User:EGM6341.s11.TEAM1.WILKS/Mtg29

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_10_26_15_07_13.djvu | Mtg 29:]] Tue, 26 Oct 10

= Application: Nonhomogeneous L2-ODE-CC cont'd =

[[media: 2010_10_26_15_07_13.djvu | Page 29-1]]
[[media: 2010_10_21_14_48_38.djvu | HW5.5 p.28-3]] : continued 2.8) Now solve the nonhomogeneous L2-ODE-CC with $$ f(t)=e^{(-t^2)} \ $$, i.e., Gaussian distribution. Coefficients: $$ a_2, a_1, a_0 \ $$ : Find these coefficients such that the L2-ODE-CC [[media: 2010_10_21_14_05_59.djvu | Eq.(2)p.27-2]] accepts as characteristic equation: 2.8.1) $$ (r+1)(r-2)=0 \ $$ 2.8.2) $$ (r-4)^2=0 \ $$

END HW5.5

= Variation of parameters: Reduction-of-order method 2 (cont'd) =

Instead of doing the homogeneous case in [[media: 2010_10_21_14_48_38.djvu | Eq.(1)p.28-4]] (see F09 Mtgs [[media: Egm6321.f09.mtg17.djvu | 17]], [[media: Egm6321.f09.mtg18.djvu | 18]], [[media: Egm6321.f09.mtg19.djvu | 19]], and King ), let's consider directly the non-homogeneous case, then recover the homogeneous as a particular case.

So consider: $$ y''+a_1(x)y'+a_0(x)y=f(x) \ $$ Given: $$ u_1(x) \ $$ homogeneous solution Find: $$ u_2(x) \ $$ 2nd homogeneous solution

[[media: 2010_10_26_15_07_13.djvu | Page 29-2]]
$$ y_P(x) \ $$ particular solution

Method
where $$ U(x) \ $$ is unknown and $$ u_1(x) \ $$ is known Add the following equations: $$ \left [ y=Uu_1 \right ] \times a_0(x) \ $$ $$ \left [ y'=Uu_1'+U'u_1 \right ] \times a_1(x) \ $$ $$ \left [ y=Uu_1+2U'u_1'+U''u_1 \right ] \times a_2(x) \ $$, where $$ a_2(x)=1 \ $$ After summing: $$ a_0y+a_1y'+y= U \left [ a_0 u_1 + a_1 u'_1 + u_1 \right ] + U' \left [ a_1u_1+2u_1' \right ] +U''u_1 = f(x) \Rightarrow \ \ $$ L2-ODE-VC with missing dependent variable $$ U \ $$, because $$ \left [ a_0u_1+a_1u'_1+u_1'' \right ] = 0 \ $$ since $$ u_1 \ $$ is a homogeneous solution. Order-reduction method 0: $$ Z:=U' \ $$ $$ \tilde{a}_1(x) Z'+\tilde{a}_0(x) Z = \frac{f(x)}{u_1(x)} \ $$ where $$ \tilde{a}_1(x)=1 \ $$ and $$ \tilde{a}_0(x)= \frac{ \left [ a_1(x)u_1(x)+2u'_1(x) \right ] }{u_1(x)} \ $$.

[[media: 2010_10_26_15_07_13.djvu | Page 29-3]]
Solve for $$ Z(x) \ $$ given by [[media: 2010_09_14_15_00_52.djvu | Eq.(6)p.10-3]] with $$ h(x) \ $$ given by [[media: 2010_09_14_15_00_52.djvu | Eq.(1)p.10-3]]. [[media: 2010_09_14_15_00_52.djvu | Eq.(1)p.10-3]] : $$ h(x) = \exp \left[ \int \tilde{a}_0(x) \, dx \right] = \exp \left[ \int \left\{ a_1(x) + \frac{2u_1'(x)}{u_1(x)} \right\} \, dx \right] \ $$

and

$$ \int \frac{2u_1'(x)}{u_1(x)}\,dx = 2 \log {u_1(x)}  = \log {u_1^2(x)} \ $$

Thus

From [[media: 2010_09_14_15_00_52.djvu | Eq.(6)p.10-3]]:

$$ Z(x) = \frac{1}{h(x)} \left[ k_2 + \int h(x) \frac{f(x)}{u_1(x)} \, dx \right] \ $$ ,

Since

$$ Z(x)=U'(x) \ $$

we have

$$ U(x) = k_1+ \int_{}^{} Z(x)\, dx \ $$

From [[media: 2010_10_26_15_07_13.djvu | Eq.(1)p.29-2]]: $$ y(x) = U(x)u_1(x) \ $$

where $$ u_2(x) = u_1(x) \int \frac{1}{h(x)} \, dx \ $$.