4.2 Reduction of Order

We can use a known solution of a homogeneous linear second order ODE to reduce a second order ODE to a 1st order ODE.

Substitute 
	y=(Subscript[y, 1])[x]u[x] 
into the standard form ODE 
  y''[x]+P[x]y'[x]+Q[x]y[x]=f[x].

So 
	({{y, ->, (Subscript[y, 1])u}, {y', ->, (Subscript[y, 1])'u+(Subscript[y, 1])u'}, {y'', ->, (Subscript[y, 1])''u+2(Subscript[y, 1])'u'+(Subscript[y, 1])u''}})

Plug in and organize
({{({{(Subscript[y, 1])''u    , +, P (Subscript[y, 1])'u, +, Q (Subscript[y, 1])u}, {2 (Subscript[y, 1])'u', +, P (Subscript[y, 1])u',  ,  }, {(Subscript[y, 1])u''        ,  ,  ,  ,  }}), =, f}})
If (Subscript[y, 1]) is a solution of 
(Subscript[y, 1])’’+P (Subscript[y, 1])’ +Q (Subscript[y, 1]) = 0
 the top row is zero.
({{(Subscript[y, 1])u''    , +, (2(Subscript[y, 1])'+P (Subscript[y, 1]))u', +, 0u, =, f}})
Which is a 1st order linear ODE for w=u'.
({{(Subscript[y, 1])w'    , +, (2(Subscript[y, 1])'+P (Subscript[y, 1]))w, =, f}})
On exams I will ALWAYS ask you for the reduced equation.

4.3 Constant Coefficients

You already know (we did it as a HW at the very start of the class) 
how to find solns of constant coefficient linear homogeneous ODEs 
(Subscript[a, n])(y^((n)))+…+(Subscript[a, 1])y'+(Subscript[a, 0])y=0 
Substitute y=(e^(m t))
(Subscript[a, n])(m^n)(e^(m t))+…+(Subscript[a, 1])(m^1)(e^(m t))+(Subscript[a, 0])(e^(m t))=0 
and solve the polynomial (called the Auxiliary Polynomial) for m.
(Subscript[a, n])(m^n)+…+(Subscript[a, 1])(m^1)+(Subscript[a, 0])=0 
Each distinct m gives you a LI solution (e^(m t)).  If you have the right number of distinct roots you are done!

Complex Roots: m=α±I β

A complex conjugate pair of roots m=α±I β  gives two complex solutions
	(e^((α ± I β) t))=(e^(α t))(Cos[β t]±I Sin[β t])
Just like in Chapter 8 most engineers prefer the real solutions
	(e^(α t))Cos[β t]  and (e^(α t))Sin[β t].
If α<0 then these are decaying osscilations.
If α>0 then these are growing osscilations.
Peak to Peak delay is T=2π/β
Sin[β t]=Sin[β t+2π]=Sin[β(t+2π/β)]

Repeated Roots

Repeated roots are simple.   If I have a root say (Subscript[m, 1]) repeated three times then my three solutions for that root are 
	(e^((Subscript[m, 1])t)), t (e^((Subscript[m, 1])t)), and (t^2) (e^((Subscript[m, 1])t)).
If it is repeated more I just keep going. You show this using reduction of order.

Plot

{γ,ω}={12,12.1}
Plot[ Cos[γ t] - Cos[ω t],
{t, 0, 12}]
Plot[ Cos[γ t] - Cos[ω t],
{t, 0, 40}]
Plot[ Cos[γ t] - Cos[ω t],
{t, 0, 123}]

4.4 Undetermined Coefficients

The solution of 
(Subscript[a, n])(y^((n)))+…+(Subscript[a, 1])y'+(Subscript[a, 0])y=f[t] 
is y=(Subscript[y, c])+(Subscript[y, p])  where we find (Subscript[y, c])using the Auxiliary polynomial. 
The easiest way to find (Subscript[y, p]) is a process with the fancy name of Undetermined Coefficients.  In plain language you guess the form and work out the numbers by plugging into the equations.

f[t]=(Subscript[f, 0 ])(e^(ω t))

If I told you I was thinking of a function that when I added up a bunch of its derivatives multiplied by constants I got (Subscript[f, 0]) (e^(ω t)) you would guess the function was an exponential. We throw an unknown number in front.
	7(e^(ω t))⟶(Subscript[y, p])=A (e^(ω t))

f[t]=(Subscript[f, 0])+(Subscript[f, 1])t+(Subscript[f, 2])(t^2)+ …+(Subscript[f, n]) (t^n)

