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Intro Scientific Comp

by: Vern Jacobi

Intro Scientific Comp CS 340

Vern Jacobi
GPA 3.54

Erin McNelis

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Erin McNelis
Class Notes
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This 3 page Class Notes was uploaded by Vern Jacobi on Thursday October 29, 2015. The Class Notes belongs to CS 340 at Western Carolina University taught by Erin McNelis in Fall. Since its upload, it has received 10 views. For similar materials see /class/230968/cs-340-western-carolina-university in ComputerScienence at Western Carolina University.

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Date Created: 10/29/15
1 Section 61 Linear Systems of Equations 11 Applications e Let h hepheeehht the ouhheht hhh empehee end 7 the heehetehhoe eh ohme then the voltege dhop E foh eeoh heehetoh he gweh by Ohm s law E 2R Kuchoff s cuuent and voltage xu1 Indicate 1 L w l h h h he the ehhm of e11 ouhhehte ehtehhhg e node muet bezexo 22 7 0 2 Ahohhhhd every closed loop the elgehheho ehhm of the voltege muet equel the elgehheho ehhm of the voltege dhope h e the elgehheho ehhhn of the potehthel ohffehehoee thet he voltege ohehgee hhh ehy loop muet equel zeho 257 DR 7 0 x um 711 1 K511 e kym R In Khhohoee ouhheht hhh1e he epphed et eeoh node to yxeld 212 152za2 Apphoethoh of the voltege hhh1e to each of the two loops ghvee 715151 7 5m 710 1015 72m65 710 5m 200 Thexefoxe the pxoblem emohhhhte to eohhhg the followmg 1 1 0 0 0 t 0 0 71 0 1 71 0 t5 0 0 0 71 0 0 1 m 7 0 0 0 1 71 155 0 0 10 710 0 715 75 151 0 5 710 0 72 0 0 the 200 e Belehohhg Chemhoel Equethohe Example eco2 0211202 7 302 AC H os yields the following system of equations CC 1 64 O 21 22 23 64 H2 2mg 124 12 Gaussian Elimination with Backward Substitution Algorithm Purpose To solve the n x 71 linear system E1 3 111961 0112962 H aina n a1n1 E2 3 121961 0122962 127957 a2n1 E3 3 a31961 0132962 013mm a3n1 En 3 M1951 0172902 ammn ann1 121 Three Allowable Row Operations in Gaussian Elimination H Multiplying an equation by a nonzero constant denoted a El Multiplying an equation by a nonzero constant then added to a second equation to replace the second equation denoted Ej a E1 to Swapping rows denoted lt gt OJ 2 Differential Equations Consider solVing the initial value problem mow lt1 ylt0gt 1 lt2 Because this is a separable problem7 we can solve it by rst collecting the y and t terms and integrating 7 4m M372 1 7y 4t dt y3 i y We need partial fraction decomposition to rewrite f i 243 1 y 37y 1 A3yBy 1 7ABy3A Equating coefficients we get the system 7AB 0 1 3A 1 A 7 5 3 and B A 1 3 Thus returning to equation our integrals d7y 4tdt 3373 13 13 LLdy 2t2C y 3 3 1 1 2 glnlyliglnl3iyl 7 2t 0 Using the initial value to solve for C we nd 1 g11111141113711 202C 1 73 ln2 0 Thus 1 y 7 2 1 1 ln Biy 7 2t 3ln2 y 7 2 ln 7 6t 7 ln2 34 If we assume 0 lt y lt 3 then y gt 0 so 3 y With this in mind we can exponentiate y i y i 34 both sides of the above equation and nd 24 Bettina 562 571119 1562 3 7 y 2 1 6t2 y 3 i 2055 3 y 7569 7 56t2 3 y 53t2 5562 1 3 24 1 556 556 Thus 356t2 y 2 56t2


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