EGM6321 - Principles of Engineering Analysis 1, Fall 2009 Mtg 17: Thur, 10Oct09 Linearity ⇒     {\displaystyle \Rightarrow \ \ } superposition y = y H + y P   {\displaystyle y=y_{H}+y_{P}\ } Homogeneous Solution y H   {\displaystyle y_{H}\ } - Euler Equations - Trial solution (undefined coefficient) - Reduction of order method 2: Undetermined factor Homogeneous L2_ODE_VC: cf. Eq. (1) P.3-1 Where a 0 ( x )   {\displaystyle a_{0}(x)\ } can be substituted for a 1 ( x )   {\displaystyle a_{1}(x)\ } or a 0 y   {\displaystyle a_{0}y\ } in Eq(1) Given one homogeneous solution u 1 ( x )   {\displaystyle u_{1}(x)\ } known Find second homogeneous solution u 2 ( x )   {\displaystyle u_{2}(x)\ } such that Where k 1 , k 2   {\displaystyle k_{1},k_{2}\ } are constants Assume full homogeneous solution Where U ( x )   {\displaystyle U(x)\ } is an unknown to be determined Where u 1 ( x )   {\displaystyle u_{1}(x)\ } is known "Full" = includes u 2 ( x )   {\displaystyle u_{2}(x)\ } Add the following: a 0 ( x ) [ y = U u 1 ]   {\displaystyle a_{0}(x)\left[y=Uu_{1}\right]\ } and a 1 ( x ) [ y ′ = U u 1 ′ + U ′ u 1 ] {\displaystyle a_{1}(x)\left[y'=Uu_{1}'+U'u_{1}\right]} and [ y ′′ = U u 1 ′′ + 2 U ′ u 1 ′ + U ′′ u 1 ] {\displaystyle \left[y''=Uu_{1}''+2U'u_{1}'+U''u_{1}\right]} To get a 0 y + a 1 y ′ + y ′′ = U [ a 0 u 1 + a 1 u 1 ′ + u 1 ′′ ] + U ′ [ a 1 u 1 + 2 u 1 ′ ] + U ′′ u 1 = 0   {\displaystyle a_{0}y+a_{1}y'+y''=U\left[a_{0}u_{1}+a_{1}u_{1}'+u_{1}''\right]+U'\left[a_{1}u_{1}+2u_{1}'\right]+U''u_{1}=0\ } by Eq(1) p17-1 Reduce to u 1 ′′ + a 1 u 1 ′ + a 0 u 1 = 0   {\displaystyle u_{1}''+a_{1}u_{1}'+a_{0}u_{1}=0\ } Since u 1   {\displaystyle u_{1}\ } is a homogeneous solution ⇒   0 = U ′ ( a 1 u 1 + 2 u 1 ′ ) + U ′′ u 1   {\displaystyle \Rightarrow \ 0=U'(a_{1}u_{1}+2u_{1}')+U''u_{1}\ } , NOTE missing dependent variable U in front of U ′   {\displaystyle U'\ } term Let Z := U ′ ⇒     {\displaystyle Z:=U'\Rightarrow \ \ } homogeneous L1_ODE_VC for Z Solve for Z, - integration factorial method (HW) - Direct integration (because Eq(1) is homogeneous) ⇒   Z ′ Z + ( a 1 + 2 u 1 ′ u 1 ) = 0   {\displaystyle \Rightarrow \ {\frac {Z'}{Z}}+(a_{1}+{\frac {2u_{1}'}{u_{1}}})=0\ } Where a 1 , 2 u 1 ′ u 1   {\displaystyle a_{1},{\frac {2u_{1}'}{u_{1}}}\ } are known Integrate log ⁡ | Z | + 2 log ⁡ | u 1 | + ∫ x a 1 ( s ) d s = k {\displaystyle \log \left|Z\right\vert +2\log \left|u_{1}\right\vert +\int _{}^{x}a_{1}(s)\,ds=k} , where k is a constant where c = e k   {\displaystyle c=e^{k}\ } ⇒   U ( x ) = ∫ x c ( u 1 ( t ) ) 2 e − ∫ t a 1 ( s ) d s d t + c ~   {\displaystyle \Rightarrow \ U(x)=\int _{}^{x}{\frac {c}{(u_{1}(t))^{2}}}e^{-\int _{}^{t}a_{1}(s)\,ds}\,dt+{\tilde {c}}\ } where a  ̄   1 ( t ) = ∫ t a 1 ( s ) d s   {\displaystyle {\bar {a}}\ _{1}(t)=\int _{}^{t}a_{1}(s)\,ds\ } Homogeneous solution where c ~ = k 1   {\displaystyle {\tilde {c}}=k_{1}\ } and c = k 2   {\displaystyle c=k_{2}\ } HW: obtain Eq. (2) P.17-3 Z ( x )   {\displaystyle Z(x)\ } using the integrating factor method
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