If I told you I was thinking of a function that when I added up a bunch of its derivatives multiplied by constants I got a polynomial you would guess the function was a polynomial.  As always we throw an unknown number in front of each term.  
	({{t, ⟶, (Subscript[y, p]), =, (Subscript[A, 1])t+(Subscript[A, 0])}, {(t^2)+1, ⟶, (Subscript[y, p]), =, (Subscript[A, 2])(t^2)+(Subscript[A, 1])t+(Subscript[A, 0])}})

f[t]=(Subscript[f, 0])cos(ω t)

If I told you I was thinking of a function that when I added up a bunch of its derivatives multiplied by constants I got a cos(ω t) you would guess the function was something like cos(ω t). Then you would realize that it had to have both cos(ω t) and sin(ω t). As always we throw in undetermined parameters.  
	({{3 Cos[ω t], ⟶, (Subscript[y, p]), =, A Cos[ω t]+B Sin[ω t]}})

Sums of Terms

The solution of 
	(Subscript[a, n])(y^((n)))+…+(Subscript[a, 1])y'+(Subscript[a, 0])y=(Subscript[f, 1])[t]+(Subscript[f, 2])[t] 
is y=(Subscript[y, c])+(Subscript[y, (Subscript[p, 1])])+(Subscript[y, (Subscript[p, 2])])  where 
(Subscript[y, (Subscript[p, 1])])is a particular solution for the (Subscript[f, 1]) and 
(Subscript[y, (Subscript[p, 2])])is a particular solution for (Subscript[f, 2]). 
In plain language you can split up the right hand side if it is easier.

Special Case: (Subscript[y, p]) in (Subscript[y, c])

If any term you guess for the (Subscript[y, p]) is in (Subscript[y, c]) then you multiply all the terms of the same flavor by t untill there are no longer any terms in (Subscript[y, c]).

4.6 Variation of Parameters

If we have solutions (Subscript[y, 1])and (Subscript[y, 2]) to
 y''+P[x] y'+Q[x]y=0
 then we can find a solution to 
   y''+P[x] y'+Q[x]y=f[x]
   by substituting y=(Subscript[u, 1])(Subscript[y, 1])+(Subscript[u, 2])(Subscript[y, 2]) into the ODE.

If we have solutions (Subscript[y, 1])and (Subscript[y, 2]) then
 ({{y', =, (Subscript[u, 1])(Subscript[y, 1])'+(Subscript[u, 2])(Subscript[y, 2])', +, (Subscript[u, 1])'(Subscript[y, 1])+(Subscript[u, 2])'(Subscript[y, 2])}})
Plugging into the ODE is going to give us one Eqn. We have two things to solve for.  We can pick one equation however we want.
 We want to keep things simple so we pick  
  ({{(Subscript[u, 1])'(Subscript[y, 1])+(Subscript[u, 2])'(Subscript[y, 2]), =, 0}}) 
 so
  ({{y', =, (Subscript[u, 1])(Subscript[y, 1])'+(Subscript[u, 2])(Subscript[y, 2])', +, 0}, {y'', =, (Subscript[u, 1])(Subscript[y, 1])''+(Subscript[u, 2])(Subscript[y, 2])'', +, (Subscript[u, 1])'(Subscript[y, 1])'+(Subscript[u, 2])'(Subscript[y, 2])'}})
 and the ODE
   y''+P[x] y'+Q[x]y=f[x]
becomes
  ({{y'', |, (Subscript[u, 1])(Subscript[y, 1])''+(Subscript[u, 2])(Subscript[y, 2])'', +, (Subscript[u, 1])'(Subscript[y, 1])'+(Subscript[u, 2])'(Subscript[y, 2])'}, {+P[x]y', |, P[x](Subscript[u, 1])(Subscript[y, 1])'+P[x](Subscript[u, 2])(Subscript[y, 2])', +, 0}, {+Q[x] y, |, Q[x](Subscript[u, 1])(Subscript[y, 1])+Q[x](Subscript[u, 2])(Subscript[y, 2]), +, 0}})
  Since (Subscript[y, 1]) and (Subscript[y, 2]) satisfy the homogeneous ODE the equations are 
	({{(Subscript[u, 1])' (Subscript[y, 1]), +, (Subscript[u, 2])' (Subscript[y, 2]), =, 0}, {(Subscript[u, 1])' (Subscript[y, 1])', +, (Subscript[u, 2])' (Subscript[y, 2])', =, f}})
Solving the linear system for (Subscript[u, 1])'  and (Subscript[u, 2])' then integrating and substituting back into the original form 
	 y=(Subscript[u, 1])(Subscript[y, 1])+(Subscript[u, 2])(Subscript[y, 2])  
solves y''+P[x] y'+Q[x]y=f[x]