I have the following definition of Linear Differential Equation:

A linear differential equation is any differential equation that can be written in the following form.


The important thing to note about linear differential equations is that there are no products of the function, $y(t)$, and its derivatives and neither the function or its derivatives occur to any power other than the first power.

But then I see this one:


And it's a non-linear differential equation because the function $y(t)$ is evaluated by sine which is a non-linear function (I think that's the reason, correct me if I'm wrong).

So, is this true: a differential equation is non-linear if there are (in the equation) products of the function $y(t)$ and its derivatives or, $y(t)$ or its derivatives are evaluated by a non-linear function.

  • $\begingroup$ There is a very precise idea what "linear" or as here "affine linear" means. Be it in finite dimensional vector spaces or in function spaces. Everything else is non-linear. You may be looking at it more from a syntactic perspective, given a finite computation formula, a computational graph/AST, decide on linearity. Then a valid idea is that no "elementary operation node" that is non-linear itself may be connected back to the $y$ via its input, and in multiplications only one input may be connected to $y$. Note that this may still result in false negatives, where the non-linearities cancel. $\endgroup$ – LutzL Apr 29 at 7:23
  • $\begingroup$ @LutzL Yeah, my question itself is confusing, that's my fault. To be honest, I only wanted to know if the reason for the above equation being non-linear was because $y^{(4)}(t)$ is evaluated by sine, which is a non-linear function. Is that right? $\endgroup$ – J. D. Gambín Apr 29 at 7:55
  • 1
    $\begingroup$ Yes. You could also formally test, a function $L$ is affine linear if $L(ax+(1-ay))=aL(x)+(1-a)L(y)$ for all scalars $a$ and vectors/functions $x,y$. $\endgroup$ – LutzL Apr 29 at 8:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.