# 1st order system of ODEs - don't understand intro

#### uint

The chapter I'm reading is actually about numerics but this introductory material is calculus I think:

http://i.imgur.com/HhFhhqF.png

I don't understand how/why they obtain (2).. Okay, it is possible to say y_1 = y',
y_2 = y'', etc., but why would you? Doesn't that result in a different problem?

And doesn't (4) disregard the other f_1, f_2, f_3, etc. functions?

I'm just overall confused. Hope someone can help me out.

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#### Country Boy

Math Team
I'm not sure I would say it "results in a different problem" but if it does, it is a problem with the same solution!

Perhaps it would help to look at a specific problem: y'''(x)= x^2+ 2y- xy'+ (y'')^2 (I just made that up and I have no intention of trying to solve it!)
If we let y1= y, y2= y', and y3= y'' then we can write y3'= x^2+ 2y1+ xy2+ y3^2. That way, we have the three equations y1'= y2. y2'= y3, and y3= x^3+ 2y1+ xy2+ y3^2. The advantage? We now have three first order equation rather than a single third order equation. Or we can write Y as the vector function $$\displaystyle Y= \begin{pmatrix}y1 \\ y2\\ y3\end{pmatrix}$$ and so write the equation as the single first order vector equation $$\displaystyle Y'= f(x, Y)$$ where $$\displaystyle f(x, Y)= x^2+ 2y1+ xy2+ y3^3$$. In that way we can easily extend the fundamental "existence and uniqueness" theorem for first order equations to higher order equations.

1 person

#### skipjack

Forum Staff
Is this from the eBook Advanced Engineering Mathematics (10th Edition) by Erwin Kreyszig?

#### uint

Yes, skipjack.

Country Boy:
Thank you for the reply. However it's still a bit unclear to me.

You started from (2) in the book, and went to (4). It has become clearer to me how these two have the same set of solutions.
But how does the book go from (1) to (2)? It says that (1) "includes initial problems for single mth-order ODEs". But I don't see it.

Finally, you say Y' = f(x, Y), but it seems like Y' becomes a scalar, as f(x,Y) is a scalar. I thought Y' = (y1', y2', y3').

#### Country Boy

Math Team
Yes, I did indicate that, although "backwards" - going from (2) to (1). Equation (1) is the "vector form" I indicated. y is the vector function having $y_1$, $y_2$, ..., $y_n$ as components.

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#### zylo

I'm not sure I would say it "results in a different problem" but if it does, it is a problem with the same solution!

Perhaps it would help to look at a specific problem: y'''(x)= x^2+ 2y- xy'+ (y'')^2 (I just made that up and I have no intention of trying to solve it!)
If we let y1= y, y2= y', and y3= y'' then we can write y3'= x^2+ 2y1+ xy2+ y3^2. That way, we have the three equations y1'= y2. y2'= y3, and y3= x^3+ 2y1+ xy2+ y3^2. The advantage? We now have three first order equation rather than a single third order equation. Or we can write Y as the vector function $$\displaystyle Y= \begin{pmatrix}y1 \\ y2\\ y3\end{pmatrix}$$ and so write the equation as the single first order vector equation $$\displaystyle Y'= f(x, Y)$$ where $$\displaystyle f(x, Y)= x^2+ 2y1+ xy2+ y3^3$$. In that way we can easily extend the fundamental "existence and uniqueness" theorem for first order equations to higher order equations.
Your answer is not in vector form because of the non-linear term y''^2. If you had y'', you could write the answer as

y'=Ay+g(x)

EDIT
Unless you meant:
y'=f(x,y) where
f1=y2
f2=y3
f3= f(x, Y)= x^2+ 2y1+ xy2+ y3^3

though I'm not sure what that gets you.

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#### uint

There's something I still don't understand.
We have a problem
y_1' = f_1(x, y_1,..., y_m)
y_2' = f_2(x, y_1,..., y_m)
...
y_m' = f_m(x, y_1,..., y_m)

And somehow it's valid to say that y_1' = y_2, etc? Doesn't that change the solution set of the problem? At least making it smaller.

Edit: Thinking more and reading examples, I realized: The problem comes in the form (2). We want it in the form (1) in order to use the numeric methods for first-order single ODEs (Euler, RK). Transforming (2) to (1), the result is (4).

I don't know why this was so hard for me to understand. Well, please tell me if I got this incorrectly.

Note to self: if you don't understand something, read on, don't stall...

(also, I seem to be unable to give "Thanks" to posts. Sorry about that)

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