Recall initial value problems for linear differential equations are those of the form
 (23.9) 
where p
Proof: Let P^{′}

and so, integrating both sides from t_{0} to t,

and so

which shows that if there is a solution to 23.9, then the above formula gives that solution. Thus there is at most one solution. Also, you see the above formula makes perfect sense on the whole interval. Since the steps are reversible, this shows y
It is not so simple for a nonlinear initial value problem of the form


This means there exists an interval, I such that t_{0} ∈ I ⊆
A much more general theorem will be proved later. Also, in the above, it suffices to say that f is continuous on the given rectangle and that for y,z ∈

for some K > 0. This is called a Lipschitz condition. For now, note that it is reasonable to believe the conclusion of this theorem. Start with the point
Example 23.8.3 Solve y^{′} = 1 + y^{2},y
This satisfies the conditions of Theorem 23.8.2. Therefore, there is a unique solution to the above initial value problem defined on some interval containing 0. However, in this case, we can solve the initial value problem and determine exactly what happens. The equation is separable.

and so arctan
Theorem 23.8.2 does not say that the local solution can never be extended beyond some small interval. Sometimes it can. It depends very much on the nonlinear equation. For example, the initial value problem

turns out to have a solution on ℝ. Here ε is a small positive number. You might think about why this is so. It is related to the fact that in this new equation, the extra term prevents y^{′} from becoming unbounded.
Also, you don’t know whether the interval of existence is symmetric about the point at which the initial condition is given.
Example 23.8.4 Solve y^{′} =
The equation is separable

Then integrating and using the initial condition,

Thus

which makes sense on
The next one looks a lot like the above, but has a solution on the whole real line.
Example 23.8.5 Consider y^{′} = 1 − y^{2}, y
You can verify that

is the solution and it makes sense for all t.
Hopefully, this has demonstrated that all sorts of things can happen when you are considering nonlinear equations. However, it gets even worse.
If you assume less on f in the above theorem, you sometimes can get existence but not uniqueness for the initial value problem. In the next example
The equation is separable so

Letting C = 0 from the initial condition, one solution is y =
Observation 23.8.7 What are the main differences between linear and nonlinear equations? Linear initial value problems have an interval of existence which is the same as the interval on which the functions in the equation are continuous. Nonlinear initial value problems sometimes don’t. Solutions to linear initial value problems are unique. This is not always true for nonlinear equations although if in the nonlinear equation, f and ∂f∕∂y are both continuous, then you at least get uniqueness as well as existence on some possibly small interval of undetermined length.