Linear substitution
About this app
ODEs of the form y′ = f(ax + by + c) in five steps: recognise u = ax + by + c, derive u′ = a + b·f(u), separate and integrate, substitute back, insert the initial value. Two cards x-y and x-u; turning the plane makes the lines u = constant horizontal. Examples y′ = 2x − y (asymptote y = 2x − 2), y′ = (x + y)², y′ = 1/(x − y) + 1, y′ = x + y + 1; games “Find u” and “Where does it go?”.
With explain mode: an AI tutor walks you through the app, points out what matters in the scene and answers your questions.
When linear substitution helps
A first-order differential equation of the form y′ = f(ax + by + c) depends on the right only through one linear term in x and y. Set u = ax + by + c. The chain rule gives u′ = a + b·y′ = a + b·f(u). This new equation contains only u; it is autonomous and therefore separable. After separating and integrating, solve u = ax + by + c for y and insert the initial value. Constant solutions u* with a + b·f(u*) = 0 are lost when dividing and must be noted separately.
What the app shows
Two cards sit side by side. On the left the x-y plane with the direction field of the ODE and the family of lines u = constant: along each of these lines all line elements have the same slope. On the right the x-u plane with the direction field of u′ = a + b·f(u), where the slope depends only on the height u. The five-step solution with numbers inserted runs alongside. For the main example y′ = 2x − y this gives u′ = 2 − u, the solution y = 2x − 2 + C·e⁻ˣ and the asymptote y = 2x − 2.
What you can try
Drag the initial point in either plane and follow how u₀ and the constant C change. “Turn the plane” maps the x-y plane continuously into the x-u plane, and the slanted lines become horizontal. Four examples show different behaviour: approaching an asymptote, a pole for y′ = (x + y)², an end on the line y = x, and an unstable line. In the game you pick the right u for an ODE or predict where the solution goes.
Frequently asked questions
- How do you solve y′ = 2x − y by substitution?
- Set u = 2x − y. Then u′ = 2 − y′ = 2 − u. Separating gives du/(2 − u) = dx, so u = 2 − C·e⁻ˣ. Back with y = 2x − u gives y = 2x − 2 + C·e⁻ˣ. With y(0) = 1, C = 3.
- Why does u′ = a + b·f(u) hold?
- u = ax + by + c depends on x through y(x). Differentiating with respect to x gives u′ = a + b·y′. For y′ insert the differential equation y′ = f(ax + by + c) = f(u). Writing just u′ = f(u) forgets the inner derivative.
- Which ODEs can be solved this way?
- All equations y′ = f(ax + by + c) with b ≠ 0 whose right-hand side depends on x and y only through one linear term, such as y′ = (x + y)², y′ = cos(x + 2y + 1) or y′ = e^(x + y). Whether the resulting separable equation can be integrated in closed form depends on f.
- What can I adjust in the app?
- You choose one of four examples, drag the initial point, step through the solution, turn the plane from x-y to x-u and show or hide the direction field, lines and family of solutions. A game and an explain mode are included.
Subject: Differential equations
More from Differential equations
- Homogeneous ODEs y′ = F(y/x)
- Initial value problem: dropping a ball
- ODE: the line-up
- Linear first-order ODEs
- Direction field y'(x) = -x/y
- Separation of variables
- Worlds of differential equations
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