Solving a third order Euler-Cauchy ODE

If you set $2x+3=e^t$, then in $u(t)=y(x)$ you get $u(t)=y(\frac{e^t-3}2)$. Thus computing the derivatives gives $$ u'(t)=y'(x)\frac{e^t}2\\ u''(t)=y''(x)\frac{e^{2t}}4+y'(x)\frac{e^t}2\\ u'''(t)=y'''(x)\frac{e^{3t}}8+y''(x)\frac{3e^{2t}}4+y'(x)\frac{e^t}2 $$ This can also be solved for the derivatives of $y$ to get $$ y'(x)=2e^{-t}u(t)\\ y''(x)=4e^{-2t}(u''(t)-u'(t))\\ y'''(x)=8e^{-3t}(u'''(t)-3u''(t)+2u'(t)) $$ This means in your initial calculations you did not consider the inner derivative/linear coefficient $2$ in $e^t=2x+3$. You could have chosen to set $e^t=x+\frac32$, then the powers of $2$ originate in the polynomial coefficients.