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?

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- Aug 31st 2006, 02:52 AMa4sweA 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 AMCaptainBlackQuote:

Originally Posted by**a4swe**

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 AMa4swe
Excuse me, that should be

$\displaystyle

y=e^{ax}cos(bx)

$ - Aug 31st 2006, 04:28 AMCaptainBlackQuote:

Originally Posted by**a4swe**

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 AMa4swe
Why didn't I think about this?

Thank you very much, this helps a lot. - Aug 31st 2006, 04:47 AMThePerfectHackerQuote:

Originally Posted by**a4swe**

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,

$\displaystyle a=1/2 \mbox{ and }b=\frac{\sqrt{11}}{2}$