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March 20th, 2017, 09:01 AM  #1 
Newbie Joined: Mar 2017 From: Thailand Posts: 4 Thanks: 0 Math Focus: Calculus  Please enlighten me.
Hello Redeemer, I got stuck at $\displaystyle (d^4y/dx^4)+3(d^2y/dx^2)^5+5y=0$. I'm not sure what method I should do. Thanks a lot for your help. TheNewDiff 
March 20th, 2017, 02:48 PM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 511 Thanks: 79 
Try$\displaystyle \;$ $\displaystyle y=e^{kx}$ 
March 20th, 2017, 03:09 PM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,157 Thanks: 878 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
March 20th, 2017, 09:17 PM  #4 
Newbie Joined: Mar 2017 From: Thailand Posts: 4 Thanks: 0 Math Focus: Calculus  
March 21st, 2017, 01:14 AM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,157 Thanks: 878 Math Focus: Wibbly wobbly timeywimey stuff.  
June 18th, 2017, 09:37 AM  #6 
Newbie Joined: Dec 2016 From: Austin Posts: 16 Thanks: 1 
Was this an initial value problem? If so, please post the initial conditions. Thanks!

December 9th, 2017, 03:27 PM  #7 
Newbie Joined: Mar 2017 From: Thailand Posts: 4 Thanks: 0 Math Focus: Calculus  
December 9th, 2017, 05:34 PM  #8 
Senior Member Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics 
The vector field for this is analytic so the formal power series solution will be the unique solution.

December 19th, 2017, 08:06 AM  #9 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  That is a standard method for linear equations, but this is not a linear equation. With $y= e^{kx}$, the equation becomes $k^4e^{kx}+ 3k^{10}e^{5k}+ 5e^{kx}= 0$. You can divide by $e^{kx}$, but that leaves $k^4+ 3k^{10}e^{4k}+ 5= 0$. How can you solve that?
Last edited by skipjack; December 19th, 2017 at 07:26 PM. 

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