Stuck with the general solution now :/ [ { ( D^2 ) + 2 } ^ 2 ] * y = 0

[itsmusa7]

[ ( D^2 + 2 ) ^ 2 ] * y = 0, where D=dy/dx ------ (1)

let, y=e^mx be the trial solution of the equation (1), then we get -

[ ( D^2 + 2 ) ^ 2 ] * e^mx = 0

A.E = ( m^2 + 2 ) ^ 2 = 0
=> m^2 + 2 = 0
=> m^2 = -2
=> m^2 = 2( i^2 )
=> m = ± √2i

I'm stuck till here, the question is, what will be the general solution of this?

This? --> y = c1 cos √2x + c2 sin √2x
Or This? --> y = c1 cos √2x + c2x sin √2x
Or Something Else??

 

[Deleted member 4993]

[ ( D^2 + 2 ) ^ 2 ] * y = 0, where D=dy/dx ------ (1)

let, y=e^mx be the trial solution of the equation (1), then we get -

[ ( D^2 + 2 ) ^ 2 ] * e^mx = 0

A.E = ( m^2 + 2 ) ^ 2 = 0
=> m^2 + 2 = 0
=> m^2 = -2
=> m^2 = 2( i^2 )
=> m = ± √2i

I'm stuck till here, the question is, what will be the general solution of this?

This? --> y = c1 cos √2x + c2 sin √2x
Or This? --> y = c1 cos √2x + c2x sin √2x
Or Something Else??

The best way to check your solutions is to:

differentiate the proposed solutions (find D and D²) and substitute those in your original DE - look which one of those satisfy the DE.

 

[HallsofIvy]

This isn't so much a problem in differential equations as a problem in basic algebra. A fourth degree equation will have four solutions (possibly duplicates). The characteristic equation here, \(\displaystyle (m^2+ 2)^2= 0\) is a fourth degree equation. What are its four solutions?

Getting back to the differential equation, the set of all solutions to an "n" order linear, homogenous, differential equation form a n dimensional vector space. That means that the general solution can be written as a linear combination of n independent solutions. Here, n= 4. What are those four independent solutions.

 

[DrPhil]

[ ( D^2 + 2 ) ^ 2 ] * y = 0, where D=dy/dx ------ (1)

let, y=e^mx be the trial solution of the equation (1), then we get -

[ ( D^2 + 2 ) ^ 2 ] * e^mx = 0

A.E = ( m^2 + 2 ) ^ 2 = 0 This is 4th order - expect 4 roots
=> ±(m^2 + 2) = 0........Repeated roots
=> m^2 = -2
=> m^2 = 2( i^2 )
=> m = ± √2i

I'm stuck till here, the question is, what will be the general solution of this?

This? --> y = c1 cos √2x + c2 sin √2x
Or This? --> y = c1 cos √2x + c2x sin √2x
Or Something Else??

Probably have cosine, sine, AND x*cosine and x*sine