My Math Forum Differential Equation Problem

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 February 7th, 2018, 08:56 AM #1 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
 February 7th, 2018, 09:32 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,552 Thanks: 1402 mathematica returns an absurdly long answer. are you sure there are no typos? Thanks from dylanxelle
 February 7th, 2018, 07:33 PM #3 Banned Camp   Joined: Apr 2017 From: durban Posts: 22 Thanks: 0 Math Focus: Algebra hmmm
 February 7th, 2018, 08:24 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra The principal problem is that the characteristic (homogeneous) equation doesn't have pleasant roots. Thanks from dylanxelle
February 8th, 2018, 05:30 AM   #5
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Quote:
 Originally Posted by romsek 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

 February 8th, 2018, 09:47 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 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 $y'' + y' + y = 0$.
 February 8th, 2018, 03:26 PM #7 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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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