# 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.

## Answers and Replies

lzkelley
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

gamesguru
According to mathematica, the most general answer is:
$$y= e^{-\frac{x}{\sqrt{2}}} \left(\left(e^{\sqrt{2} x} C[1]+C[2]\right) \text{Cos}\left[\frac{x}{\sqrt{2}}\right]+\left(C[3]+e^{\sqrt{2} x} C[4]\right) \text{Sin}\left[\frac{x}{\sqrt{2}}\right]\right)$$.
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.

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

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 separate equations, not a system.

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

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)

michujo
Apply Laplace transformation.