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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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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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