# find the general solution when 1 solution is given

• Mar 4th 2009, 01:46 PM
mandy123
find the general solution when 1 solution is given
I am suppose to find the general solution when one solution is given:

9y''' + 11y'' + 4y' -14y = 0

when y= e^(-x)sinx

Thanks
• Mar 4th 2009, 03:07 PM
Jester
Quote:

Originally Posted by mandy123
I am suppose to find the general solution when one solution is given:

9y''' + 11y'' + 4y' -14y = 0

when y= e^(-x)sinx

Thanks

Sure, if you let $\displaystyle y = u\,e^{-x} \sin x$, the your ODE will becomes another ODE that has no u and if you let $\displaystyle v = u'$, you'll get a separable ODE for v.
• Mar 4th 2009, 04:57 PM
mandy123
???
HUH?? I don't get what you are saying???? Can you break it down a little more for me please???

thanks
• Mar 4th 2009, 08:53 PM
TheEmptySet
Quote:

Originally Posted by mandy123
HUH?? I don't get what you are saying???? Can you break it down a little more for me please???

thanks

So what you are going to do is assume that there is a function of the form

$\displaystyle y=ue^{-x}\sin(x)$ where $\displaystyle u$ is a function of x.

Now you will need to take three derivatives of this new function and plug them into the ODE.

I.e here is the first one

$\displaystyle y'=u'(e^{-x}\sin(x))+u(-e^{-x}\sin(x)+e^{-x}\cos(x))$

Dont forget to use the product rule.

You will end up with an ODE in u and make the substituion that Danny recommended and solve for what u needs to be.

Good luck.
• Mar 4th 2009, 09:09 PM
CaptainBlack
Quote:

Originally Posted by mandy123
I am suppose to find the general solution when one solution is given:

9y''' + 11y'' + 4y' -14y = 0

when y= e^(-x)sinx