Second-Order Linear ODE (8) - Variation of Parameters

15. Variation of Parameters

 

$ y''+py'+qy=r $

$ y(x)=y_{h}(x)+y_{p}(x)=(c_{1}y_{1}(x)+c_{2}y_{2}(x))+y_{p}(x) $

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

 

$ y_{p}=uy_{1}+vy_{2} $

$ y_{p}'=u'y_{1}+uy_{1}'+v'y_{2}+vy_{2}'=uy_{1}'+vy_{2}'(u'y_{1}+v'y_{2}=0) $

$ y_{p}''=u'y_{1}'+uy_{1}''+v'y_{2}'+vy_{2}'' $

$ (u'y_{1}'+uy_{1}''+v'y_{2}'+vy_{2}'')+p(uy_{1}'+vy_{2}')+q(uy_{1}+vy_{2}) $

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

   $ \Rightarrow u'y_{1}'+v'y_{2}'=r $: ㄱ

   $ u'y_{1}+v'y_{2}=0 $: ㄴ

ㄱ($ -y_{2} $)+ㄴ($ y_{2}' $): $ u'(y_{1}y_{2}'-y_{2}y_{1}')=-y_{2}r \rightarrow u'W=-y_{2}r $

ㄱ($ y_{1} $)+ㄴ($ -y_{a}'' $): $ u-v'(y_{1}y_{2}'-y_{2}y_{1}')=y_{1}r \rightarrow v'W=y_{1}r $

$ \Rightarrow u=-\int \frac{y_{2}r}{W}dx, v=\int \frac{y_{1}r}{W}dx $

$ \therefore y_{p}=uy_{1}+vy_{2}=-y_{1}\int \frac{y_{2}r}{W}dx+y_{2}\int \frac{y_{1}r}{W}dx $

 

ex) $ x^{2}y''+xy'-9y=48x^{5} $

$  y=y_{h}+y_{p} $

i) $ y_{h} $: $ y=c_{1}x^{3}+c_{2}x^{-3} $

ii) $ y_{p} $: $ W(x^{3},x^{-3})=\begin{vmatrix}
x^{3} & x^{-3}\\ 
3x^{2} & -3x^{-4}
\end{vmatrix}=-3x^{-1}-3x^{-1}=-6x^{-1} $

   $ y''+\frac{1}{x}y'-\frac{9}{x^{2}}y=48x^{3}, r=48x^{3} $

   $ y=uy_{1}+vy_{2} $

   $ u=-\int \frac{y_{2}r}{W}dx=-\int \frac{x^{-3}\cdot 48x^{3}}{-6x^{-1}}dx=8\int x dx=4x^{2} $

   $ v=\int \frac{y_{1}r}{W}dx=\int \frac{x^{3}\cdot 48x^{3}}{-6x^{-1}}dx=-8\int x^{7} dx=-x^{8} $

   $ \Rightarrow y_{p}=(4x^{2})x^{3}+(-x^{8})x^{-3}=3x^{5} $

$ \therefore y=y_{h}+y_{p}=c_{1}x^{3}+c_{2}x^{-3}+3x^{5} $