# A simple DE...for some.

• Aug 31st 2006, 02:52 AM
a4swe
A simple DE...for some.
If $\displaystyle y=e^{ax}cos(bx)$ and $\displaystyle y'''+2y'+3y=0$ what is a and b?
That is the problem to solve and I don't even know what to do.
Can anyone help me out a little and give me a suggestion or two for the solution of this one?
• Aug 31st 2006, 03:05 AM
CaptainBlack
Quote:

Originally Posted by a4swe
If $\displaystyle y=e^{ax}$ and $\displaystyle y'''+2y'+3y=0$ what is a and b?
That is the problem to solve and I don't even know what to do.
Can anyone help me out a little and give me a suggestion or two for the solution of this one?

You don't appear to have a "b".

But never mind. What you do is calculate the derivatives from your given
form for y(x), and substitute them into the DE. That will give you an
algebraic equation (after dividing the common factor $\displaystyle e^{ax}$ which will occur
on the left hand side of the equation), this will allow you to determine what
a (and b?) have to be.

RonL
• Aug 31st 2006, 03:13 AM
a4swe
Excuse me, that should be

$\displaystyle y=e^{ax}cos(bx)$
• Aug 31st 2006, 04:28 AM
CaptainBlack
Quote:

Originally Posted by a4swe
Excuse me, that should be

$\displaystyle y=e^{ax}cos(bx)$

The basic idea is to keep differentiating, and at each stage substitute for
complicated expression simpler ones base on what you already know.

Finally you should arrive at an equation of the required form, then you can
equate coefficients between the given DE and what you have found from
repeated differentiateion to get some algebraic equations for a and b.

(You will need to check the algebra in what follows)

If:

$\displaystyle y=e^{ax}\cos(bx)$

Then:

$\displaystyle y'=ae^{ax}\cos(bx)- be^{ax}\sin{bx} $$\displaystyle =a y - be^{ax}\sin{bx} , and so: \displaystyle y''=a y' - bae^{ax}\sin{bx} -b^2e^{ax}\cos(bx)$$\displaystyle =ay'-b^2y- bae^{ax}\sin{bx}=ay'-b^2y+ay'-a^2y$,

hence:

$\displaystyle y''=2ay'-(a^2+b^2)y$

so:

$\displaystyle y'''=2ay''-(a^2+b^2)y'=4a^2y'-2a(a^2+b^2)y-(a^2+b^2)y$

which you should now rearrange into the form $\displaystyle y'''+Ay'+By=0$ then equate
coefficients with the given form to get the algebraic equations for $\displaystyle a$ and
$\displaystyle b$.

RonL
• Aug 31st 2006, 04:46 AM
a4swe
Thank you very much, this helps a lot.
• Aug 31st 2006, 04:47 AM
ThePerfectHacker
Quote:

Originally Posted by a4swe
If $\displaystyle y=e^{ax}cos(bx)$ and $\displaystyle y'''+2y'+3y=0$ what is a and b?
That is the problem to solve and I don't even know what to do.
Can anyone help me out a little and give me a suggestion or two for the solution of this one?

Here is another way.
Since this is a homogenous linear differencial equation of order three with constand coefficients look at the charachachteriistic polynomial.
---
$\displaystyle k^3+2k+3=0$
Trivially, $\displaystyle k=-1$ is a solution.
Synthetic division,
$\displaystyle k^3+2k+3\div k+1=k^2-k+3$
The solution of,
$\displaystyle k^2-k+3=0$ are,
$\displaystyle k=\frac{1}{2}\pm i\frac{\sqrt{11}}{2}$
Therefore the general solution of this differential equation is,
$\displaystyle C_1e^{-x}+C_2e^{1/2 x}\sin \left( \frac{\sqrt{11}}{2} x\right)+C_3e^{1/2 x}\cos \left( \frac{\sqrt{11}}{2} x\right)$
The particular solution you have is when $\displaystyle C_1=C_3=0,C_2=1$
Thus,
$\displaystyle e^{1/2 x}\sin \left( \frac{\sqrt{11}}{2} x \right)=e^a \sin (bx)$
Thus,
$\displaystyle a=1/2 \mbox{ and }b=\frac{\sqrt{11}}{2}$