Answer
$r=\dfrac{1}{t}+2t-2$
Work Step by Step
The anti-derivative for $\dfrac{d^2r}{dt^2}=\dfrac{2}{t^3}$ is
$\dfrac{dr}{dt}=\dfrac{-1}{t^2}+c$; $r=\dfrac{1}{t}+ct+c'$
Now, apply Initial conditions $r'(1)=1 ; r(1)=1$
This implies , $-1+c=1$ or, $c=2 $ and $1+c+c'=1$ and $c'=-2$
Hence, $r=\dfrac{1}{t}+2t-2$
