Calculus Honors · AP-aligned study guide

6.1 · Differential equations, slope fields, and Euler’s method

Interpret a differential equation as local slope information, verify solutions, and approximate a solution through successive tangent steps.

AP 7.1–7.4; Euler method: BC 7.5 · Reading: Larson & Edwards §6.1

Learning objectives

Read and construct slope-field segments from a differential equation.
Verify a proposed solution by substitution.
Apply Euler’s method using the previous step’s slope.

A differential equation specifies a rate

An equation $\displaystyle y'=F(x,y)$ assigns a slope to each allowed point. A slope field places short line segments with those slopes; a solution curve follows the local directions. An initial condition selects a curve through a specified point.

For $\displaystyle y'=y$ , slopes depend on height alone: zero along $\displaystyle y=0$ , positive above, and negative below. For $\displaystyle y'=x-y$ , zero slopes occur on $\displaystyle y=x$ .

Verify rather than guess a solution

To verify $\displaystyle y=Ce^x$ solves $\displaystyle y'=y$ , differentiate the proposed formula and compare with the right side. Also check any initial condition separately.

A slope field suggests the shape but does not supply an exact formula. A proposed curve must satisfy the differential equation throughout its stated interval, not just at the initial point.

Euler’s method follows short tangent steps

Start at $\displaystyle (x_n,y_n)$ , evaluate the slope there, then use $\displaystyle y_{n+1}=y_n+hF(x_n,y_n)$ and $\displaystyle x_{n+1}=x_n+h$ . Recompute the slope at every new point. Euler’s method is a BC extension in the AP framework.

Smaller steps usually improve a numerical approximation for a smooth well-behaved solution, but they do not turn it into an exact value. Clearly label the step size and the requested endpoint.

$y_{n+1}=y_n+hF(x_n,y_n)$

Worked examples

Verify a formula

Check whether $\displaystyle y=3e^x$ solves $\displaystyle y'=y$ , $\displaystyle y(0)=3$ .

Differentiation gives $\displaystyle y'=3e^x=y$ .
At $\displaystyle 0$ , the function value is $\displaystyle 3$ .
Both the equation and initial condition hold.

Two Euler steps

Approximate $\displaystyle y(1)$ for $\displaystyle y'=y$ , $\displaystyle y(0)=1$ , using $\displaystyle h=1/2$ .

The first slope is $\displaystyle 1$ , so $\displaystyle y_1=1+(1/2)1=3/2$ .
The second slope is $\displaystyle 3/2$ , so $\displaystyle y_2=3/2+(1/2)(3/2)=9/4$ .
The approximation is $\displaystyle y(1)\approx9/4$ .

Common mistakes

Using the initial slope for every Euler step.
Treating a numerical approximation as an exact solution.
Checking an initial value but not the differential equation.

Try it yourself

No calculator required.

1. For $\displaystyle y'=x-y$ , identify zero, positive, and negative slopes.

Slope segments are horizontal on y = x, positive below that line, and negative above it. — Compare segment directions above and below the diagonal.

Show worked solution

Slopes are zero on $\displaystyle y=x$ , positive below that line where $\displaystyle x-y>0$ , and negative above it. A solution curve must follow these directions.

Slope segments are horizontal on y = x, positive below that line, and negative above it. The dashed diagonal has zero slopes; the blue example solution follows the field through (0, 0). — The dashed diagonal has zero slopes; the blue example solution follows the field through (0, 0).

BC extension

2. Use two Euler steps of $\displaystyle h=1/2$ for $\displaystyle y'=x+y$ , $\displaystyle y(0)=1$ , to estimate $\displaystyle y(1)$ .

Show worked solution

At $\displaystyle (0,1)$ the slope is $\displaystyle 1$ , so $\displaystyle y_1=3/2$ . At $\displaystyle (1/2,3/2)$ the slope is $\displaystyle 2$ , so $\displaystyle y_2=5/2$ . Thus $\displaystyle y(1)\approx5/2$ .

3. Verify $\displaystyle y=2e^{-x}$ as a solution of $\displaystyle y'=-y$ , $\displaystyle y(0)=2$ .

Show worked solution

The derivative is $\displaystyle -2e^{-x}=-y$ and the initial value is $\displaystyle 2$ , so both conditions are satisfied.

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