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

116.5 KB Views: 6
If you differentiate your ODE twice you getI have the system below.
Any help on how I go about begining to solve this would be much appreciated. Thanks
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??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 \(\displaystyle U\).
That's correct. I got: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 ??
Thanks for the help. I get it now.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)
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.
Careful on your integration of \(\displaystyle e^{1/4 \xi^2}\) realize it's not \(\displaystyle e^{1/4 \xi}\).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