Lecture Notes in Differential Equations - Bruce E. Shapiro

6 LESSON 1. BASIC CONCEPTS
Example 1.7. The solution of the IVP given by example 1.6 is given by
equation 1.12, which you should verify. In fact, this solution is unique,
in the sense that it is the only function that satisfies both the differential
equation and the initial value problem.
Example 1.8. Solve the initial value problem dy/dt = t/y , y(1) = 2.
We can rewrite the differential equation as
ydy = tdt (1.30)
and then integrate,
∫ ∫
ydy = tdt (1.31)
1
2 y2 = 1 y t2 + C (1.32)
When we substitute the initial condition (that y = 2 when t = 1) into the
general solution, we obtain
Hence C = 3/2.
through by 2,
Taking square roots,
1
2 (2)2 = 1 2 (1)2 + C (1.33)
Substituting back into equation 1.32 and multiplying
y 2 = t 2 + 3 (1.34)
y = √ t 2 + 3 (1.35)
which we call the solution of the initial value problem. The negative square
root is excluded because of the initial condition which forces y(1) = 2.
Not all initial value problems have solutions. However, there are a large
class of IVPs that do have solution. In particular, those equations for
which the right hand side is differentiable with respect to y and the partial
derivative is bounded. This is because of the following theorem which we
will accept without proof for now.
Theorem 1.5 (Fundamental Existence and Uniqueness Theorem).
Let f(t, y) be bounded, continuous and differentiable on some neighborhood
R of (t 0 , y 0 ), and suppose that ∂f/∂y is bounded on R. Then the initial
value problem 1.29 has a unique solution on some open neighborhood of
(t 0 , y 0 ).