ODE: the line-up
About this app
Differential equations as a detective story: the culprit is a function, and all we know is something about its derivative. Guessing and checking made visible: the real tangent against the slope the ODE demands. y′ = 5y with eˣ, e⁵ˣ, e⁵ˣ + C and C·e⁵ˣ (family, trajectory), and y′·y = −x: the tangent is perpendicular to the radius, so only circles solve it.
With explain mode: an AI tutor walks you through the app, points out what matters in the scene and answers your questions.
What a solution of a differential equation is
A differential equation does not reveal the unknown function itself, only something about its derivative, for example y′ = 5y: the slope is everywhere five times the function value. A solution is therefore not a number but a function. Whether a candidate fits is shown by the check: substitute it into the left and right side. If both sides agree for every x, it is a solution. For y′ = 5y, e⁵ˣ and in general C·e⁵ˣ for every real C pass, while eˣ and e⁵ˣ + C do not.
What the app shows
The app makes the check visible. At the point x₀ on the suspect’s curve there is the real tangent with slope y′(x₀) and, dashed, the slope the equation requires at that point. For a solution both coincide everywhere, otherwise there is a gap. The Check panel computes the left and right side with numbers and shows the verdict. In the second case y′·y = −x the position vector of the point shows that the tangent of a solution is perpendicular to the radius.
What you can try
Switch between the suspects eˣ, e⁵ˣ, e⁵ˣ + C and C·e⁵ˣ and watch the curve morph smoothly while the tangent glides along. Drag the point along the curve: for the real solution the gap stays closed everywhere. Set C, also negative or zero, and show the whole family of solutions. In the circle case you compare the line y = x with half circles √(r² − x²) and change the radius.
Frequently asked questions
- How do you check whether a function solves a differential equation?
- Differentiate the candidate and substitute the function and its derivative into both sides of the equation. If the left and right side agree for every x in the domain, it is a solution. For y = e⁵ˣ and y′ = 5y we get y′ = 5e⁵ˣ = 5y.
- Why is e⁵ˣ + C not a solution of y′ = 5y?
- The derivative of e⁵ˣ + C is 5e⁵ˣ; the C disappears. The right-hand side, however, is 5y = 5e⁵ˣ + 5C. Both sides agree only for C = 0. An additive constant fits for antiderivatives, not for this differential equation; here the constant belongs in front as a factor: C·e⁵ˣ.
- Why are the solutions of y′·y = −x circles?
- y′·y = −x can be written as x + y·y′ = 0, i.e. as the dot product (1, y′)·(x, y) = 0. So the tangent direction is perpendicular to the position vector of the point everywhere. That holds exactly for circles around the origin: y = ±√(r² − x²) for every r > 0.
- What can I adjust in the app?
- You choose one of the two cases and a suspect, drag the point x₀ along the curve, set the constant C or the radius r, and show the required slope and the whole family of solutions. An explain mode walks through both cases.
Subject: Differential equations
More from Differential equations
- Initial value problem: dropping a ball
- Direction field y'(x) = -x/y
- Separation of variables
- Worlds of differential equations
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