I'm trying to do this problem:
Let the operators $L_1, L_2$ be defined by $$L_1u=au_x+bu_y+cu ~~~; L_2u=du_x+eu_y+f_u$$
where $a,b,c,d,e,f$ are constants with $ac-bd\neq 0$. Prove that
- The equations $L_1u=w_1, L_2u=w_2$ have a common solution $u$, if $L_1w_2= L_2w_1$
(Hint: By linear transformation reduce to the case $a = e = 1, h = d = 0.$)
- The general solution of $L_1L_2 u=0$ has the form $u=u_1+u_2$, where $L_1u_1=0, L_2u_2=0.$
In (1), given $u_1, u_2$ solutions of $Lu=w_1, Lu=w_2$, respectively, then starting with $L_1w_2=L_2w_1$ and replacing with $L_1u_1=w_1, L_2u_2=w_2$ i got
$$(ae+bd)({u_2}_x-{u_1}_x)+ad({u_2}_{xx}-{u_1}_{xx})+(af+cd)({u_2}_x-{u_1}_x)+(bf+ce)({u_2}_y-{u_1}_y)+be({u_2}_{yy}-{u_1}_{yy})+cf(u_2-u_1)=0$$
i don't know what else to do.
In part (2), i don't know where to start
I would need some hint.