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Constant Coefficient Homogeneous High Order ODEs
Constant Coefficient Homogeneous High Order ODEs


Ex <math>L(y) = ay" + by' +cy = 0, a, b, c \in \mathbb{R}</math>
Ex <math>Ly = a y'' + b y' + c y = 0</math>, <math>a \in \mathbb{R}, b \in \mathbb{R}, c \in \mathbb{R}</math>


Or generally <math>L(y) = \sum_k=0^n a_k y^{(k)} = 0, a_k \in \mathbb{R}</math>
Or generally <math>Ly = \sum_k=0^n a_k y^{(k)} = 0, a_k \in \mathbb{R}</math>


<math>L:{functions on \mathbb{R} \rightarrow {functions on \mathbb{R}</math> is a linear transformation ("linear operator").
<math>L:\{f: \mathbb{R} \rightarrow \mathbb{R}\} \rightarrow \{f: \mathbb{R} \rightarrow \mathbb{R}\}</math> is a linear transformation ("linear operator").


What do we expect from <math>{y:L(y) = 0} = ker(L)</math>? We expect an n-dimensional vector space.
What do we expect from <math>\{y: Ly = 0\} = ker(L)</math>? We expect an n-dimensional vector space.


Take <math> y"+y'-6y = 0</math>, guess <math> y = c, y' = \alpha e^{\alpha x}, y" = \alpha^2e^{\alpha x}</math>
Take <math> y''+y'-6y = 0</math>, guess <math> y = c, y' = \alpha e^{\alpha x}, y'' = \alpha^2e^{\alpha x}</math>


<math> \alpha^2 e^{\alpha x} + \alpha e^{\alpha x} - 6 e^{\alpha x} = 0</math>
<math> \alpha^2 e^{\alpha x} + \alpha e^{\alpha x} - 6 e^{\alpha x} = 0</math>

Latest revision as of 14:33, 24 October 2012

Week 6, Lecture 3

Constant Coefficient Homogeneous High Order ODEs

Ex [math]\displaystyle{ Ly = a y'' + b y' + c y = 0 }[/math], [math]\displaystyle{ a \in \mathbb{R}, b \in \mathbb{R}, c \in \mathbb{R} }[/math]

Or generally [math]\displaystyle{ Ly = \sum_k=0^n a_k y^{(k)} = 0, a_k \in \mathbb{R} }[/math]

[math]\displaystyle{ L:\{f: \mathbb{R} \rightarrow \mathbb{R}\} \rightarrow \{f: \mathbb{R} \rightarrow \mathbb{R}\} }[/math] is a linear transformation ("linear operator").

What do we expect from [math]\displaystyle{ \{y: Ly = 0\} = ker(L) }[/math]? We expect an n-dimensional vector space.

Take [math]\displaystyle{ y''+y'-6y = 0 }[/math], guess [math]\displaystyle{ y = c, y' = \alpha e^{\alpha x}, y'' = \alpha^2e^{\alpha x} }[/math]

[math]\displaystyle{ \alpha^2 e^{\alpha x} + \alpha e^{\alpha x} - 6 e^{\alpha x} = 0 }[/math]

[math]\displaystyle{ (\alpha^2 +\alpha - 6) e^{\alpha x} = 0 }[/math]

[math]\displaystyle{ (\alpha +3)(\alpha - 2) = 0 }[/math]

So we have [math]\displaystyle{ y = c_1e^{-3x} +c_2 e^{2x} }[/math] as the general solution.

Say we have complex [math]\displaystyle{ \alpha }[/math]. Then what?