# Uniqueness and Existence Theorem

## Homework Statement

for the differential equation

t^2y''-2ty'+2y=0 with the general solutions y=C(t) + D(t^2) where C and D are constants. given the inital solution y(0)=1 and y'(0)=1 there are no solutions that exist. Why does this not contradict the Existence and Uniqueness Theorem?

## The Attempt at a Solution

This theorem says that all linear, homogeneous equations have a solution. Since t^2, -2t and 2 are all continuous I don't understand why there is no solution?

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HallsofIvy
Homework Helper
Well, there's your problem: the "existence and uniqueness theorem" says nothing of the sort!

It says "If there exist a neighborhood of $(t_0, y_0)$ such that f(t,y) is continuous in the neighborhood and f is "Lipschitz in y" in the neighborhood (that, essentially means that $|f(t, y_1- f(t,y_2)|\le C|y_1- y_2$ but that is implied by f being differentiable with respect to y so many elementary texts use that stronger condition) then dy/dt= f(t,y), with condition $y(t_0)= y_0$, has a unique solution in some neighborhood of $(t_0, y_0)$

That's stated for first order equations but if we let x= y', we can write this equation as
$$t^2\frac{dx}{dt}- 2tx+ 2y= 0$$
or
$$\frac{dx}{dt}= \frac{2}{t}x- \frac{2}{t^2}y$$
and together with x'= y, we can write that as the first order matrix equation:
$$\frac{\begin{bmatrix}x \\ y\end{bmatrix}}{dt}= \begin{bmatrix}\frac{2}{t}x- \frac{2}{t^2}y\\ x\end{bmatrix}$$

Now, the conditios of "continuous" and "Lipgarbagez in y" (or differentiable with respect to y) must apply to both components of the matrix on the right. For linear equations, the requirement that f be "Lipschitz in y" (or "differentiable with respect to y") is automatic but you can see that we have a problem with continuity at t= 0!