Second-Order Linear ODE (6) - Nonhomogeneous ODE

10. Definition and Theorems

 

$ y''+p(x)y'+q(x)y=r(x) $

 

Definition)

A general Solution of $ y''+p(x)y'+q(x)y=r(x) $

$ \Rightarrow y=y_{h}+y_{p}=(c_{1}y_{1}+c_{2}y_{2})+y_{p}  $ ($ y_{h} $: solution of $ y''+py'+qy=0 $)

 

Theorem 1)

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

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

i) a의 해 + b의 해 = b의 해

ii) b의 해 - b의 해 = a의 해

 

$ Y_{1} $: a의 해 $ \rightarrow Y_{1}''+pY_{1}'+qY_{1}=0 $

$ Y_{2} $: b의 해 $ \rightarrow Y_{2}''+pY_{2}'+qY_{2}=r $

$ Y_{3} $: b의 해 $ \rightarrow Y_{3}''+pY_{3}'+qY_{3}=r $

$ Y_{1}+Y_{2} \rightarrow y''+py'+qy=r $

$ \Rightarrow (Y_{1}+Y_{2})''+p(Y_{1}+Y_{2})'+q(Y_{1}+Y_{2})=(Y_{1}''+pY_{1}'+qY_{1})+(Y_{2}''+pY_{2}'+qY_{2})=r $

$ Y_{2}-Y_{3} \rightarrow y''+py'+qy=r $

$ \Rightarrow (Y_{2}-Y_{3})''+p(Y_{2}-Y_{3})'+q(Y_{2}-Y_{3})=(Y_{2}''+pY_{2}'+qY_{2})-(Y_{3}''+pY_{3}'+qY_{3})=0 $

 

Theorem 2)

p, q, r: continuous on I

Then exery solution of $ y''+p(x)y'+q(x)y=r(x) $ ← assigning suitable values to $ c_{1}, c_{2} $ in $ y=y_{h}+y_{p}=(c_{1}y_{1}+c_{2}y_{2})+y_{p} $

 

 

11. Method of Undetermined Coefficients

 

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

$  y=y_{h}+y_{p} $

 

1) Basic Rule: $ y_{p} $ is related to $ r(x) $

$ r(x) $	$ y_{p} $
$ ke^{\gamma x} $	$ ce^{\gamma x} $
$ kx^{n} (n=0,1,...) $	$ c_{n}x^{n}+c_{n-1}x^{n-1}+...+c_{1}x+c_{0} $
$ k\cos \omega x $	$ A\cos \omega x+B\sin \omega x $
$ k\sin \omega x $
$ ke^{\alpha x}\cos \omega x $	$ ke^{\alpha x}(A\cos \omega x+B\sin \omega x) $
$ ke^{\alpha x}\sin \omega x $

 

2) Modification Rule: If $ y_{p} $ is included in $ y_{h} $, then multiply $ y_{p} $ by $ x $ or $ x^{2} $

 

3) Sum Rule: If $ r(x) $ is a sum of functions, then choose for $ y_{p} $ the sum

 

ex) $ y''+3y'+2.25y=-10e^{-1.5x} $

$ y=y_{h}+y_{p} $

i) $ y_{h} $: $ y=(c_{1}+c_{2}x)e^{-1.5x} $

ii) $ y_{p} $: $ y=\alpha x^{2}e^{-1.5x} $

   $ y'=2\alpha xe^{-1.5x}-1.5\alpha x^{2}e^{-1.5x} $

   $ y''=2\alpha e^{-1.5x}-3\alpha xe^{-1.5x}-3\alpha xe^{-1.5x}+2.25\alpha x^{2}e^{-1.5x} $

   $ y''+3y'+2.25y $

   $ =e^{-1.5x}(2\alpha -6\alpha x+2.25\alpha x^{2})+e^{-1.5x}(6\alpha x-4.5\alpha x^{2})+e^{-1.5x}(2.25\alpha x^{2}) $

   $ =2\alpha e^{-1.5x}=-10e^{-1.5x} $

   $ \rightarrow \alpha =-5 $

   $ \rightarrow y_{p}=-5x^{2}e^{-1.5x} $

$ \therefore y=y_{h}+y_{p}=(c_{1}+c_{2}x)e^{-1.5x}-5x^{2}e^{-1.5x} $