Approximating solutions using Euler's method (BC only) — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: This chapter covers Euler’s method recursive formula, step size calculation, approximating solutions to first-order initial value problems, over/under-approximation identification, and step size and error behavior.

You should already know: First-order initial value problems, point-slope form of a line, derivative tests for concavity of functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Approximating solutions using Euler's method (BC only)?

Euler's method is a numerical iterative technique for approximating values of the unknown solution $y(x)$ to a first-order initial value problem (IVP) of the form $\frac{dy}{dx}=f(x,y)$, $y(x_0)=y_0$. Unlike analytical methods such as separation of variables that give an exact closed-form solution, Euler's method works even when an exact solution cannot be found easily, which is why it is tested exclusively on the AP Calculus BC exam. Per the AP Calculus Course and Exam Description (CED), this topic makes up approximately 2-3% of the total exam score. It typically appears in both multiple-choice (MCQ), often as a 1-point question asking for an approximation after 2-3 steps, and free-response (FRQ), usually paired with other differential equation topics like slope fields or population growth. The core intuition is that if we know the slope of the solution at a point, we can use a small linear step along that slope to approximate the solution at the next x-value, repeating this process until we reach the desired coordinate. It is sometimes called the Euler tangent line method, since each step relies on tangent line approximation.

2. The Euler's Method Recursive Formula

The core of Euler's method is breaking the interval from the initial x-value $x_0$ to the target x-value $x_t$ into $n$ equal steps, each of size $h=\frac{x_t-x_0}{n}$, where $h$ is called the step size. Starting from our known initial point $(x_0,y_0)$, the slope of the solution at this point is given directly by the differential equation: $\frac{dy}{dx}{|}_{(x_0,y_0)}=f(x_0,y_0)$. To get from $x_0$ to $x_1=x_0+h$, we approximate the solution curve between these two points with the tangent line at $(x_0,y_0)$. Using point-slope form: $y-y_0=f(x_0,y_0)(x-x_0)$. At $x=x_1=x_0+h$, this gives $y_1=y_0+h\cdot f(x_0,y_0)$, where $y_1$ is our approximation for $y(x_1)$. We repeat this process recursively for every subsequent step, leading to the general formulas: $$x_k=x_0+k\cdot h$$ $$y_k=y_{k-1}+h\cdot f(x_{k-1},y_{k-1})$$ This recursion continues until we reach the target x-value, which gives us our final approximation.

Worked Example

Given the initial value problem $\frac{dy}{dx}=2x-y$, with $y(0)=1$, use Euler's method with step size $h=0.5$ to approximate $y(1.0)$.

1. Identify parameters: Initial $x_0=0$, $y_0=1$, step size $h=0.5$, target $x=1.0$, so we need 2 total steps.
2. First step to $x_1=0+0.5=0.5$: Calculate the slope at $(x_0,y_0)$: $f(0,1)=2(0)-1=-1$. Then $y_1=1+(0.5)(-1)=0.5$, so $y(0.5)\approx 0.5$.
3. Second step to $x_2=0.5+0.5=1.0$ (our target): Calculate the slope at $(x_1,y_1)$: $f(0.5,0.5)=2(0.5)-0.5=0.5$.
4. Calculate the final approximation: $y_2=0.5+(0.5)(0.5)=0.75$, so $y(1.0)\approx 0.75$.

Exam tip: Always write down the number of steps you need before starting calculations. It is extremely common to do one fewer step than required, especially when step size is a fraction and target x is a whole number.

3. Identifying Over- and Under-Approximation

After calculating an approximation with Euler's method, AP exam questions frequently ask whether the approximation is greater than or less than the exact value of the solution. This depends entirely on the concavity of the solution over the interval of approximation, because each Euler step uses the tangent line at the start of the step to approximate the entire interval. If the solution is concave up ($y^{\prime \prime }>0$ everywhere on the interval), the tangent line will lie entirely below the solution curve, so the approximation will be too low (an under-approximation). If the solution is concave down ($y^{\prime \prime }<0$ everywhere on the interval), the tangent line will lie entirely above the solution curve, so the approximation will be too high (an over-approximation). To find the sign of $y^{\prime \prime }$, we differentiate the original differential equation implicitly: starting from $y^{\prime }=f(x,y)$, we get $y^{\prime \prime }=\frac{d}{dx}f(x,y)=f_x+f_y\cdot y^{\prime }$, then substitute $y^{\prime }=f(x,y)$ to get $y^{\prime \prime }$ in terms of $x$ and $y$, then check its sign over the entire interval.

