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While studying differential equations I stumble upon the term planar ODEs. It is used without introduction in the lecture slides and neither the course literature nor the internet can give me a definition. Is it as simple as that planar ODE is just another term for a system of linear first order differential equations? If not, then what is a planar ODE and how do i visualize it?

The example given in the lecture slides is given as follows

$ x'(t) =f (t, x(t), y(t)), \\ y'(t) =g(t, x(t), y(t)), \\t ∈ T .$

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That the system is "planar" means that its phase space is 2-dimensional, hence can be viewed as a plane. In the case you wrote down in your post, imagine fixing some initial condition to the ODE and plotting $y(t)$ vs. $x(t)$, where you let $t$ range over an interval. You end up with a trajectory where each point on the curve is equal to $(x(t),y(t))$ for some value of $t$. These points may be plotted in the plane, hence the ODE can be called "planar".

More info and images here: https://en.wikipedia.org/wiki/Phase_space

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    So if i understand correctly, it is called planar since the range of all possible solutions lies on a plane? Can i conclude that does imply that a system of 3 first order linear ODEs is in fact *not* planar since the phase space would be 3-dimentional? – 2017-01-03
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    Well, it could be that the solutions of a system of 3 first order linear ODEs lie in a plane anyways (perhaps the solution could be e.g. ($x(t),y(t),1$), if maybe $z'(t)=0$ and the initial condition is $z(0)=1$ ?)…but yes, if the phase portrait would be 3-dimensional, the system is definitely not planar. – 2017-01-03
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    Ahh, I mean above that it could be that every solution of a 3-equation system, regardless of initial conditions, could happen to lie in the same plane (I realized my example in the comment doesn't work for that), but you get the point… – 2017-01-03