Solving high order equations

[Syringer]

Okay so in differential equations we have..

16''''+24y''+9y = 0

You can rewrite that as an algebra problem to

16m^4 + 24m^2 + 9 = 0

How do you solve for this though? The roots involve imaginary numbers..

 

[falconx80]

Originally posted by: Syringer
Okay so in differential equations we have..

16''''+24y''+9y = 0

You can rewrite that as an algebra problem to

16m^4 + 24m^2 + 9 = 0

How do you solve for this though? The roots involve imaginary numbers..

isnt this just...
(4m^2+3)(4m^2+3)

 

[Keego]

I've forgotten, and I did that stuff LAST TERM!!


wow, I suck

 

[raptor13]

You should have paid attention in class.




Anyway, replacing your m with r where r is y', you get:


16r^3 + 24r + 9


Solve for the roots, examine them, and apply the proper rules to get your answer. It doesn't matter what the roots are as there is a general form for the result regardless if the roots are imaginary, real, repeated, or any combination thereof.

 

[Syringer]

Originally posted by: falconx80

Originally posted by: Syringer
Okay so in differential equations we have..

16''''+24y''+9y = 0

You can rewrite that as an algebra problem to

16m^4 + 24m^2 + 9 = 0

How do you solve for this though? The roots involve imaginary numbers..

isnt this just...
(4m^2+3)(4m^2+3)

Goddamnit you're right!

 

[arynn]

Basically, the form of the roots dictates the form of your solution.

If you have a real root, i.e. r1 = a1 - x1 = C1*exp(a1*t)
If you have a repeated real root, i.e. r2 = a1 - x2 = C2*t*exp(a1*t)
If you have an imaginary pair of roots, i.e. r3a = a3 + b3i; r3b = a3 - b3i - x3 = exp(a3*t)*(C3*cos(b3*t) + C4*sin(b3*t))

The form of the imaginary roots comes from the translation from x = C*exp((a+/- b*i)*t) (The form is translated using Euler's equation or something).