Before hunting for a solution it is worth knowing one exists. Numerical methods will happily produce output for a problem with no solution at all.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Existence
If f is continuous near the starting point, a solution exists at least on some small interval around it.
Uniqueness
If the partial derivative of f with respect to y is also continuous there, that solution is the only one.
The consequence
Where uniqueness holds, two different solution curves can never cross. A crossing point would have two futures.
When it fails
dy/dx = the square root of y at y = 0 has more than one solution. The partial derivative blows up exactly there.
Only local
The theorem promises a small interval, not all of time. A solution may still blow up in finite time.
Why existence needs proving
It is not obvious that a differential equation has any solution at all, or that it has only one. Numerical methods and qualitative arguments both assume it, so the theorem underwrites them.
The conditions
Continuity of f gives existence; continuity of its partial derivative with respect to y gives uniqueness. Both are local, so the guarantee applies near the initial point rather than everywhere.
Uniqueness can genuinely fail
dy/dx = y^(1/3) with y(0) = 0 has several solutions, because the partial derivative is undefined there. The hypothesis is not a technicality — dropping it breaks the conclusion.
Solutions can escape in finite time
A solution guaranteed to exist near the start may blow up to infinity at a finite value of x. The theorem promises a local solution, not a global one, and the distinction matters in modelling.
Step 2: Try It Yourself
Tap and try it out.
- The equationdy/dx = a·y
- Through(-2, 1)
Every short line shows the slope the equation demands at that point. The curve simply follows them, and the marked point is the initial condition that picks it out of the family.
Step 3: Watch an Example
One step at a time.
Watch Marcus Check a Problem
Marcus checks dy/dx = x + y with y = 2 at x = 0.
- Step 1
He examines f = x + y, which is a polynomial and therefore continuous everywhere.
Step 4: Your Turn
Practice makes it stick.
The Crossing
Problem 1 of 2
Two distinct solution curves cross at a point where uniqueness holds. How many such crossings are possible?
The Polynomial
Problem 2 of 2
f(x, y) = 3x + 2y. Is it continuous everywhere? 1 yes, 0 no.
Can You Trust It
1 of 8
f is continuous near the start. Does a solution exist? 1 yes, 0 no.
2 of 8
Continuity of f alone. Does it guarantee uniqueness? 1 yes, 0 no.
3 of 8
Which derivative must also be continuous for uniqueness? Enter 1 for the partial by y, 2 for the partial by x.
4 of 8
Under uniqueness, how many solutions pass through a given point?
5 of 8
f = x + y. Is the partial derivative with respect to y continuous? 1 yes, 0 no.
6 of 8
Does the theorem guarantee a solution for all time? 1 yes, 0 no.
7 of 8
Match each condition to what it guarantees.
Tap a card on the left to start.
8 of 8
A numerical method run on a problem with no solution. Does it still produce numbers? 1 yes, 0 no.
Step 5: Quick Check
Show what you know.
Question 1 of 2
Under uniqueness, how many solution curves pass through one point?
Question 2 of 2
Why can solution curves not cross where uniqueness holds?
What You Learned
- Continuity of f near the starting point guarantees a solution exists there.
- Continuity of the partial derivative with respect to y makes that solution unique.
- Where uniqueness holds, solution curves can never cross.