Second order DE help

**Jones** · January 1st 2010, 08:23 AM

Hi,

I have the following

$y'' + 2y' +3y = 0$

So the auxilary equation would look like $r^2 +2r +3 = 0$

The solution to this equation is on the form $\frac{-2}{2}\pm \sqrt{\frac{4-12}{2}}$

which evaluates to $-1 \pm \sqrt{\frac{-8}{2}} \rightarrow \sqrt{-4}$

$= -1\pm2i$

Which gives $Ae^{-t}cos(2t) + Be^{-t}sin(2t)$

but apparently thats incorrect, the answer should be $-1\pm\sqrt{2i}$

whats wrong?

Edit: Darn i just remebered there's a forum for DE's. I can't delete my post
Forgive me

**Danny** · January 1st 2010, 10:10 AM

Originally Posted by Jones

Hi,

I have the following

$y'' + 2y' +3y = 0$

So the auxilary equation would look like $r^2 +2r +3 = 0$

The solution to this equation is on the form $\frac{-2}{2}\pm \sqrt{\frac{4-12}{2}}$ (1)

which evaluates to $-1 \pm \sqrt{\frac{-8}{2}} \rightarrow \sqrt{-4}$

$= -1\pm2i$

Which gives $Ae^{-t}cos(2t) + Be^{-t}sin(2t)$

but apparently thats incorrect, the answer should be $-1\pm\sqrt{2i}$

whats wrong?

Edit: Damn i just remebered there's a forum for DE's. I can't delete my post
Forgive me

Look at step (1) in blue. Not sure why the 2 is in the square root.

**Abu-Khalil** · January 1st 2010, 10:18 AM

$r=\frac{-2\pm\sqrt{4-12}}{2}=\frac{-2\pm2\sqrt{2}i}{2}=-1\pm \sqrt{2}i$

$y(t)=Ae^{-t}\cos\left(\sqrt{2}t\right)+Be^{-t}\sin\left(\sqrt{2}t\right)$

**Sudharaka** · January 1st 2010, 05:04 PM

Dear Jones,

I have attached the solution to your question. Hope it will help you to find where you went wrong. But I would greatly appreciate your reply in whether you were able to read my attachment clearly. If it is not clear or if you have any doubt regarding this question please don't hesitate to mail me.

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