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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.

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

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

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

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

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

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