Section 4.2 - How to solve
Look for solutions of the form
In general, given the second-order equation
There are now three cases:
- If
then the characteristic equation has two distinct, real-valued roots . The solution to the ODE is
for arbitrary constants. - If
then the characteristic equation has a repeated root . The solution to the ODE is
for arbitrary constants. - If
then the characteristic equation has complex roots where and . The solution to the ODE is
for arbitrary constants.
Sections 4.4 & 4.5 - The Method of Undetermined Coefficients (a.k.a. Making educated guesses)
Undetermined Coefficients Method
Consider the non-homogenous second order linear equation with constant coefficients
Then the solution to this non-homogeneous equation,, can be expressed in the form
is soln to the corresponding homogeneous eqn (what we’ve been doing this week)
: any particular solution to the non-hom. eqn
For certain RHS, we can make a guess for whatwill be.
Heremust be one of:
- Exponential:
- Polynomial:
and/or - any product/combo of the above 3
: indep. variable
: arbitrary parameters/constants
Our guess for:
- Solve corresponding hom. eqn. for
. Here, we solve for
Chara. eqn.:which has soln
So- Guess that
, plug in the ODE and solve for .
and
plug inin LHS:
Meanwhile RHS
So
Thus,- General soln to the non-homo. eqn. is
- Based on RHS, we guess that
. But there is a “conflict” between & so we make another one, by multiplying our current guess by . New guess:
Now there’s no more “conflict”, we can useand plug it into the ODE to solve for .
LHS =
With RHS =, we have so thus,
General soln: