y² = t² + 4 is a relation, and the function it gives depends on which sign of the square root the initial value selects. The lab solves the relation for y when it is linear or quadratic in y, or a reciprocal, prints the branch through the starting point, and states the interval of t on which that branch exists. One of the presets exists for all t, one on an open interval with both ends excluded, and one ends at a finite t where the solution blows up.
Core ideas
Three ideas, one for each thing the lab prints after the relation: the formula, the sign, and the interval.
y² = t² + 4 allows two values of y for each t. Each choice of a sign, continued across all the t where it makes sense, is a function, and both are solutions of y·y′ = t. The relation names the pair, and only an initial value names one of them.
A solution passes through (t₀, y₀). The positive root is above the axis and the negative root below it, so y₀ > 0 selects the positive root and y₀ < 0 the negative one. The branch you did not choose is still a solution, but not of this initial value problem.
The explicit formula is defined where the square root has a non-negative radicand and where a denominator is not zero. The solution is then the piece of that set that contains t₀. It may be all t, a bounded open interval, or a half-line that ends where the solution runs off to infinity.
The material
Solving for y, choosing the branch, reading off the interval, and three presets that differ in what the interval turns out to be.
Take the relation y² = t² + 4. As an equation in y it is a quadratic with no linear term, so y = ±√(t² + 4). The lab handles a relation that is linear in y, or quadratic in y with coefficients that are polynomials in t, by the quadratic formula, and the reciprocal form −1/y = H(t) + C by inverting. Anything else it reports as implicit only, and says so.
The radicand t² + 4 is positive for every t, so both branches exist everywhere and never reach y = 0. That matters: the branch through (0, 2) never gets to the axis, and no solution of y′ = t/y can cross it, because at y = 0 the equation gives no slope to follow. The lab's first preset prints y = −√(t² + 4) and all t.
Now change the right side to −t: y·y′ = −t with y(0) = 1. Then y²/2 = −t²/2 + C, and C = 1/2, so y² = 1 − t². The branch through (0, 1) is y = √(1 − t²), and it exists only where 1 − t² ≥ 0. The lab prints the domain as −1 < t < 1, with both ends excluded.
The ends are excluded because the solution is a function that satisfies the differential equation, and at t = ±1 the value is y = 0, where y′ = −t/y has no value. The solution reaches the axis there and cannot be continued as a solution past it; the formula survives at the ends, the equation does not.
For (1/y²)·y′ = 1 with y(0) = 1 the relation is −1/y = t − 1, and inverting gives y = 1/(1 − t). The formula is defined for every t except t = 1, but the solution through (0, 1) is only the branch on the side of t = 1 that contains t = 0, which is t < 1. As t approaches 1 from below, y grows without bound. This is the blow-up met in “Blow-Up and the Interval of Existence”, now with a formula attached.
y = 1/(1 − t) is also defined for t > 1, where it is negative. That piece is a solution of the equation, but it is not a continuation of the one through (0, 1): the two are separated by the point where the formula has no value. A solution is a function on one interval, and the interval is the piece that holds the initial t.
The lab prints three tiles for this: the relation, the explicit solution and the domain. When a relation cannot be solved for y the explicit tile reads implicit only, and the domain is not given. That is the honest answer for cubics and beyond: the relation is the solution, and it can still be differentiated and checked.
Interactive
Both sides integrated exactly
Method
Four moves after the relation is written. The third and fourth are the ones that are usually skipped.
Linear in y: one function. Quadratic in y: the quadratic formula gives two. If neither, say implicit only, and stop here.
Compare the sign of y₀ with the signs of the two roots at t₀. Exactly one matches. If y₀ is zero the two roots coincide, neither branch is differentiable there, and y′ = h(t)/g(y) has no slope to give; the lab reports implicit only.
A square root needs a non-negative radicand; a denominator must not be zero. Write the set of t where both hold.
A solution lives on one interval. State it with its ends open or closed as the equation dictates, and say what happens at an open end.
The same equation as before, with the starting value on the other side of the axis. The relation is unchanged, and the branch and the interval are what the initial value decides.
The positive branch is a perfectly good solution of the equation. It simply never passes through (0, −2), and since no solution of y′ = t/y can cross the axis, a start below it stays below it for every t.
Try the second preset next and compare the domain: the formula √(1 − t²) is defined on a closed interval, but the solution is defined only on the open one, because at each end the slope −t/y has no value. The domain tile prints −1 < t < 1.
You should be able to solve a quadratic relation for y, pick the branch by the sign of the initial value, find where the radicand or denominator stops the formula, and state the interval of existence with its ends correctly open.
Something to change about this lesson? Write it down, tick it off when it is handled, and export the list when you want to act on it. Kept in this browser only: nothing is sent anywhere, and nobody is notified.
No recommendations yet for this lesson.