diff eq. transform the given equation into an equivalent system of first order differential equations y'' + (y')^2 + 25y = sin(x) diff eq. transform the given equation into an equivalent system of first order differential equations y'' + (y')^2 + 25y = sin(x) @Mathematics

anything?

i got nothing

feel free to throw up an airball

nothing i can be sure about, im just learning to solve these

y'' = -y'(x)^2 -25y(x) + sin(x) is all i got

\[\int y'' + (y')^2 + 25y\ dx= \int sin(x)dx\] \[y' + \frac{1}{3y'}(y')^3 + 25xy= -cos(x)\] that doesnt look like anything useful

kinda looks like the question again lol

where's JamesJ

its a mess is what it is. ive been reading these things for a week and only retain a small portion of it. james is around, he just answered a q of mine

yea i got a test on wednesday...i need to get crackin at understanding this

Diff eqns are honey for me; I'm surprised I didn't smell this one out 20 minutes ago.

well then great...i got 1 more after this one :)

So, let y1 = y and y2 = y'. Then y1' = y2 ---- (*) and y2' = y'' hence y'' + (y')^2 + 25y = sin(x) is equivalent to y2' = - y2^2 - 25y1 + sin x ---- (**) Therefore the system, combining (*) and (**) is ...

\[\left(\begin{matrix}y_1 \\ y_2\end{matrix}\right)' = \left(\begin{matrix}y_2 \\ -y_2^2 - 25y_1 + \sin x\end{matrix}\right)\]

It doesn't gain you a lot writing it in column vector form as I just have above. But most of the time when you're looking at systems they will be linear, and it is very useful to think of the system this way as a vector ODE, so I'm writing it that way here with one eye to that fact.

can i write the answer like u have written?

Sure. In fact, I recommend it, so you get use to this sort of notation

wanna do another?

i'll creat another post so i can medal u up again

Yes ... post it on the left that's ok.

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