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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

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  • $\begingroup$ For $T$ to be a linear transformation, it must be that $T(y+z)=T(y)+T(z)$ $\endgroup$ – J. W. Tanner Feb 3 at 18:23
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    $\begingroup$ " The transformation still seems linear to me": can you explain your understanding of linear ? $\endgroup$ – Yves Daoust Feb 3 at 18:24
  • $\begingroup$ 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 $\endgroup$ – Siddharth Prakash Feb 3 at 18:35
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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 $α$.

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