Take the change of variable \(\displaystyle{z}={\ln{{y}}}\). So you would end up with the following relations

\(\displaystyle{y}={e}^{{z}}\)

\(\displaystyle{y}'={z}'{e}^{{z}}\)

\(\displaystyle{y}{''}={e}^{{z}}{z}{''}+{\left({z}'\right)}^{{2}}{e}^{{z}}\)

Upon substitution (all \(\displaystyle{e}^{{z}}\)'s will cancel out), leaving you with a 2-nd order differential equation, which is easy to solve:

\(\displaystyle{z}{''}={x}\)

\(\displaystyle{y}={e}^{{z}}\)

\(\displaystyle{y}'={z}'{e}^{{z}}\)

\(\displaystyle{y}{''}={e}^{{z}}{z}{''}+{\left({z}'\right)}^{{2}}{e}^{{z}}\)

Upon substitution (all \(\displaystyle{e}^{{z}}\)'s will cancel out), leaving you with a 2-nd order differential equation, which is easy to solve:

\(\displaystyle{z}{''}={x}\)