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Question Number 125693 by fajri last updated on 13/Dec/20
$${find}\:: \\ $$$$ \\ $$$$\frac{{d}^{\mathrm{4}} {y}}{{dx}^{\mathrm{4}} }\:+\:\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:−\:\mathrm{7}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:−\:\frac{{dy}}{{dx}\:\:}\:+\:\mathrm{6}{y}\:=\:\mathrm{0} \\ $$$$ \\ $$$${for}\:{y}\left(\mathrm{0}\right)\:=\:\mathrm{1},\:{y}'\left(\mathrm{0}\right)\:=\:\mathrm{0},\:{y}''\left(\mathrm{0}\right)\:=\:−\mathrm{2}\:,\:{y}'''\left(\mathrm{0}\right)\:=\:−\mathrm{1} \\ $$
Answered by bobhans last updated on 13/Dec/20
$${characteristic}\:{eq}\: \\ $$$${r}^{\mathrm{4}} +{r}^{\mathrm{3}} −\mathrm{7}{r}^{\mathrm{2}} −{r}+\mathrm{6}=\mathrm{0} \\ $$$$\left({r}−\mathrm{1}\right)\left({r}^{\mathrm{3}} +\mathrm{2}{r}^{\mathrm{2}} −\mathrm{5}{r}−\mathrm{6}\right)=\mathrm{0} \\ $$$$\left({r}−\mathrm{1}\right)\left({r}+\mathrm{1}\right)\left({r}^{\mathrm{2}} +{r}−\mathrm{6}\right)=\mathrm{0} \\ $$$$\left({r}−\mathrm{1}\right)\left({r}+\mathrm{1}\right)\left({r}−\mathrm{2}\right)\left({r}+\mathrm{3}\right)=\mathrm{0} \\ $$$${General}\:{solution} \\ $$$${y}_{{h}} =\:{Ae}^{{x}} +{Be}^{−{x}} +{Ce}^{\mathrm{2}{x}} +{De}^{−\mathrm{3}{x}} \\ $$$${we}\:{can}\:{find}\:{A},{B},{C}\:{and}\:{D} \\ $$