Variation of parameters

[willmoore21]

\(\displaystyle y''\)\(\displaystyle + w^2y=E_{0}sin(wt)\) where w and E_o constants. (E0 is supposed to be E₀ but can't get subs to work in Latex)

I need to find the general solution.

\(\displaystyle GS=PI+CF\)

First I need to find the complimentary function, I have done this by:

Try \(\displaystyle y=exp(at)\) ( where a = alpha)
\(\displaystyle a^2+w^2=0\)
2 complex distinct solutions, \(\displaystyle iw,-iw\)
Therefore CF = \(\displaystyle Aexp(iwt)+Bexp(-iwt)\)

This is one part where I am confused. I can simplify this further in the form of exp(at)((Ccos(at)+Dsin(at)), I think it is this exp(wx)((Ccos(x)+Dsin(-x)). Even if this is wrong I can continue do solve the particular integral.

So now I need the particular integral (I think this is a Wronskian as it is under out variation of parameters part of our course)

Apologies (can't get y1 and y2 subs to work in Latex)

\(\displaystyle so: PI=\)\(\displaystyle u(t)y1(t)\)\(\displaystyle +v(t)y2(t)\)

\(\displaystyle u(t)=-int(y2(t)F(t)/W(y1,y2)dt\)

\(\displaystyle v(t)=int(y2(t)F(t)/W(y1,y2)dt\)

\(\displaystyle W(y1,y2)=y1y2'-y2y1'\)

From here do I just procede as normal? I haven't come across a complex wronskian before so wasn't sure? I am happy with yes/no answers unless any of the previous is wrong. I will work out what I think it is as and when I get a reply.

I am NOT asking for a solution here, just a verification and a "go ahead" to carry on.

Thanks.

 

[galactus]

The auxiliary equation is \(\displaystyle m^{2}+w^{2}=0\)

\(\displaystyle m=iw, \;\ m=-iw\)

\(\displaystyle y_{1}=cos(wt), \;\ y_{2}=sin(wt)\)

\(\displaystyle y_{c}=C_{1}cos(wt)+C_{2}sin(wt)\)

\(\displaystyle W=w, \;\ W_{1}=-E_{0}sin^{2}(wt), \;\ W_{2}=E_{0}sin(wt)cos(wt)\)

Now, find \(\displaystyle u_{1}^{'}=\frac{W_{1}}{W}, \;\ u_{2}^{'}=\frac{W_{2}}{W}\)

Then, integrate to find \(\displaystyle u_{1}, \;\ u_{2}\)

The particular solution is \(\displaystyle y_{p}=u_{1}\cdot y_{1}+u_{2}\cdot y_{2}\)

The general solution is \(\displaystyle y=y_{c}+y_{p}\)

 

[willmoore21]

Thanks galactacus, will have a go at this tomorrow and finish it off. Will post my solution to finish the thread.

 

[willmoore21]

I don't understand how W=w.

If W= y₁y₂'-y₂y₁'

then W= \(\displaystyle (-sin^2(wt)-cos^2(wt))/w\)

I know \(\displaystyle sin^2(t)+cos^2(t)=1\)

so does \(\displaystyle sin^2(wt)+cos^2(wt)=2w?\)

It's the only way I can get W=w.

Thanks.

 

[willmoore21]

Actually, that can't equal 2W. I don't know where I'm going wrong.