.</p>

</div>

<p class="prompt">4. If a slope field has vertical segments (undefined slope) along the line x = 0, which d.e. below is most consistent with this?</p>

</div>

<p><span class="label">Explanation:</span> dy/dx = 1/x is undefined at x = 0, matching a slope field with vertical (undefined-slope) behavior along that line.</p>

</div>

<p class="prompt">5. When sketching a solution curve on a slope field starting from a given initial point, what should you do?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p2-5',true)">A) Move from segment to segment, keeping the curve tangent to each segment</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-5',false)">B) Connect the initial point directly to the nearest axis with a straight line</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-5',false)">C) Ignore the segments and draw any smooth curve through the point</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-5',false)">D) Only plot the single point without connecting a curve</button>

</div>

<p><span class="label">Explanation:</span> The segments approximate the solution curve's tangent direction at each point, so tracing tangentially from segment to segment reconstructs the actual solution curve.</p>

</div>

<h2 id="separable">3. Separable Differential Equations</h2>

<p>A first-order d.e. is <strong>separable</strong> if it can be written as dy/dx = f(x)/g(y), with all y-terms and all x-terms isolated to opposite sides. Solve by separating, then integrating both sides independently.</p>

<p><strong>Worked Example:</strong> Solve dy/dx = −x/y given y(0) = 2.</p>

<p>Separate: y dy = −x dx. Integrate: y²/2 = −x²/2 + k, so x² + y² = C.</p>

<p class="step-math">Using y(0) = 2: 0 + 4 = C, so x² + y² = 4 (with y > 0)</p>

</div>

<p class="prompt">1. Find the general solution of du/dv = e<sup>v−u</sup>.</p>

</div>

<p><span class="label">Explanation:</span> Rewrite as e<sup>u</sup> du = e<sup>v</sup> dv. Integrating: e<sup>u</sup> = e<sup>v</sup> + C, so u = ln(e<sup>v</sup> + C).</p>

</div>

<p class="prompt">2. If ds/dt = √(st) and s = 1 when t = 0, separate the variables.</p>

</div>

<p><span class="label">Explanation:</span> Since √(st) = √s · √t, dividing both sides by √s gives ds/√s = √t dt.</p>

</div>

<p class="prompt">3. Using ds/√s = √t dt, find the general solution after integrating.</p>

</div>

<p><span class="label">Explanation:</span> ∫s<sup>−1/2</sup> ds = ∫t<sup>1/2</sup> dt gives 2s<sup>1/2</sup> = <span class="frac"><span class="num">2</span><span class="den">3</span></span>t<sup>3/2</sup> + C.</p>

</div>

<p class="prompt">4. Solve dy/dx = 6x²y, given y(0) = 5.</p>

</div>

<p><span class="label">Explanation:</span> dy/y = 6x² dx gives ln|y| = 2x³ + C, so y = Ce<sup>2x³</sup>. Using y(0) = 5: C = 5.</p>

</div>

<p class="prompt">5. A first-order differential equation is called "separable" if it can be written in which form?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p3-5',true)">A) dy/dx = f(x)/g(y), with x-terms and y-terms separable to each side</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p3-5',false)">D) Any equation involving x, y, and dy/dx</button>

</div>

<p><span class="label">Explanation:</span> This is the defining form of a separable d.e., all y-dependence isolated on one side, all x-dependence on the other.</p>

</div>

<h2 id="growth-decay">4. Exponential Growth and Decay</h2>

<p>If a quantity y changes at a rate proportional to the amount present, dy/dt = ky, then y = ce<sup>kt</sup>, where c is the initial amount. If k > 0, this is exponential growth; if k < 0, exponential decay. The <strong>half-life</strong> is the time required for a decaying quantity to reduce by half.</p>

<p><strong>Worked Example:</strong> A population grows at a rate proportional to its size, at 3% per year. How long will it take to double?</p>

<p>dP/dt = 0.03P, so P = P₀e<sup>0.03t</sup>. Setting 2P₀ = P₀e<sup>0.03t</sup>:</p>

<p class="step-math">ln 2 = 0.03t, so t = <span class="frac"><span class="num">ln 2</span><span class="den">0.03</span></span> ≈ 23.1 years</p>

