solving ODE

NYC

Apr 2010
19
1
I have the system below.
Any help on how I go about begining to solve this would be much appreciated. Thanks
 

Attachments

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
I have the system below.
Any help on how I go about begining to solve this would be much appreciated. Thanks
If you differentiate your ODE twice you get

\(\displaystyle U^{(4)} + \frac{1}{2} \xi U''' = 0\) (it's linear now)

from which you can solve for \(\displaystyle U\).
 
  • Like
Reactions: shawsend and NYC

NYC

Apr 2010
19
1
If you differentiate your ODE twice you get

\(\displaystyle U^{(4)} + \frac{1}{2} \xi U''' = 0\) (it's linear now)

from which you can solve for \(\displaystyle U\).
I obtain a \(\displaystyle U''\) term when I differentiate twice. I used the product rule in differentiating the term \(\displaystyle \xi U'\) is that right??
 

NYC

Apr 2010
19
1
OK it works if I don't use the product rule for that term. Thanks for the help.
 

NYC

Apr 2010
19
1
But surely you need to use the product rule for \(\displaystyle \xi U'\)
as the term \(\displaystyle U'\) can be differentiated with respect to \(\displaystyle \xi\) just as \(\displaystyle U\) and \(\displaystyle U''\) can be ??
 

shawsend

MHF Hall of Honor
Aug 2008
903
379
But surely you need to use the product rule for \(\displaystyle \xi U'\)
as the term \(\displaystyle U'\) can be differentiated with respect to \(\displaystyle \xi\) just as \(\displaystyle U\) and \(\displaystyle U''\) can be ??
That's correct. I got:

\(\displaystyle y'-1/2xy''-1/2y'=y'''\)

and then:

\(\displaystyle y''-1/2(xy'''+y')-1/2y''=y''''\)

which reduces to the equation posted by Danny.

(sorry, I'm using y and x. It's simpler)
 
  • Like
Reactions: NYC

NYC

Apr 2010
19
1
That's correct. I got:

\(\displaystyle y'-1/2xy''-1/2y'=y'''\)

and then:

\(\displaystyle y''-1/2(xy'''+y')-1/2y''=y''''\)

which reduces to the equation posted by Danny.

(sorry, I'm using y and x. It's simpler)
Thanks for the help. I get it now.
How do I now solve ODE danny posted? The only thing I can think to do is integrate twice but that would take me back to my orginal ODE
 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
Let \(\displaystyle V = U'' \) so you have \(\displaystyle V' + \frac{1}{2} \xi V = 0\) (this separates)

\(\displaystyle
\frac{dV}{V} = - \frac{1}{2} \xi
\)

\(\displaystyle
\ln V = -\frac{1}{4} \xi^2 + ln c_1
\)

\(\displaystyle V = c_1 e^{-1/4 \xi^2}\)

so

\(\displaystyle
U''' = c_1 e^{-1/4 \xi^2}.
\)

Now integrate three times. Note. Once you have your final answer, substitute back into the original ODE to determine that it is fact a solution and determine whether there's any restriction on your contants of integration. Realize there should only be two arbitrary contants since the original ODE is second order.
 
  • Like
Reactions: NYC

NYC

Apr 2010
19
1
Let \(\displaystyle V = U'' \) so you have \(\displaystyle V' + \frac{1}{2} \xi V = 0\) (this separates)

\(\displaystyle
\frac{dV}{V} = - \frac{1}{2} \xi
\)

\(\displaystyle
\ln V = -\frac{1}{4} \xi^2 + ln c_1
\)

\(\displaystyle V = c_1 e^{-1/4 \xi^2}\)

so

\(\displaystyle
U''' = c_1 e^{-1/4 \xi^2}.
\)

Now integrate three times. Note. Once you have your final answer, substitute back into the original ODE to determine that it is fact a solution and determine whether there's any restriction on your contants of integration. Realize there should only be two arbitrary contants since the original ODE is second order.

Upon integrating you get

\(\displaystyle U=-8 c_1 e^{-1/4 \xi^2} +c_2 1/2 \xi^2 +c_3 \xi + c_4\)

\(\displaystyle U(0)=0\) gives \(\displaystyle -8 c_1 + c_4 =0\)

also \(\displaystyle U( \infty)=0\) gives \(\displaystyle c_4 + \infty =0\)

However I know that \(\displaystyle U= \xi^2 + 2\) is a soln

hence the \(\displaystyle c_1\) and \(\displaystyle c_3\) terms need to vanish

Im now stuck to achieve this.

Im not sure if as the modulus of xi tends to infinity then U = o(1/xi)
where o is ''little o''

and if it does whether this helps
 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
Upon integrating you get

\(\displaystyle U=-8 c_1 e^{-1/4 \xi^2} +c_2 1/2 \xi^2 +c_3 \xi + c_4\)

\(\displaystyle U(0)=0\) gives \(\displaystyle -8 c_1 + c_4 =0\)

also \(\displaystyle U( \infty)=0\) gives \(\displaystyle c_4 + \infty =0\)

However I know that \(\displaystyle U= \xi^2 + 2\) is a soln

hence the \(\displaystyle c_1\) and \(\displaystyle c_3\) terms need to vanish

Im now stuck to achieve this.

Im not sure if as the modulus of xi tends to infinity then U = o(1/xi)
where o is ''little o''

and if it does whether this helps
Careful on your integration of \(\displaystyle e^{-1/4 \xi^2}\) realize it's not \(\displaystyle e^{-1/4 \xi}\).
 
  • Like
Reactions: NYC