# why are bernoulli's equation non-linear

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

• For $T$ to be a linear transformation, it must be that $T(y+z)=T(y)+T(z)$ – J. W. Tanner Feb 3 at 18:23
• " The transformation still seems linear to me": can you explain your understanding of linear ? – Yves Daoust Feb 3 at 18:24
• T(aX+bY)=aT(X)+bT(Y) I understand the definition of linear transformation. What I am trying to understand is the difference between a linear dfifferential equation and Bernoulli's differential equation.In the first equation T(y)=Q we want to solve for y which after the transformation T is mapped to Q. In the second equation T(y)= Qy^n we are looking for y which after the transformation T is mapped to Qy^n.How does that makes the transformation T non-linear? Doesn't Q*y^n lie in the range space of T. I hope I made my doubt clear – Siddharth Prakash Feb 3 at 18:35

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 $$α$$.