Help w/ Lineariz-ing Differential Equation

Question

mrhoola · Answer

attachment

taylor0402 · Answer

frick man im doing this to and i dont understand it at all

mrhoola · Answer

Dang  LOL.

mrhoola · Answer

@AloneS

mrhoola · Answer

@zepdrix , I have come to some sort of conclusion.

mrhoola · Answer

to linearize D.E.  I have to find a small excursion to the point about 0 . So what I did was 
change in output  = slope  * change in input 

15x - 15(0) = d/dx (15x)| x=0  + (f(x)-f(x= 0))

mrhoola · Answer

$$\Delta 15x - 15(0) = 15*(3e^(-5x) - 3e^0)$$

mrhoola · Answer

now just solve for delta x ?? and then plug that in the D.E. ??

mrhoola · Answer

only example from book :

mrhoola · Answer

concept:

zepdrix · Answer

Hmm I'm not sure moose :(

@kainui

mrhoola · Answer

@zepdrix Thanks for taking the time to look at it .

irishboy123 · Answer

form Maclaurin $e^{z}=1+z+{\frac {z^{2}}{2!}}+{\frac {z^{3}}{3!}}+\cdots $

you get

$x''' + 10 x'' + 20 x' + 15 x = 3(1-5x)$

3rd order but now linear about x = 0

you can re-write it as 

$\delta x''' + 10 \delta x'' + 20 \delta x' + 15 \delta x = 3(1-5\delta x)$

to keep in with the book

kainui · Answer

Just to add on a bit where @Mrhoola got stuck, basically this is what you do, for small values we know this is approximately true; it's the definition of the derivative:

$$f'(x) \approx \frac{f(x+\delta x) - f(x)}{\delta x}$$
Now specifically they wanted the 'excursion' from x=0 so plugging that in we have:
$$f'(0)\approx \frac{f(\delta x)-f(0)}{\delta x}$$
Then we rearrange to solve for $f(\delta x)$,

$$f(\delta x) \approx f(0)+\delta x f'(0)$$

So this is just exactly the same as taking the first couple terms off the Taylor series but I think maybe this makes it a little clearer how to solve for the linear approximation from scratch.