Mtg 29: Tue, 26 Oct10

HW5.5 p.28-3 : continued

2.8) Now solve L2_ODE_CC with ${\displaystyle f(t)=e^{(-t^{2})}\ }$ Gaussian distribution

Coefficients: ${\displaystyle a_{2},a_{1},a_{0}\ }$ : Find these such that L2_ODE_CC Eq.(2)p.27-2 accept as characteristic equation:

2.8.1) ${\displaystyle (r+1)(r-2)=0\ }$

2.8.2) ${\displaystyle (r-4)^{2}=0\ }$ END HW5.5

Variation of Parameters: (continued)

Instead of doing the homogeneous ( F09 and King ) cases Eq.(1)p.28-4 , Let's consider directly the non-homogeneous case, then recover the homogeneous as a particular case. So consider:

${\displaystyle y''+a_{1}(x)y'+a_{0}(x)y=f(x)\ }$

Given: ${\displaystyle u_{1}(x)\ }$ homogeneous solution

Find: ${\displaystyle u_{2}(x)\ }$ 2nd homogeneous solution

${\displaystyle y_{P}(x)\ }$ particular solution

Method

${\displaystyle \displaystyle {\begin{aligned}y(x)=U(x)u_{1}(x)\end{aligned}}}$

(1)

Where ${\displaystyle U(x)\ }$ is unknown and ${\displaystyle u_{1}(x)\ }$ is known

Add the following equations:

${\displaystyle \left[y=Uu_{1}\right]a_{0}(x)\ }$

${\displaystyle \left[y'=Uu_{1}'+U'u_{1}\right]a_{1}(x)\ }$

${\displaystyle \left[y''=Uu_{1}''+2U'u_{1}'+U''u_{1}\right]a_{2}(x)\ }$ , where ${\displaystyle a_{2}(x)=1\ }$

After summing: ${\displaystyle a_{0}y+a_{1}y'+y''=U\left[a_{0}u_{1}+a_{1}u'_{1}+a_{1}''\right]+U'\left[a_{1}u_{1}+2u_{1}'\right]+U''u_{1}=f(x)\Rightarrow \ \ }$ L2_ODE_VC with missing dependeant variable ${\displaystyle U\ }$

Where ${\displaystyle \left[a_{0}u_{1}+a_{1}u'_{1}+a_{1}''\right]=0\ }$ , thus ${\displaystyle u_{1}\ }$ is a homogeneous solution.

Order reduction method ${\displaystyle \theta \ \ }$ : ${\displaystyle Z:=U'\ }$

${\displaystyle {\tilde {a}}_{1}(x)z'+{\tilde {a}}_{0}(x)z={\frac {f(x)}{u_{1}(x)}}\ }$

Where ${\displaystyle {\tilde {a}}_{1}(x)=1\ }$ and ${\displaystyle {\tilde {a}}_{0}(x)={\frac {\left[a_{1}(x)u_{1}(x)+2u'_{1}(x)\right]}{u_{1}(x)}}\ }$

Solve for ${\displaystyle Z(x)\ }$ given by Eq.(6)p.10-3 with ${\displaystyle h(x)\ }$ given by Eq.(1)p.10-3 .

Eq.(1)p.10-3 : ${\displaystyle h(x)=e^{\left[\int _{}^{}{\tilde {a}}_{0}(x)\,dx\right]}=e^{\left[\int _{}^{}\left\{a_{1}(x){\frac {2u_{1}'(x)}{u_{1}(x)}}\right\}\,dx\right]}\ }$

Where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikiversity.org/v1/":): {\displaystyle \left [ \int_{}^{} \left \{ a_1(x)\frac{2u_1'(x)}{u_1(x)} \right \}\,dx \right ] = \int_{}^{} a_1(x)\,dx+2 \log {u_1(x)} \ } and ${\displaystyle 2\log {u_{1}(x)}=\log {u_{1}^{2}(x)}\ }$

${\displaystyle \displaystyle {\begin{aligned}h(x)=u_{1}^{2}(x)e^{\left[\int _{}^{}a_{1}(x)\,dx\right]}\end{aligned}}}$

(1)

From Eq.(6)p.10-3: ${\displaystyle Z(x)={\frac {1}{h(x)}}\left[k_{2}+\int _{}^{}h(x)f(x)\,dx\right]\ }$ , where ${\displaystyle Z(x)=U'(x)\ }$ and ${\displaystyle f(x)=u_{1}(x)\ }$

${\displaystyle U(x)=k_{1}+\int _{}^{}Z(x)\,dx\ }$

From Eq.(1)p.29-2: ${\displaystyle y(x)=U(x)u_{1}(x)\ }$

${\displaystyle \displaystyle {\begin{aligned}y(x)=k_{1}u_{1}(x)+k_{2}u_{1}(x)\int _{}^{}{\frac {1}{h(x)}})\,dx+y_{P}\end{aligned}}}$

(2)

Where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikiversity.org/v1/":): {\displaystyle u_1(x) \int_{}^{} \frac{1}{h(x)})\, dx=u_2(x) \ }