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October 29th, 2016, 09:36 AM  #1 
Member Joined: Feb 2014 Posts: 91 Thanks: 1  Getting the General solution
On question 5 I was told to find a particular solution of the equation. I'm having trouble just getting the general solution. Could anyone show me how to get the general solution? Last edited by skipjack; October 30th, 2016 at 09:50 AM. 
October 29th, 2016, 10:08 AM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,238 Thanks: 638 
if you have a linear, second order differential equation with constant coefficients the general solutions is as follows. Let $a y^{\prime \prime} + b y^{\prime} + c y = 0$ form the polynomial $a x^2 + b x + c$ this is known as the characteristic polynomial Find the roots of this polynomial, $r_1, r_2$ If $r_1 \neq r_2$ then the general solution is then given by $y(x) = \large A e^{r_1 x} + B e^{r_2 x}$ where $A,~B$ are arbitrary constants. if $r_1 = r_2 = r$ then the general solution is given by $y(x) = (A + B x)e^{r x} $ Last edited by romsek; October 29th, 2016 at 10:12 AM. 
October 29th, 2016, 10:21 AM  #3  
Member Joined: Feb 2014 Posts: 91 Thanks: 1  Quote:
Would it be shameful to say I'm having trouble finding the roots..... It doesn't factors nicely and quadratic formula isn't working for me.  
October 29th, 2016, 10:43 AM  #4  
Senior Member Joined: Sep 2015 From: CA Posts: 1,238 Thanks: 638  Quote:
if you can't find the roots of a polynomial you really have no business messing with differential equations yet. The quadratic formula isn't working for you? It's working for all of us every day! $a=b=c=1$ $ = \dfrac{1 \pm \sqrt{1  4}}{2} =\dfrac 1 2 \pm i\dfrac{\sqrt{3}}{2}$  
October 29th, 2016, 11:10 AM  #5  
Member Joined: Feb 2014 Posts: 91 Thanks: 1  Quote:
Thanks..... Edit: I got this root too BTW.....  
October 29th, 2016, 11:24 AM  #6  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,559 Thanks: 601 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
October 29th, 2016, 11:33 AM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,784 Thanks: 2198 Math Focus: Mainly analysis and algebra 
This question has almost nothing to do with the general solution of the homogeneous equation. You are asked to find a single particular solution to the nonhomogeneous equation. Looking at the questions, one supposes that you've been looking at the method of undetermined coefficients. The book should give you a solid idea of what solutions you should try. PS: complex roots of the characteristic equation aren't "dirty", they represent an important class of solution to the homogeneous equation. Last edited by v8archie; October 29th, 2016 at 11:36 AM. 
October 29th, 2016, 02:11 PM  #8  
Member Joined: Feb 2014 Posts: 91 Thanks: 1  Quote:
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October 30th, 2016, 10:41 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 17,221 Thanks: 1294 
It helps to replace $\sin^2\!x$ with the equivalent $(1  \cos(2x))/2$.


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