4. 44 Let $m$ be a real number, solve the equation for $x$: $\left|x^{2}-1\right|+\left|x^{2}-4\right|=m x$.

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[Solution] Since the two terms on the left side of the equation $\left|x^{2}-1\right|+\left|x^{2}-4\right|=m x$ cannot be zero simultaneously, the left side is always positive, so $m \neq 0$.
First, consider the case where $m>0$, in which case $x>0$.
If $0<x<1$, then by (1) we get $5-m x=0$.
If $1 \leq x \leq 2$, then by (1) we get $3-m x=0$.
If $x>2$, then by (1) we get $2 x^{2}-m x-5=0$.
Solving this, we get
$$\begin{array}{l}
x=\frac{m+\sqrt{m^{2}+40}}{4} \\
\text { thus } \frac{m+\sqrt{m^{2}+40}}{4}>2 \Longleftrightarrow m^{2}+40>(8-m)^{2} \text { or } m>8 \\
\Longleftrightarrow 5>8-2 m \text { or } m>8 \Longleftrightarrow m>\frac{3}{2} \text {. }
\end{array}$$

For the case where $m<0$, we have already discussed the case where $m>0$.

Therefore, the answer to this problem is:
When $m \geqslant 3$, $x=\frac{-m \pm \sqrt{m^{2}+40}}{4}$;
When $\frac{3}{2} \leqslant m<3$, $x=\frac{m-\sqrt{m^{2}+40}}{4}, \frac{3}{m}$;
When $m$ takes other values, equation (1) has no solution.