Second-Order Linear ODE (1) - Basic

1. Homogeneous Linear ODE of Second Order

 

$ y''+p(x)y'+q(x)y=r(x)\left\{\begin{matrix}
r=0 (homo) \\ 
r\neq 0 (non-homo)
\end{matrix}\right. $

 

$ L=\frac{\mathrm{d}^{2} }{\mathrm{d} x^{2}}+p\frac{\mathrm{d} }{\mathrm{d} x}+q $

$ L(y)=r(x) $

$ L(\alpha y_{1}+\beta y_{2})=(\alpha y_{1}+\beta y_{2})''+p(\alpha y_{1}+\beta y_{2})'+q(\alpha y_{1}+\beta y_{2}) $

$ =\alpha y_{1}''+\beta y_{2}''+p\alpha y_{1}'+p\beta y_{2}'+q\alpha y_{1}+q\beta y_{2} $

$ =\alpha (y_{1}''+py_{1}'+qy_{1})+\beta (y_{2}''+py_{2}'+qy_{2}) $

$ =\alpha L(y_{1})+\beta L(y_{2}) $: Linear

 

Initial Value: $ y(x_{0})=K_{0}, y'(x_{0})=K_{1} $

 

Theorem) Superposition Principle

For a homogeneous linear ODE (2), any linear combination of two solutions on an open interval I is again a solution of (2) on I. In particular, for such an equation, sums and constant multiples of solutions are again solutions.

 

$ y_{1}, y_{2} $: solutions of $ y''+py'+qy=0 $

$ \Rightarrow y_{1}''+py_{1}'+qy_{1}=0, y_{2}''+py_{2}'+qy_{2}=0 $

$ c_{1}y_{1}+c_{2}y_{2} $

$  \Rightarrow (c_{1}y_{1}+c_{2}y_{2})''+p(c_{1}y_{1}+c_{2}y_{2})'+q(c_{1}y_{1}+c_{2}y_{2}) $

   $ =c_{1}y_{1}''+c_{2}y_{2}''+pc_{1}y_{1}'+pc_{2}y_{2}'+qc_{1}y_{1}+qc_{2}y_{2} $

   $ =c_{1}( y_{1}''+py_{1}'+qy_{1})+c_{2}( y_{2}''+py_{2}'+qy_{2})=0 $

 

 

2. Linearly Independent, Basis

 

Linearly Independent: $ k_{1}y_{1}+k_{2}y_{2}=0 \Leftrightarrow k_{1}=0 $ and $ k_{2}=0 $ ($ \frac{y_{2}}{y_{1}}= $ x에 대한 식)

Linearly Dependent: $ k_{1}y_{1}+k_{2}y_{2}=0 \rightarrow  k_{1}\neq 0 $ or $ k_{2}\neq0 $ ($ \frac{y_{2}}{y_{1}}= $ 상수)

 

Defenition) Basis (Fundamental System)

A Basis of soluitons of (2) on an open interval I is a pair of linearly indpendent soolutions of (2) on I.

 

 

3. Reduction of Order

 

$ y''+py'+qy=0 $

We know $ y_{1} $ and we don't know $ y_{2} $

 

$ \frac{y_{2}}{y_{1}}=u \rightarrow y_{2}=uy_{1} $

$ y_{2}'=u'y_{1}+uy_{1}', y_{2}''=u''y_{1}+2u'y_{1}'+uy_{1}'' $

$ y_{1}''+py_{1}'+qy_{1}=0, y_{2}''+py_{2}'+qy_{2}=0 $

$ \rightarrow (u''y_{1}+2u'y_{1}'+uy_{1}'')+p(u'y_{1}+uy_{1}')+quy_{1} $

   $ =u''y_{1}+u'(2y_{1}'+py_{1})+u(y_{1}''+py_{1}'+qy_{1}) $

   $ =u''y_{1}+u'(2y_{1}'+py_{1})=0 $

   $ u''+u'(\frac{2y_{1}'}{y_{1}}+p)=0 $

$ U=u' \rightarrow U'+U(\frac{2y_{1}'}{y_{1}}+p)=0 $

$ \frac{dU}{U}=-(\frac{2y_{1}'}{y_{1}}+p)dx $

$ \ln |U|=-2\ln |y_{1}|-\int p dx $

$ U=\frac{1}{y_{1}^{2}}e^{-\int p dx} $

$ u=\int U dx $

$ y_{2}=uy_{1} $

$ \therefore y=c_{1}y_{1}+c_{2}y_{2} $