The Question:
Consider the differential equation
$$x^2\frac{dy}{dx}+(2x-1)y = \frac{1}{1+x^2}e^{-1/x} \qquad y(a) = b$$
Define the rectangle
$$R = \{ (x,y) \in \Bbb R^2:|x-a|≤h \; \text{ and } \; |y-b|≤1 \}$$
Assuming that $h>0$ is sufficiently small, for which $a$ and $b$:
(i) there is a unique solution in $R$?
(ii) there are no solutions in $R$?
(iii) there are infinitely many solutions in $R$?
Picard's Theorem:
Suppose we have the differential equation
$$y'(x) = f(x,y) \qquad y(a)=b$$
and define the rectangle
$$R = \{ (x,y) \in \Bbb R^2:|x-a|≤h \; \text{ and } \; |y-b|≤k \}$$
Then there is a unique solution in $R$ if
$(P1)$ $f$ is continuous in $R$ (and hence bounded)
$(P2)$ For some upperbound $M$ of $f$ in $R$, we have $Mh≤k$
$(P3)$ $f$ satisfies a Lipschitz condition in $R$. That is,
$$\exists \text{ constant } L \; | \; |f(x,u)-f(x,v)|≤|u-v| \qquad \forall \; (x,u), (x,v) \in R$$
Question (i):
In this case, we have $k=1$ and
$$f(x,y) = \frac{1}{x^2} \bigg(\frac{1}{1+x^2}e^{-1/x} + (1-2x)y \bigg)$$
$(P1)$ $f$ is continuous as long as $x \neq 0$ in $R$. Since $h$ is as small as we like, this holds as long as $a \neq 0$ (and $h<|a|$).
$(P2)$ Start with any $h<|a|$ and obtain an upperbound $M$. As we make $h$ smaller and smaller, $M$ is still an upperbound (since the smaller interval is contained in the original one), so we can just make $h<1/M$ to satisfy $Mh≤1$.
$(P3)$ We compute the partial derivative of $f$ with respect to $y$:
$$\bigg| \frac{\partial f}{\partial y} \bigg| = \bigg|\frac{1-2x}{x^2} \bigg|$$
and we can always find $L$ that bounds this expression, so long as we fix a closed interval not containing $x=0$. Hence by the MVT, the Lipschitz condition is satisfied.
We conclude that the solution is unique, as long as $a \neq 0$.
The Problem:
The problem arises when I go on to do questions (ii) and (iii). I don't quite understand how you can get infinitely many solutions?
Have I done something wrong with part (i)?