If at a specific value of x in a non-linear ODE a term is cancelled so that it looks just like the linear ODE, should the result be the same?

Question

If at a specific value of x in a non-linear ODE a term is cancelled so that it looks just like the linear ODE that I am comparing to, should the result be the same (graphic at bottom)?

I am simulating the deflection of beams.I have a linear and a non linear equation and I used the MATLAB program to iteratively integrate and solve with some boundary conditions. The picture below is the non-linear ODE:

And below now is the linear ODE (dy/dx is small for a specific condition so it´s equal to zero). (dy is now named dv in this picture)

Now, due to physical conditions, in the middle of the x-axis dy/dx is 0 for both solutions. As seen below:

So my question is very simple: I can see in the middle of the x-axis that dy/dx=0 in both cases, but should they have the same value at this point, since the cancelation of dy/dx in the non-linear leads to the linear?

Thanks a lot!

Answer 1

The answer is not (in general). Because you solve the different bondary-value problems.

For example, if you don't have any boundary conditions, then $y + C$ is also a solution for every constant $C$. Therefore, you can find such a constant $C$ that the values of linear and nonlinear equations will be the same at the point where $\frac{dy_1}{dx} = \frac{dy_2}{dx} = 0$. But this trick doesn't hold for boundary value problems.