LinearForm(ODE, y(x)) outputs an equivalent linear ODE, if that is possible. If not, an error is produced.

$\mathrm{with}\left({\mathrm{Student}}_{\mathrm{ODEs}}\right)\:$

$\mathrm{ode1}≔\frac{t^2\left(z\left(t\right)\+1\right)}{z\left(t\right)}\+\left(t-1\right)\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)\=t^2\left(z\left(t\right)\+\frac{1}{z\left(t\right)}\right)$

$\mathrm{ode1}≔\frac{t^2\left(z\left(t\right)+1\right)}{z\left(t\right)}+\left(t-1\right)\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)=t^2\left(z\left(t\right)+\frac{1}{z\left(t\right)}\right)$

$\mathrm{LinearForm}\left(\mathrm{ode1}\,z\left(t\right)\right)$

$\frac{\ⅆ}{\ⅆt}z\left(t\right)=\frac{t^{2}z\left(t\right)}{t-1}-\frac{t^2}{t-1}$

$\mathrm{ode2}≔t^2\left(z\left(t\right)\+1\right)\+{z\left(t\right)}^2\left(t-1\right)\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)\=0$

$\mathrm{ode2}≔t^2\left(z\left(t\right)+1\right)+{z\left(t\right)}^2\left(t-1\right)\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)=0$

$\mathrm{LinearForm}\left(\mathrm{ode2}\,z\left(t\right)\right)$

Error, (in Student:-ODEs:-LinearForm) the ODE, t^2*(z(t)+1)+z(t)^2*(t-1)*diff(z(t),t) = 0, does not appear to be linearizable, at least without further simplification.

The Student[ODEs][LinearForm] command was introduced in Maple 2021.

For more information on Maple 2021 changes, see Updates in Maple 2021.