Differential equation to be solved (CHALLENGING...URGENT...Pl help!)

**unisa_students** · May 29th 2008, 08:14 PM

Please open the attachment...the question can be found there.

PLEASE HELP US...

IT'S A HUGE THING!!!

PLEASE!

**mr fantastic** · May 29th 2008, 08:35 PM

Originally Posted by unisa_students

Please open the attachment...the question can be found there.

PLEASE HELP US...

IT'S A HUGE THING!!!

PLEASE!

The question is solve $\frac{dy}{dx} = y - 0.3 \sin \left( \frac{\pi x}{3} \right)$ .

$\frac{dy}{dx} - y = -0.3 \sin \left( \frac{\pi x}{3} \right)$ is readily solved using the integrating factor method.

You will need to apply integration by parts twice to get $\int e^{-x} \sin \left( \frac{\pi x}{3} \right) \, dx$ .

Note: It's good form to take the few minutes it requires to type your question if you expect help.

**Mathstud28** · May 29th 2008, 08:37 PM

Originally Posted by unisa_students

Please open the attachment...the question can be found there.

PLEASE HELP US...

IT'S A HUGE THING!!!

PLEASE!

I think it reads either

$y'=y-.3\sin\bigg(\frac{\pi{x}}{3}\bigg)$ in which case use an integrating factor

or

$y'=y-.3\sin\bigg(\frac{2\pi}{3}\bigg)$

in which case use the charcteristic equation

$\lambda^2=\lambda-0.3\sin\bigg(\frac{2\pi}{3}\bigg)$

**CaptainBlack** · May 29th 2008, 10:59 PM

Originally Posted by Mathstud28

or

$y'=y-.3\sin\bigg(\frac{2\pi}{3}\bigg)$

in which case use the charcteristic equation

$\lambda^2=\lambda-0.3\sin\bigg(\frac{2\pi}{3}\bigg)$

Why does that ODE have that characteristic equation?

This is non-homogeneous ODE, so you should be looking at the characteristic equation of the homogeneous version:

$y'-y=0$

which is:

$\lambda-1=0$

which has solution $\lambda=1$ , so the general solution of the homogeneous equation is:

$y=Ae^{\lambda x}=Ae^x$

and a particular integral of the inhomogeneous equation would be:

$y=0.3 \sin \left(\frac{2\pi}{3}\right)$

So the general solution is:

$y=Ae^x + 0.3 \sin \left(\frac{2\pi}{3}\right)$

Which is all academic because the problem in the attachment is clearly not this one (but the real question could be sloved this way as well if you know what the particular integral will look like).

RonL

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