</div>

<p class="prompt">1. Bacteria in a culture increase at a rate proportional to the number present. If the number doubles in 5 hours, how many times the original amount will be present in 15 hours?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p4-1',true)">A) 8 times</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-1',false)">B) 6 times</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-1',false)">C) 4 times</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-1',false)">D) 16 times</button>

</div>

<p><span class="label">Explanation:</span> Doubling every 5 hours means 15 hours is exactly 3 doubling periods: 2³ = 8 times the original amount.</p>

</div>

<p class="prompt">2. A radioactive substance has a half-life of 20 years. What fraction of the original amount remains after 60 years?</p>

</div>

<p><span class="label">Explanation:</span> 60 years is exactly 3 half-lives, so the remaining fraction is (1/2)³ = 1/8.</p>

</div>

<p class="prompt">3. At a yearly rate of 4% compounded continuously, set up the equation to find how long it takes for an investment to double.</p>

</div>

<p><span class="label">Explanation:</span> A = Pe<sup>0.04t</sup>; setting A = 2P and dividing by P gives 2 = e<sup>0.04t</sup>.</p>

</div>

<p class="prompt">4. Using 2 = e<sup>0.04t</sup> from the previous problem, solve for t (to the nearest year).</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p4-4',true)">A) ≈ 17 years</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-4',false)">B) ≈ 25 years</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-4',false)">C) ≈ 4 years</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-4',false)">D) ≈ 50 years</button>

</div>

<p><span class="label">Explanation:</span> t = ln 2 / 0.04 ≈ 17.3, so about 17 years.</p>

</div>

<p class="prompt">5. If a quantity satisfies dy/dt = ky with k < 0, what does this tell us about the quantity's behavior over time?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p4-5',true)">A) The quantity is decreasing (decaying) exponentially</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-5',false)">B) The quantity is increasing exponentially</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-5',false)">C) The quantity remains constant</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-5',false)">D) The quantity oscillates</button>

</div>

<p><span class="label">Explanation:</span> A negative proportionality constant k means the quantity shrinks over time, exponential decay.</p>

</div>

<p class="prompt">6. A fossil contains 40% of its original carbon-14. Using Q = Q₀e<sup>−0.00012t</sup>, set up the equation to find its age t.</p>

</div>

<p><span class="label">Explanation:</span> 0.40Q₀ = Q₀e<sup>−0.00012t</sup>; dividing both sides by Q₀ gives 0.40 = e<sup>−0.00012t</sup>.</p>

</div>

<h2 id="cooling">5. Newton's Law of Cooling</h2>

<p>Newton's Law of Cooling states that a hot object cools at a rate proportional to the difference between its own temperature and that of its surroundings. This is a case of <strong>restricted growth</strong>: T(t) = A + ce<sup>−kt</sup>, where A is the (constant) ambient temperature the object approaches, and c is determined by the initial condition.</p>

<p><strong>Worked Example:</strong> A cup of coffee at 190°F is placed in a 70°F room. Using T(t) = 70 + ce<sup>−kt</sup>, find c.</p>

</div>

<p class="prompt">1. Continuing the coffee example, T(t) = 70 + 120e<sup>−kt</sup>. Using T(10) = 150, find e<sup>−10k</sup>.</p>

</div>

<p><span class="label">Explanation:</span> 150 = 70 + 120e<sup>−10k</sup>, so 120e<sup>−10k</sup> = 80, giving e<sup>−10k</sup> = 80/120 = 2/3.</p>

</div>

<p class="prompt">2. A cold drink at 40°F is placed in a 75°F room. Using Newton's Law of Warming, T(t) = 75 − ce<sup>−kt</sup>, find c using T(0) = 40.</p>

</div>

<p><span class="label">Explanation:</span> 40 = 75 − c, so c = 35.</p>

</div>

<p class="prompt">3. In the restricted growth model f'(t) = k[A − f(t)] with f(t) increasing, what does A represent as t → ∞?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p5-3',true)">A) The upper limit that f(t) approaches as t → ∞</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-3',false)">B) The initial value of f(t)</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-3',false)">C) The rate of growth</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-3',false)">D) The value of f(t) at t = 0 only</button>

</div>

<p><span class="label">Explanation:</span> As the exponential term decays to 0, f(t) approaches A, making A a horizontal asymptote, the limiting value.</p>

