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Suppose a is a nonzero constant. Solve the initial value problem
$$y_{+2 a}^{\prime \prime} y^{\prime}+\left(a^{2}+1\right) y=0, y(0)=1, y^{\prime}(0)=0$$

$$f(m)=m^{2}+2 a m+a^{2}+1=0$$

$$=m^{2}+2 a m+a^{2}=-1$$

$$(m+a)(m+a)=(m+a)^{2}=-1$$

$$=\sqrt{(m+a)^{2}}=\sqrt{-1}$$

$$m=-a \pm i$$

$$y=e^{-a x}\left[c_{1} \cos x+c_{2} \sin x\right]$$

$$y(0)=1 \rightarrow c_{1}+0=1 \rightarrow\left[c_{1}=1\right]$$

$$y^{\prime}=-a e^{-a x}\left[c_{1} \cos x+c_{2} \sin x\right]+\left[-c_{1} \sin x+c_{2} \cos x\right] e^{-a x}$$

$$y^{\prime}(0)=0 \rightarrow-a\left[c_{1}+0\right]+1\left[0+c_{2}\right]=0 \rightarrow\left[c_{2}=a\right)$$

$$y=e^{-a x}[\cos x+a\sin x]$$

Given that $$y_{1}(t)=t^{2}$$ and $$y_{2}(t)=t^{-1}$$ are two solutions of the DE $$t^{2} y^{\prime \prime}-2 y=0$$ for $$t>0$$
then show that $$y=c_{1} t^{2}+c_{2} t^{-1}$$ is a solution for this equation.

$$t^{2} y^{\prime \prime}-2 y=0$$

$$y=c_{1} t^{2}+c_{2} t^{-1}$$

$$y^{\prime}=2 c_{1} t-c_{2} t^{-2}$$

$$y^{\prime \prime}=2 c_{1}+2 c_{2} t^{-3}$$

$$t^{2}\left[2 c_{1}+2 c_{2} t^{-3}\right]-2\left[c_{1} t^{2}+c_{2} t^{-1}\right]$$

$$=2 c_{1} t^{2}+2 c_{2} t^{-1}-2 c_{1} t^{2}-2 c_{2} t^{-1}=0$$

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