# find the general solution when 1 solution is given

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• March 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

dont know where to start can someone please help me??
Thanks
• March 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

dont know where to start can someone please help me??
Thanks

Sure, if you let $y = u\,e^{-x} \sin x$, the your ODE will becomes another ODE that has no u and if you let $v = u'$, you'll get a separable ODE for v.
• March 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
• March 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

$y=ue^{-x}\sin(x)$ where $u$ 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

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

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.
• March 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

dont know where to start can someone please help me??
Thanks

The form of the given solution tells you that the charateristic equation has two complex roots and one real root.

This tells you that the two complex roots of the characteristic equation are -1+i, and -1-i (since the characteristic equation has real coefficients complex roots occur in conjugate pairs and only a linear combinatiation of the solutions corresponding to these roots will give the solution that you have been given).

This should be enough information to allow you to find the third root.

CB