General solution to a 6th order DE

[MathewsMD]

Hi,

Problem: Given the DE: y^(6) + y = 0, find the general solution.

Solution. I found the roots to be: (sqrt3 - i)/2, (sqrt3 + i)/2, i, -i, (-sqrt3 + i)/2, (-sqrt3 - i)/2

Thus, my general solution led to:

y = c1cost + c2sint + c3exp(t/2*(sqrt3))*cos(t/2) + c4exp(t/2*(sqrt3))*sin(t/2) + c5exp(t/2*-(sqrt3))*cos(t/2) + c6exp(t/2*(-sqrt3))*sin(t/2)

I don't see any errors in this, but please let me know if there are.

When I looked at the solution, the answer is:

y = exp(t/2*(sqrt3))*(c1cos(t/2) + c2sin(t/2)) + c3cost + c4sint*exp(t/2*-(sqrt3))*(c5cos(t/2) + c6sin(t/2)

How exactly did they derive the solution? Is my original answer wrong? The solutions in my original answer seem linearly dependent, and I don't quite see how they derived the solution they did...any help is great!

 

[Mark44]

I think there's an error in their solution. Notice that they have a cos(t) term but no sin(t) term. You don't get one without the other. You could check your solution, although that's going to be a lot of work.

 

[MathewsMD]

I think there's an error in their solution. Notice that they have a cos(t) term but no sin(t) term. You don't get one without the other. You could check your solution, although that's going to be a lot of work.

Yes, exactly! I just wasn't sure if they used a trig identity of sorts to find linear dependency or not b/w the terms.
It also seems odd (I may be mistaken) that they included c4 and c5 and c6 in their solution despite those terms being multiplied to yield only 2 terms, making the extra constant factor redundant, no?
Regardless, thank you for your comment.