Question Number 183211 by Mastermind last updated on 23/Dec/22
$$\mathrm{y}^{\left(\mathrm{iv}\right)} +\mathrm{16y}^{\left(\mathrm{iii}\right)} +\mathrm{9y}^{\left(\mathrm{ii}\right)} +\mathrm{256y}^{\left(\mathrm{i}\right)} +\mathrm{256y}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$
Answered by aleks041103 last updated on 23/Dec/22
$${y}={e}^{{rx}} \\ $$$$\Rightarrow{r}^{\mathrm{4}} +\mathrm{16}{r}^{\mathrm{3}} +\mathrm{9}{r}^{\mathrm{2}} +\mathrm{256}{r}+\mathrm{256}=\mathrm{0} \\ $$$${r}={r}_{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}} \\ $$$$\Rightarrow{y}\left({x}\right)={Ae}^{{r}_{\mathrm{1}} {x}} +{Be}^{{r}_{\mathrm{2}} {x}} +{Ce}^{{r}_{\mathrm{3}} {x}} +{De}^{{r}_{\mathrm{4}} {x}} \\ $$