Here's an example to get you started:
$$u^{(3)}(t)+t^3u''(t)+5u'(t)+\sin(t)u=e^{6t}$$
with initial values $u''(0)=1$, $u'(0)=2$, and $u(0)=3$
First, give new names to $u$ and its derivatives (stopping one short of the order of the ODE): $u=x_1$, $u'=x_2$, $u''=x_3$.
Substituting back into the DE (keeping in mind that $u^{(3)}(t)=x'_3(t)$) we get:
$$x'_3(t)+t^3x_3(t)+5x_2(t)+\sin(t)x_1(t)=e^{6t}$$
Thus we have the equivalent system:
$$\begin{array}{ccrrrr} x'_1 & = & & x_2 & & \\ x'_2 & = & & & x_3 & \\ x'_3 & = & -\sin(t)x_1 & -5x_2 & -t^3x_3 & +e^{6t} \end{array}$$
Also, $x_1(0)=3$, $x_2(0)=2$, and $x_3(0)=1$.
The main tool is still Picard; we just have to keep controlling the solutions all the way to $t=\infty$, not letting it escape via a vertical asymptote. For example, if $f:\mathbb R\to\mathbb R$ is globally Lipschitz, then the solution of $x'=f(x)$ is global, because the intervals we get from Picard's theorem have infinite total length. (Sketch: starting with $(t_0,x_0)$, we have safety intervals of size roughly $\sim |f(x_0)|^{-1}$, within which the solution will exist and will not increase by more than $1$. Repeat and use the fact that the harmonic series diverges.)
The parenthetical is so sketchy because it does not apply to your situation: your $f$ is not globally Lipschitz. The function $f(x)=-3x+x^2\log x$ looks like this:
If you started with $x_0$ such that $f(x_0)>0$, the solution would keep on increasing, and since $f$ grows faster than $x^p$ for some $1<p<2$, it would blow up in finite time.
But, you start with $3/2$, where $f$ is negative. So, the solution moves to the left on the $x$-axis. Now we should be worried about hitting $x=0$. But this will not happen. Indeed, extending $f$ by $0$ to the negative semiaxis, we have a Lipschitz function on $(-\infty,3/2]$. There is an equilibrium solution $x(t)\equiv 0$. By the uniqueness theorem, a nonzero solution cannot reach this equilibrium in finite time; if it did, uniqueness would fail going backward in time.
Thus, the solution never reaches $0$, staying positive and decreasing all the time.
Best Answer
Consider any ODE that satisfies the standard existence and uniqueness theorem in some rectangle $a < x < b$, $c < y < d$. Pick some $x_0 \in (a,b)$. For each $y_0 \in (c,d)$, the existence and uniqueness theorem says there is a unique solution satisfying the initial condition $y(x_0) = y_0$, defined on some interval containing $x_0$. There are infinitely many possible $y_0$, so that's infinitely many solutions right there.