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-   -   Getting the General solution (http://mymathforum.com/differential-equations/336936-getting-general-solution.html)

The_Ys_Guy October 29th, 2016 09:36 AM

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?

http://i.imgur.com/hTeCUGD.jpg

romsek October 29th, 2016 10:08 AM

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} $

The_Ys_Guy October 29th, 2016 10:21 AM

Quote:

Originally Posted by romsek (Post 551848)
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.

romsek October 29th, 2016 10:43 AM

Quote:

Originally Posted by The_Ys_Guy (Post 551852)
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! :D

$a=b=c=1$

$ = \dfrac{-1 \pm \sqrt{1 - 4}}{2} =-\dfrac 1 2 \pm i\dfrac{\sqrt{3}}{2}$

The_Ys_Guy October 29th, 2016 11:10 AM

Quote:

Originally Posted by romsek (Post 551855)
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! :D

$a=b=c=1$

$ = \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..... :(

topsquark October 29th, 2016 11:24 AM

Quote:

Originally Posted by The_Ys_Guy (Post 551859)
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

v8archie October 29th, 2016 11:33 AM

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.

The_Ys_Guy October 29th, 2016 02:11 PM

Quote:

Originally Posted by v8archie (Post 551862)
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?
http://i.imgur.com/nlEo4E3.jpg

skipjack October 30th, 2016 10:41 AM

It helps to replace $\sin^2\!x$ with the equivalent $(1 - \cos(2x))/2$.

integration April 19th, 2018 06:35 AM

which book is this?


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