Show an IVP admits two solutions over an interval Consider the initial value problem
$x(2x-3)y''+(6-6x)y'+6y = 0$,  $y(1) = 1$, $y'(1) = 3$
$y_1=x^3$, $y_2=x-1$ are given as solutions the differential equation
Prove the following: if $b$ is any number less than zero, then the interval $(b, 3/2)$ admits at least two solutions
My attempt: solving the IVP I get a single solution of $y = x^3$. Is moving the interval below zero meant to include $y = 0$ as a solution?
 A: Your intuition was correct. For any $b \lt 0$, the map $y$ defined by
$$y(x)=\begin{cases}
0 & x \le 0\\
x^3 & x \ge 0\end{cases}$$ is indeed a twice differentiable map that is solution to the IVP on $(b,3/2)$. $x \mapsto x^3$ is another solution.
This is however not contradicting Picard-Lindelöf theorem.
Answer to the additional question in the comment: find $Y_0,Y_1 \in \mathbb R$ such that the IVP with initial conditions $y(1)=Y_0$ and $y^\prime(1)=Y_1$ has several solutions on the interval $(0,b)$.
To do that, let $y(x) = a x^3 + b(x-1)$ be a solution defined around $1$. We must have $a = Y_0$ and
$$y^\prime(1) = 3a + b=Y_1.$$ Therefore $b=Y_1-3Y_0$ and
$$y(x) = Y_0 x^3 + (Y_1-3Y_0)(x-1).$$ Now, let see how we can have $x(2x-3)y^ {\prime\prime}+(6-6x)y^\prime+6y = 0$ at $x = \frac{3}{2}$. For this we need to have
$$-\frac{1}{2}y^\prime\left(\frac{3}{2}\right) + y\left(\frac{3}{2}\right)=0.$$
I let you follow on the computation. You'll find a linear equation in $Y_0,Y_1$. If this linear equation is satisfied by other values $\overline{Y_0},\overline{Y_1}$ and the other solution has the same value and first derivative than $y$ at $\frac{3}{2}$... bingo! You're able to build a second solution.
