# Forth Order DiffEq

[SOLVED] Forth Order DiffEq

I've recently come across the following differential equations. y''''+y=0 and y''''-y=0. Can differential equations such as these be solved with any technique other than guessing for the particular solutions? They seem very simular to trig's equation but are still not quite the same.

the only solution that i can see is y=0.
from those equations y = -y and y''''=-y'''' which can only happen if y=0

According to mathematica, the most general answer is:
$$y= e^{-\frac{x}{\sqrt{2}}} \left(\left(e^{\sqrt{2} x} C+C\right) \text{Cos}\left[\frac{x}{\sqrt{2}}\right]+\left(C+e^{\sqrt{2} x} C\right) \text{Sin}\left[\frac{x}{\sqrt{2}}\right]\right)$$.
The way I would solve this is to write down the characteristic equation to it.
$$r^4+1=0$$
$$r=(-1)^{1/4}$$
I think you know how to do it from here.

I don't see how thats a solution to both of the differential equations (although it is a solution to one of them)...

lurflurf
Homework Helper
the only solution that i can see is y=0.
from those equations y = -y and y''''=-y'''' which can only happen if y=0

I hope that was meant to be two seperate equations, not a system.

It's only the solution to the first one (y''''+y=0).

lurflurf
Homework Helper
I've recently come across the following differential equations. y''''+y=0 and y''''-y=0. Can differential equations such as these be solved with any technique other than guessing for the particular solutions? They seem very simular to trig's equation but are still not quite the same.

factor

y''''+y=0
(D^4+1)y=0
(D^2+2cos(pi/4)D+1)(D^2-2cos(pi/4)D+1)y=0

y''''-y=0
(D^4-1)y=0
(D+1)(D-1)(D^2+1)y=0

Then either do partial fraction decomposition or note that since the factors are independent the solution of the product is the sum of the solutions of the factors
ie
D(D-1)y=0
Du=0->u=A
(D-1)v=0->v=Bexp(x)
y=u+v=A+Bexp(x)

Apply Laplace transformation.