Worked Example

For the IVP from the previous example ($\frac{dy}{dx}=2x-y$, $y(0)=1$), the approximation for $y(1.0)$ is 0.75. Is this an over- or under-approximation of the exact value?

1. Differentiate the differential equation to find $y^{\prime \prime }$: $y^{\prime }=2x-y⟹y^{\prime \prime }=2-y^{\prime }$.
2. Substitute $y^{\prime }=2x-y$ into the expression for $y^{\prime \prime }$: $y^{\prime \prime }=2-(2x-y)=2-2x+y$.
3. Check the sign of $y^{\prime \prime }$ over the interval $[0,1]$: $y(x)$ is positive between $x=0$ and $x=1$, so $2-2x+y$ is always positive: at $x=0$, $y^{\prime \prime }=2+1=3>0$; at $x=1$, $y^{\prime \prime }=2-2+y=y>0$.
4. Since $y^{\prime \prime }>0$ everywhere on $[0,1]$, the solution is concave up, so the approximation 0.75 is an under-approximation, and the exact value of $y(1.0)$ is greater than 0.75.

Exam tip: You can only conclude over/under approximation if $y^{\prime \prime }$ has constant sign over the entire interval. If concavity changes, you cannot make a general claim about the final approximation.

4. Step Size and Error Behavior

Euler's method is classified as a first-order numerical method, which means the total (global) error after n steps to reach the target x is approximately proportional to the step size $h$. Formally, this means $E=k\cdot h$ for some constant $k$ that depends on the specific IVP and target x-value, but does not depend on $h$. This relationship is a common point tested in multiple-choice questions: if you know the error for one step size, you can predict the error for any other step size. A key takeaway is that smaller step size always produces a more accurate approximation, because the tangent line approximation is more accurate over shorter intervals. On the AP exam, you will never need more than 3-4 steps for an Euler's method question, even with small step size, so computation is always manageable.

Worked Example

An Euler approximation of $y(2)$ with step size $h=0.4$ has a total error of 0.12. Assuming error follows the expected proportionality for Euler's method, what is the expected error when using step size $h=0.1$ to approximate $y(2)$?

1. Recall that for Euler's method, global error $E=k\cdot h$, where $k$ is a constant for this IVP and target.
2. Solve for $k$ using the known error: $0.12=k(0.4)⟹k=\frac{0.12}{0.4}=0.3$.
3. Calculate the expected error for $h=0.1$: $E=k\cdot h=0.3(0.1)=0.03$.
4. The expected error for step size 0.1 is 0.03.

Exam tip: Do not confuse Euler's first-order error with second-order error for higher-order methods (like Runge-Kutta) that are not tested on AP BC. Only Euler's linear proportionality to step size is tested.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Stopping one step early because you count the starting point as step 1. For example, approximating $y(1)$ with $h=0.5$ starting at $x_0=0$, stopping after 1 step at $x=0.5$. Why: Students forget the initial point is step 0, not step 1, so they miscount the number of steps needed. Correct move: Before starting any calculation, explicitly write $n=\frac{x_{\text{target}}-x_0}{h}$ to get the number of steps required, then cross off each step as you complete it.
• Wrong move: Using $f(x_k,y_k)$ instead of $f(x_{k-1},y_{k-1})$ when calculating $y_k$. Why: Students mix up the recursive order, since $y_k$ is unknown when calculating the step. Correct move: Always remember you use the slope at the start of the step (the known point you already have) to calculate the y-value at the end of the step.
• Wrong move: Claiming a concave up solution gives an over-approximation, or a concave down solution gives an under-approximation. Why: Students mix up the position of the tangent line relative to the solution curve. Correct move: Draw a 2-second sketch: concave up curves bend upward above their tangent lines, so approximation is low; concave down curves bend below their tangent lines, so approximation is high.
• Wrong move: Stating that cutting the step size in half cuts the error by one-quarter for Euler's method. Why: Students confuse Euler's first-order error with second-order error for untested higher-order methods. Correct move: Remember Euler's method is first-order: error scales linearly with step size, so halving $h$ halves the error.
• Wrong move: Calculating step size as $h=\frac{n}{x_{\text{target}}-x_0}$ instead of $h=\frac{x_{\text{target}}-x_0}{n}$. Why: Students mix up the formula when given a fixed number of steps $n$. Correct move: Step size is total distance to the target divided by number of steps, so the difference in x goes in the numerator.
• Wrong move: Leaving $y^{\prime }$ in the expression for $y^{\prime \prime }$ when checking concavity, so you cannot determine the sign of $y^{\prime \prime }$. Why: Students stop after implicit differentiation and forget to substitute the original differential equation. Correct move: After finding $y^{\prime \prime }=...+...y^{\prime }$, always substitute $y^{\prime }=f(x,y)$ to get $y^{\prime \prime }$ in terms of $x$ and $y$ before checking its sign.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Let $\frac{dy}{dx}=x+y$, with initial condition $y(0)=1$. Using Euler's method with two steps to approximate $y(1)$, what is the resulting approximation? A) $2.00$ B) $2.25$ C) $2.50$ D) $2.75$

