why are bernoulli's equation non-linear

Question

So i was studying linear algebra and I learned about linear transformations and their definitions. In the applications of linear transformations i learnt that in linear differential equation the differential operator can be treated as a linear transformation from a vector space of differentiable functions in a given interval to itself T(y)=Q where 'T' is the differential operator & Q is a function of x. But a Bernoulli's equation which is of the form T(y)=Q*y^n is not considered a linear equation. Why is that? The transformation still seems linear to me. The only difference I see is that we are looking for a function which is being mapped to its own nth power. How does that makes the transformation non-linear

Answer 1

Indeed, if you write the Bernoulli equation $y'=Py+Qy^n$ as $$T[y]=Qy^n~~\text{ with }~~T[y]=y'-Py,$$ the left side $T[y]$ is linear. However, the full equation is non-linear, as the power on the right side is not linear.

If it were linear, then with any solution $y$, also any multiple $\alpha y$ would be a solution. However that would imply that $α^{n-1}=1$, which is not valid for almost all $α$.