Reduce the differential equation to first order

**Apprentice123** · November 9th 2010, 09:51 AM

How reduce the differential equation to first order ?

For example:
$\frac{d^5y}{dx^5}+c_1 \frac{d^4y}{dx^4}+c_2 \frac{d^3y}{dx^3}+c_3 \frac{d^2y}{dx^2}+c_4 \frac{dy}{dx} = f(x)$

and write the resulting system of equations in the form:
$\vec{z} = A \vec{z} + Bf(x)$

**Ackbeet** · November 9th 2010, 10:05 AM

Wrong forum, I think. Anyway. What ideas have you had so far?

**Apprentice123** · November 9th 2010, 10:54 AM

Originally Posted by Ackbeet

Wrong forum, I think. Anyway. What ideas have you had so far?

That's the problem. I do not know the formula to reduce to the first order

**Ackbeet** · November 9th 2010, 10:56 AM

See here for a very similar problem. Note that in your case, the function itself is not present.

**Apprentice123** · November 9th 2010, 12:35 PM

Originally Posted by Ackbeet

See here for a very similar problem. Note that in your case, the function itself is not present.

I will calculate at hand.
That link you now, I did not get any formula to reduce the differential equation

**Ackbeet** · November 9th 2010, 01:04 PM

It's not really a formula. It's a procedure. You let higher-order derivatives be components of the vector z. Here's another link.

**HallsofIvy** · November 10th 2010, 04:14 AM

$\vec{z}= \begin{pmatrix}y \\ \frac{dy}{dx} \\ \frac{d^2y}{d^2x} \\\frac{d^3y}{dx^3}\\ \frac{d^4y}{dx^4}\end{pmatrix}$

**Ackbeet** · November 10th 2010, 04:38 AM

You actually don't need the first component of y there, since there is no y in the original DE. Only the derivatives show up. So you can get by with a 4 x 4 system.

**HallsofIvy** · November 10th 2010, 04:52 AM

Right. If you write $u= \frac{dy}{dx}$ then the differential equation can be written
$\frac{d^4u}{dx^4}+c_1 \frac{d^3u}{dx^3}+c_2 \frac{d^2u}{dx^2}+c_3 \frac{du}{dx}+c_4 u = f(x)$
However, since that was given as a "for example", I suspect the y was inadvertantly left out.

**Apprentice123** · November 10th 2010, 02:41 PM

Originally Posted by HallsofIvy

Right. If you write $u= \frac{dy}{dx}$ then the differential equation can be written
$\frac{d^4u}{dx^4}+c_1 \frac{d^3u}{dx^3}+c_2 \frac{d^2u}{dx^2}+c_3 \frac{du}{dx}+c_4 u = f(x)$
However, since that was given as a "for example", I suspect the y was inadvertantly left out.

Sorry, but I do not understand. Are not in the order 4 ?

**Ackbeet** · November 11th 2010, 01:53 AM

Technically, your original DE is order 5. However, since you left out the y term (the one with no derivative), you can use the substitution HallsofIvy mentioned, and immediately get a 4th order DE in u. Then, once you've found u, integrate once more to get y.

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