Variation of Parameters to solve a second order ODE

[Bonnie]

Homework Statement

The question I am working on is the one in the file attached.

Homework Equations

y = u₁y₁ + u₂y₂ :

u₁'y₁ + u₂'y₂ = 0
u₁'y₁' + u₂'y₂' = g(t)

The Attempt at a Solution

I think I have got part (i) completed, with y₁ = e^3it and y₂ = e^-3it. This gives a general solution to the homogeneous equation to be y = C₁cos3t + C₂sin3t.
For part (ii) I know that for variation of parameters you need to substitute y₁ and y₂ and their derivatives into the above system of equations to solve for u₁' and u₂', then integrate these to find u₁ and u₂, from which you can get the desired solution of
y = y_p + y_h
But I find that when I try to do that by substituting y₁ = e^3it and y₂ = e^-3it I just end up with very complicated expressions involving lots of e's and i's which the desired solution does not contain. I know it involves combining the information about the general soln from (i) with the method I have described, but I just don't know how to make it work. Any help would be appreciated, thank you in advance!

 

[BvU]

I just end up with very complicated expressions involving lots of e's and i's which the desired solution does not contain

So the thing to do is avoid these exponential $y$ and take the trigonometric $y$ that you see in front of you when you write $y = C_1\cos 3t + C_2\sin3t\$ !

http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

 

[Ray Vickson]

Homework Statement

The question I am working on is the one in the file attached.

Homework Equations

y = u₁y₁ + u₂y₂ :

u₁'y₁ + u₂'y₂ = 0
u₁'y₁' + u₂'y₂' = g(t)

The Attempt at a Solution

I think I have got part (i) completed, with y₁ = e^3it and y₂ = e^-3it. This gives a general solution to the homogeneous equation to be y = C₁cos3t + C₂sin3t.
For part (ii) I know that for variation of parameters you need to substitute y₁ and y₂ and their derivatives into the above system of equations to solve for u₁' and u₂', then integrate these to find u₁ and u₂, from which you can get the desired solution of
y = y_p + y_h
But I find that when I try to do that by substituting y₁ = e^3it and y₂ = e^-3it I just end up with very complicated expressions involving lots of e's and i's which the desired solution does not contain. I know it involves combining the information about the general soln from (i) with the method I have described, but I just don't know how to make it work. Any help would be appreciated, thank you in advance!

Have you forgotten that $e^{\pm i \theta} = \cos \theta \pm i \sin \theta \,?$

 

[Bonnie]

Have you forgotten that $e^{\pm i \theta} = \cos \theta \pm i \sin \theta \,?$

Yes I had, thank you. I'll have another try!