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 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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 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. Tags differential, equation, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post BonaviaFx Calculus 6 July 19th, 2015 08:45 AM greg1313 Differential Equations 11 July 11th, 2011 11:38 PM sivela Differential Equations 1 January 21st, 2011 05:53 PM David McLaurin Differential Equations 3 July 8th, 2009 07:50 AM Crumboo Differential Equations 0 November 20th, 2007 03:35 AM

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