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Question Number 183210 by Mastermind last updated on 23/Dec/22

(d^3 y/dx^3 )+4(d^2 y/dx^2 )+(dy/dx)−6y=0 M.m

$$\frac{\mathrm{d}^{\mathrm{3}} \mathrm{y}}{\mathrm{dx}^{\mathrm{3}} }+\mathrm{4}\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }+\frac{\mathrm{dy}}{\mathrm{dx}}−\mathrm{6y}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Answered by aleks041103 last updated on 23/Dec/22

y=e^(rx) ⇒(r^3 +4r^2 +r−6)y=0 ⇒r^3 +4r^2 +r−6=0 r^3 −r^2 +5r^2 −5r+6r−6=0 r^2 (r−1)+5r(r−1)+6(r−1)=0 ⇒(r−1)(r^2 +5r+6)=0 (r−1)(r^2 +2r+3r+6)=0 (r−1)(r(r+2)+3(r+2))=0 (r−1)(r+2)(r+3)=0 ⇒r=−3,−2,1 ⇒y(x)=Ae^(−3x) +Be^(−2x) +Ce^x

$${y}={e}^{{rx}} \Rightarrow\left({r}^{\mathrm{3}} +\mathrm{4}{r}^{\mathrm{2}} +{r}−\mathrm{6}\right){y}=\mathrm{0} \\ $$$$\Rightarrow{r}^{\mathrm{3}} +\mathrm{4}{r}^{\mathrm{2}} +{r}−\mathrm{6}=\mathrm{0} \\ $$$${r}^{\mathrm{3}} −{r}^{\mathrm{2}} +\mathrm{5}{r}^{\mathrm{2}} −\mathrm{5}{r}+\mathrm{6}{r}−\mathrm{6}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} \left({r}−\mathrm{1}\right)+\mathrm{5}{r}\left({r}−\mathrm{1}\right)+\mathrm{6}\left({r}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({r}−\mathrm{1}\right)\left({r}^{\mathrm{2}} +\mathrm{5}{r}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\left({r}−\mathrm{1}\right)\left({r}^{\mathrm{2}} +\mathrm{2}{r}+\mathrm{3}{r}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\left({r}−\mathrm{1}\right)\left({r}\left({r}+\mathrm{2}\right)+\mathrm{3}\left({r}+\mathrm{2}\right)\right)=\mathrm{0} \\ $$$$\left({r}−\mathrm{1}\right)\left({r}+\mathrm{2}\right)\left({r}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{r}=−\mathrm{3},−\mathrm{2},\mathrm{1} \\ $$$$\Rightarrow{y}\left({x}\right)={Ae}^{−\mathrm{3}{x}} +{Be}^{−\mathrm{2}{x}} +{Ce}^{{x}} \\ $$

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