Determine all possible solutions to the subsequent IVP.
y' = y?
y(0) = 0
Solution: First, see that this differential equation does NOT satisfy the conditions of the theorem.
f(y) = y1/3
df/dy = 1/(3y2/3)
Hence, the function is continuous on any interval, although the derivative is not continuous at y = 0 and thus will not be continuous at any interval containing y= 0. So as to use the theorem both should be continuous on an interval that contains yo = 0 and it is problem for us as we do have yo = 0.
Here, let's actually work the problem. This differential equation is fairly simple to solve and is separable.
∫ (y-1/3)dy = ∫dt
3/2 y2/3 = t + c
Applying the initial condition provides c = 0 and therefore the solution is,
3/2 y2/3 = t
y2/3 = (2/3)t
y2 = ((2/3)t)3
y(t) = + ((2/3)t)3/2
Therefore we've got two possible solutions now, both of which satisfy the differential equation and the initial condition. Here is also a third solution to the Initial Value Problem. y(t) = 0 is satisfies the initial condition and is also a solution to the differential equation.
In this last illustration we had an extremely simple Initial Value Problem and it only violated one of the conditions of the theorem, even it had three diverse solutions. All the illustrations we've worked in the earlier sections satisfied the conditions of this theorem and had a particular unique solution to the Initial Value Problem. This illustration is a useful reminder of the information that, in the field of differential equations, things don't all the time behave nicely. It's simple to forget this as most of the problems which are worked in a differential equations class are nice and behave in a nice, predictable way.