d^2y/dx^2=1-dy/dx
am i allowed to rearrange this to d^2y/dx^2-dy/dx=1 and say this is a linear ODE
Using the notation
y"=d²y/dx², and
y'=dy/dx,
you can write
y"=1-y'
as
y"+y'=1, or
y"+y'-1=0
It is a linear ODE because the terms of the dependent variable, y, y', y", ... are linear in y, i.e. no terms contains (y")², or y³, etc.