</div>

<p class="prompt">4. A falling object's velocity approaches a limiting value L due to air resistance, following V(t) = L(1 − e<sup>−kt</sup>). What is V(0)?</p>

</div>

<p><span class="label">Explanation:</span> V(0) = L(1 − e⁰) = L(1 − 1) = 0, meaning the object starts at rest, as expected.</p>

</div>

<p class="prompt">5. A leaking tire's pressure P(t) satisfies P(t) = O + ce<sup>−kt</sup>, where O is the outside pressure. What happens to P(t) as t → ∞?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p5-5',true)">A) P(t) approaches O, the outside pressure</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-5',false)">B) P(t) approaches 0</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-5',false)">C) P(t) increases without bound</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-5',false)">D) P(t) remains constant at its initial value</button>

</div>

<p><span class="label">Explanation:</span> As t → ∞, the exponential term vanishes, so the inside pressure equalizes with the outside pressure O.</p>

</div>

<h2 id="mistakes">Common Mistakes to Avoid</h2>

<li><strong>Forgetting the constant of integration when solving a separable d.e.</strong> Every general solution needs it, and it's exactly what an initial condition is used to pin down.</li>

<li><strong>Separating variables incorrectly.</strong> Every y-term (including dy) must end up on one side, and every x-term (including dx) on the other, double-check that no mixed terms remain before integrating.</li>

<li><strong>Confusing the sign of k in growth versus decay.</strong> A positive k means growth (y = ce<sup>kt</sup> increasing); a negative k means decay. Setting up the wrong sign flips the entire behavior of the model.</li>

<li><strong>Mixing up half-life calculations with the growth/decay constant k.</strong> Half-life and k are related but distinct, always set up the exponential equation explicitly rather than trying to shortcut with memorized half-life arithmetic alone.</li>

<li><strong>Forgetting that Newton's Law of Cooling approaches the ambient temperature, not zero.</strong> The limiting value in T(t) = A + ce<sup>−kt</sup> is A (the surrounding temperature), not 0, a very common point of confusion.</li>

</ul>

<h2 id="faq">Frequently Asked Questions</h2>

<h3>How do I know if a differential equation is separable?</h3>

<p>Check whether you can algebraically rearrange it so every y (including dy) is on one side and every x (including dx) is on the other. If a term genuinely mixes x and y in a way that can't be factored apart, the equation isn't separable by this method.</p>

</div>

<h3>What's the difference between a slope field and an actual solution curve?</h3>

<p>A slope field shows the direction (slope) a solution curve would have at many sample points, it's a picture of the differential equation itself. A solution curve is one specific function that is actually tangent to those slope segments everywhere along its path.</p>

</div>

<h3>Why do exponential growth and Newton's Law of Cooling use such similar formulas?</h3>

<p>Both come from the same family of first-order linear differential equations, just with different setups: unrestricted growth (dy/dt = ky) has no limiting value, while restricted growth (like cooling) has a rate proportional to the distance from a fixed target value, producing the extra additive constant A in the solution.</p>

</div>

<h3>Where can I practice more problems like these?</h3>

<p>The <a href="https://theschoolofmathematics.com/quiz/differential-equations-quiz-1">Differential Equations quizzes</a> in the AP Calculus AB Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested throughout the course.</p>

</div>

<a class="cta-btn cta-primary" href="https://theschoolofmathematics.com/quiz/differential-equations-quiz-1">Practice Differential Equations Free</a>

<a class="cta-btn cta-secondary" href="https://theschoolofmathematics.com/quiz/course/AP-Calculus-AB-QBank">Explore the Full AP Calculus AB Qbank</a>

</div>

</footer>

</div>

function checkAnswer(btn, problemId, isCorrect) {

var problem = document.getElementById(problemId);

var buttons = problem.querySelectorAll('.option-btn');

buttons.forEach(function(b) { b.disabled = true; });

if (isCorrect) {

btn.classList.add('correct');

} else {

btn.classList.add('incorrect');

var correctBtn = problem.querySelector('.option-btn[data-correct="true"]');

if (correctBtn) { correctBtn.classList.add('correct'); }

}

var exp = problem.querySelector('.explanation');

if (exp) { exp.classList.add('show'); }

}

</script>

</body>

</html>