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Find the general solution
$\left(D^{2}+4 D+8\right)^{2}\left(D^{4}+2 D^{3}-4 D^{2}-8 D\right) y=0$

put $f(m)=0$

$\left(m^{2}+4 m+8\right)^{2}\left(m^{4}+2 m^{3}-4 m^{2}-8 m\right) y=0$

$\left(m^{2}+4 m+8\right)^{2} m\left(m^{2}(m+2)-4(m+2)\right)=0$

$\left(m^{2}+4 m+8\right)^{2} m\left((m+2)\left(m^{2}-4\right)\right)=0$

$\left(m^{2}+4 m+8\right)^{2} m((m+2)(m+2)(m-2))=0$

$\left(m^{2}+4 m+8\right)^{2} m(m+2)^{2}(m-2)=0$

$m=\frac{-4 \pm \sqrt{16-4(1)(8)}}{2(1)}=\frac{-4 \pm 4 i}{2}=-2 \pm 2 i$ repeated complex

$-2 \pm 2 i,-2 \pm 2 i,-2,-2,2,0$ (m valnes)

$y=c_{1} e^{-2 x} \cos (2 x)+c_{2} e^{-2 x} \sin (2 x)+x\left[c_{3} e^{-2 x} \cos (2 x)\right.$

$\left.+c_{4} e^{-2 x} \sin (2 x)\right]+c_{5} e^{0}+{\left(c_{6}+c_{7} x\right) e^{-2 x}}+c_{8} e^{2 x}$

where $c_{1} \longrightarrow c_ 8$ are Constants

Solve the initial value problem $y^{\prime \prime}-y^{\prime}-2 y=0, y(0)=\alpha, y^{\prime}(0)=2$ then find $\alpha$ so that $\lim _{t \rightarrow \infty} y=0$

$f(r)=r^{2}-r-2=0 \rightarrow(r-2)(r+1)=0$

$r=2, r=-1 \rightarrow y=c_{1}e^{-t}+c_{2} e^{2 t}$

$y(0)=\alpha \rightarrow c_{1}+c_{2}=\alpha \cdots-(1)$

$y^{\prime}(0)=2 \rightarrow y^{\prime}=-c_{1} e^{-t}+2 c_{2} e^{2t} \rightarrow y^{\prime}=-c_{1}+2 c_{2}=2 \cdots (2)$

$c_{1}+c_{2}-c_{1}+2 c_{2}=\alpha+2$

$\rightarrow 3 c_{2}=\alpha+2 \quad \div 3$

$c_{2}=\frac{\alpha}{3}+\frac{2}{3}$ in $(1)$

$\rightarrow c_{1}+\frac{\alpha}{3}+\frac{2}{3}=\alpha$

$\rightarrow c_{1}=\frac{2}{3} \alpha-\frac{2}{3}$

$y=c_{1} e^{-t}+c_{2} e^{2t} \quad c_{1}=\frac{2}{3} \alpha-\frac{2}{3} \quad c_{2}=\frac{\alpha}{3}+\frac{2}{3}$

$y=\left(\frac{2}{3} \alpha-\frac{2}{3}\right) e^{-t}+\left(\frac{\alpha}{3}+\frac{2}{3}\right) e^{2 t}$

$\lim _{t \rightarrow \infty} y=0$ then

$\lim _{t \rightarrow \infty}\left[\left(\frac{2}{3} \alpha-\frac{2}{3}\right) e^{-t}+\left(\frac{\alpha}{3}+\frac{2}{3}\right) e^{2 t}\right]=0$

$e^{-\infty}=0$

$e^{2 t}=\infty$

$\frac{\alpha}{3}+\frac{2}{3}=0$

$\frac{\alpha}{3}=-\frac{2}{3}$

$\alpha=-2$