Worked Solution: First, calculate step size: we go from $x_0=0$ to $x_t=1$ in 2 steps, so $h=\frac{1-0}{2}=0.5$. Next, compute the first step: starting from $(0,1)$, the slope is $f(0,1)=0+1=1$, so $y_1=1+0.5(1)=1.5$ at $x_1=0.5$. For the second step, the slope at $(0.5,1.5)$ is $f(0.5,1.5)=0.5+1.5=2$, so $y_2=1.5+0.5(2)=2.50$. The correct answer is C.

Question 2 (Free Response)

Consider the initial value problem $\frac{dy}{dx}=3-xy$, $y(1)=2$. (a) Use Euler's method with step size $h=0.2$ to approximate $y(1.6)$. (b) Is your approximation from part (a) an over-approximation or under-approximation of the exact value of $y(1.6)$? Justify your answer. (c) If we use step size $h=0.05$ instead of $h=0.2$, by what factor would we expect the total error to decrease? Explain your reasoning.

Worked Solution: (a) We need $n=\frac{1.6-1}{0.2}=3$ steps. Step 1: $x_0=1,y_0=2$, $f(1,2)=3-(1)(2)=1$, so $y_1=2+0.2(1)=2.2$ at $x_1=1.2$. Step 2: $f(1.2,2.2)=3-(1.2)(2.2)=0.36$, so $y_2=2.2+0.2(0.36)=2.272$ at $x_2=1.4$. Step 3: $f(1.4,2.272)=3-(1.4)(2.272)=-0.1808$, so $y_3=2.272+0.2(-0.1808)=2.23584$. The approximation is $y(1.6)\approx 2.236$. (b) Differentiate to get $y^{\prime \prime }=-y-x{y}^{\prime }=-y-x(3-xy)=-y-3x+x^{2}y$. Testing points on $[1,1.6]$ shows $y^{\prime \prime }<0$ everywhere, so the solution is concave down. For concave down solutions, tangent lines lie above the solution curve, so the approximation is an over-approximation. (c) Euler's method has global error proportional to step size. Original $h=0.2$, new $h=0.05=\frac{1}{4}(0.2)$, so error is $\frac{1}{4}$ of the original error. The error decreases by a factor of 4.

Question 3 (Application / Real-World Style)

A population of bacteria in a controlled experiment grows according to the differential equation $\frac{dP}{dt}=0.2P-0.001P^2$, where $P$ is the number of bacteria measured in thousands, and $t$ is time measured in hours. At time $t=0$, the population is $P=50$ thousand bacteria. Use Euler's method with step size $h=1$ hour to approximate the bacteria population at $t=3$ hours.

Worked Solution: Initial values: $t_0=0$, $P_0=50$, $h=1$, 3 steps to $t=3$. Step 1: $\frac{dP}{dt}{|}_{(0,50)}=0.2(50)-0.001(50{)}^2=7.5$, so $P_1=50+1(7.5)=57.5$ thousand at $t=1$. Step 2: $\frac{dP}{dt}{|}_{(1,57.5)}=0.2(57.5)-0.001(57.5{)}^2=8.19375$, so $P_2=57.5+8.19375=65.69375$ thousand at $t=2$. Step 3: $\frac{dP}{dt}{|}_{(2,65.69375)}\approx 8.823$, so $P_3=65.69375+8.823\approx 74.52$ thousand. After 3 hours, Euler's method approximates the bacteria population is approximately 74.5 thousand bacteria.

7. Quick Reference Cheatsheet

8. What's Next

Mastering Euler's method is a critical prerequisite for the remaining topics in Unit 7 (Differential Equations), specifically applied modeling with differential equations, where numerical approximations are often required when exact solutions are too complex to derive. This topic builds on your prior knowledge of tangent line approximations from Unit 2, extending that local linear approximation idea to an iterative process that can reach any desired x-value. It also introduces core numerical analysis concepts that are widely used in engineering, biology, and physics for solving real-world differential equations that lack closed-form exact solutions. Without mastering the recursive step and error behavior of Euler's method, you will lose easy points on both MCQ and FRQ questions that explicitly test this BC-only topic.

• Logistic differential equations
• Slope fields and solution curves
• Separation of variables