# Is this an equivalent definition of non-linear differential equation?: $yy^{(k)}$ or $f(y), f(y^{(k)})$ where $f$ is non-linear occur in the equation

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.

$$a_{n}(t)y^{(n)}(t)+a_{n-1}(t)y^{(n-1)}(t)+...+a_1(t)y'(t)+a_0y(t)=g(t)$$

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:

$$y'''+\sin{(x+y^{(4)})}=\sin{x}$$

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.

• 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. Apr 29, 2019 at 7:23
• @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? Apr 29, 2019 at 7:55
• 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 $x,y$. Apr 29, 2019 at 8:51