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

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

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

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

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

by adding (1) and (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

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

$$e^{-\infty}=0$$

$$e^{2 t}=\infty$$

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

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

$$\alpha=-2$$

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