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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 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$\displaystyle \;$ $\displaystyle y=e^{kx}$ Thanks from topsquark March 20th, 2017, 03:09 PM   #3
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 Originally Posted by TheNewDiff 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
The best I've come up with is y(x) = 0. You'll probably find that the solution(s) can only be found by numerical methods.

-Dan March 20th, 2017, 09:17 PM   #4
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 Originally Posted by topsquark The best I've come up with is y(x) = 0. You'll probably find that the solution(s) can only be found by numerical methods. -Dan
Hi Dan,
Could you please show me the step,please? March 21st, 2017, 01:14 AM   #5
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 Originally Posted by TheNewDiff Hi Dan, Could you please show me the step,please?
There was no method. I simply noted that y = 0 is a solution. It's one of the first things I look for, especially when dealing with a non-linear equation.

-Dan 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
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 Originally Posted by thegrade Was this an initial value problem? If so, please post the initial conditions. Thanks!
Don't have any initial value problem when I browns the book. 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
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 Originally Posted by idontknow Try$\displaystyle \;$ $\displaystyle y=e^{kx}$ 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?

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