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C29 · Unit C7 Differential equations

Differential equations, verification and slope fields

CoreReview editionNot yet human-reviewed
Time
About 45 minutes, untimed
Course framework reference
AP 7.1–7.4
Builds on
Part of

Progress on this lesson counts in every pathway listed.

Understand the idea

A differential equation relates an unknown function to its derivatives. A slope field displays the prescribed derivative at each (x,y); a solution curve must follow those local directions. Verify a proposed solution by substituting its function and derivative into the equation. There may be infinitely many solutions; an initial condition can select one when suitable existence and uniqueness conditions hold.

Worked example

For y minutes equals x minus y, the slope at (1,2) is minus 1 and at (2,2) is 0. The proposed y equals x minus 1+Ce^( minus x) has derivative 1 minus Ce^( minus x), equal to x minus y. Setting y(0) equals 2 yields C equals 3.

Explore the mathematics

Differential equations, verification and slope fields

Follow local slopes with Euler steps and compare with an exact solution.

Setting 3 of 5

Scroll sideways to see the full diagram.

The slope equals the current height. Predict the first step.xy12y′=y, y(0)=1; Euler step h=0.3333At x=0: Euler y=1, exact eˣ=1

Predict before continuing

The slope equals the current height. Predict the first step.

Read the complete explanation
  1. The slope equals the current height. Predict the first step.
  2. Use the starting slope 1: new y = 1+h×1.
  3. Update the slope at each new point; a constant slope for the whole journey would be wrong.
  4. Euler uses tangent segments. Here the exact curve is convex, so forward Euler underestimates; smaller h improves this example.
Differential equations, verification and slope fields · visual worked method
Differential equations, verification and slope fields · visual worked methodFor y′=x−y, the slope at (1,2) is −1 and at (2,2) is 0. The proposed y=x−1+Ce^(−x) has derivative 1−Ce^(−x), equal to x−y. Setting y(0)=2 yields C=3.Worked example · 1/2For y′=x−y, the slope at (1,2) is−1 and at (2,2) is 0. The proposedy=x−1+Ce^(−x) has derivative1−Ce^(−x), equal to x−y.Worked example · 2/2Setting y(0)=2 yields C=3.

For y′=x−y, the slope at (1,2) is −1 and at (2,2) is 0. The proposed y=x−1+Ce^(−x) has derivative 1−Ce^(−x), equal to x−y. Setting y(0)=2 yields C=3.

Discuss: explain what each label represents, then draw a related example with different values.

A common misconception

A slope field shows slopes, not y-values. Parallel-looking segments at nearby points do not imply a globally linear solution.

Investigate and explain

Draw a 3 by 3 slope field for y minutes equals x minus y at x,y∈{ minus 1,0,1}. Sketch a solution through the origin.
Identify one step in the worked example, name the mathematical fact that permits it, and state its domain conditions.