Differential Equation Problem

dylanxelle · February 7th, 2018, 08:56 AM

Can somebody help me with this differential equation? Thank you very much

y''' + y'' + y = x^2 + 3e^3x

romsek · February 7th, 2018, 09:32 AM

mathematica returns an absurdly long answer.

are you sure there are no typos?

Ola Lawson · February 7th, 2018, 07:33 PM

hmmm

v8archie · February 7th, 2018, 08:24 PM

The principal problem is that the characteristic (homogeneous) equation doesn't have pleasant roots.

dylanxelle · February 8th, 2018, 05:30 AM

Quote:

Originally Posted by romsek $View Post$

mathematica returns an absurdly long answer.

are you sure there are no typos?

Oh, thanks mate! I checked it again. It's supposed to be y" + y' + y

skipjack · February 8th, 2018, 09:47 AM

If the equation is $y'' + y' + y = x^2 + 3e^{3x}$, a particular solution is $y = x^2 - 2x + \frac{3}{13}e^{3x}$.
Add to that the general solution of $y'' + y' + y = 0$.

Country Boy · February 8th, 2018, 03:26 PM

The characteristic equation for y''+ y'+ y= 0 is $r^2+ r+ 1= 0$. Writing that as $r^2+ r= -1$ and "completing the square", $r^2+ r+ \frac{1}{4}= (r+ \frac{1}{2})^2= -1+ \frac{1}{4}= -\frac{3}{4}$. Taking the square root of both sides $r+ \frac{1}{2}= \pm\frac{\sqrt{3}}{2}i$ and $r= -\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$.

The general solution to that homogeneous differential equation is $y(x)= e^{-x/2}\left(C_1\cos(x\sqrt{3}/2)+ C_2\sin(x\sqrt{3}/2)\right)$

February 7th, 2018, 08:56 AM	#1
dylanxelle Newbie Joined: Feb 2018 From: Philippines Posts: 2 Thanks: 0	Differential Equation Problem Can somebody help me with this differential equation? Thank you very much y''' + y'' + y = x^2 + 3e^3x
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February 7th, 2018, 09:32 AM	#2
romsek Senior Member Joined: Sep 2015 From: USA Posts: 2,039 Thanks: 1063	mathematica returns an absurdly long answer. are you sure there are no typos? Thanks from dylanxelle
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February 7th, 2018, 07:33 PM	#3
Ola Lawson Banned Camp Joined: Apr 2017 From: durban Posts: 22 Thanks: 0 Math Focus: Algebra	hmmm
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February 7th, 2018, 08:24 PM	#4
v8archie Math Team Joined: Dec 2013 From: Colombia Posts: 7,342 Thanks: 2465 Math Focus: Mainly analysis and algebra	The principal problem is that the characteristic (homogeneous) equation doesn't have pleasant roots. Thanks from dylanxelle
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February 8th, 2018, 09:47 AM	#6
skipjack Global Moderator Joined: Dec 2006 Posts: 19,287 Thanks: 1681	If the equation is $y'' + y' + y = x^2 + 3e^{3x}$, a particular solution is $y = x^2 - 2x + \frac{3}{13}e^{3x}$. Add to that the general solution of $y'' + y' + y = 0$.
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February 8th, 2018, 03:26 PM	#7
Country Boy Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894	The characteristic equation for y''+ y'+ y= 0 is $r^2+ r+ 1= 0$. Writing that as $r^2+ r= -1$ and "completing the square", $r^2+ r+ \frac{1}{4}= (r+ \frac{1}{2})^2= -1+ \frac{1}{4}= -\frac{3}{4}$. Taking the square root of both sides $r+ \frac{1}{2}= \pm\frac{\sqrt{3}}{2}i$ and $r= -\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$. The general solution to that homogeneous differential equation is $y(x)= e^{-x/2}\left(C_1\cos(x\sqrt{3}/2)+ C_2\sin(x\sqrt{3}/2)\right)$ Last edited by skipjack; February 8th, 2018 at 04:56 PM.
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