My Math Forum Getting the General solution

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 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: USA Posts: 2,549 Thanks: 1399 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 $y(x) = (A + B x)e^{r x}$ Thanks from skeeter and The_Ys_Guy Last edited by romsek; October 29th, 2016 at 10:12 AM.
October 29th, 2016, 10:21 AM   #3
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 Originally Posted by romsek 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 $y(x) = (A + B x)e^{r x}$

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
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 Originally Posted by The_Ys_Guy 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.
dude...

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
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 Originally Posted by romsek dude... 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! $a=b=c=1$ $= \dfrac{-1 \pm \sqrt{1 - 4}}{2} =-\dfrac 1 2 \pm i\dfrac{\sqrt{3}}{2}$
I'm sheltered... I got to many clean equations. Okay I guess the value for the of x of e^x won't always be some clean looking value. That's okay my expectations are now looser.

Thanks.....

Edit: I got this root too BTW.....

October 29th, 2016, 11:24 AM   #6
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Quote:
 Originally Posted by The_Ys_Guy I'm sheltered... I got to many clean equations. Okay I guess the value for the of x of e^x won't always be some clean looking value. That's okay my expectations are now looser. Thanks..... Edit: I got this root too BTW.....
In this situation I have a tendency to define two variables: $\displaystyle \lambda _+ = \frac{1}{2} + i \frac{\sqrt{3}}{2}$ and $\displaystyle \lambda _- = \frac{1}{2} - i \frac{\sqrt{3}}{2}$. It makes things a bit less messy.

-Dan

 October 29th, 2016, 11:33 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 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 non-homogeneous 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
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 Originally Posted by v8archie 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 non-homogeneous 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.
I'm back and you're right. I'm so frightened about not knowing enough. I faintly remember a statement my professor claimed about getting the general solution before going for the particular solution so I followed it.

Quote:
 The book should give you a solid idea of what solutions you should try.
I think I know what you're talking about. I saw this list and it was bought up several times. It doesn't make sense to me but I know this list is very important otherwise my professor wouldn't have mentioned it so much. I can use this to see what solutions to try?

 October 30th, 2016, 10:41 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2216 It helps to replace $\sin^2\!x$ with the equivalent $(1 - \cos(2x))/2$.
 April 19th, 2018, 06:35 AM #10 Banned Camp   Joined: Nov 2017 From: india Posts: 204 Thanks: 2 which book is this?

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