Solutions Hmwk 3

# Solutions Hmwk 3 - ME 391 Homework 3 Assigned Feb 15 2008...

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Unformatted text preview: ME 391 Homework 3 Assigned: Feb. 15, 2008 Due: Feb. 22, 2008 Problem 1: Find the General Solution of the Constant Coefﬁcient Diff. Eq.: (5 pts) , I‘ AM m1+4m+5zo 2‘ Cow « 2.x ix 9‘;C|e CDSX 4’ CLC SKIAX Problem 2: Determine the General Solution of: (5 PtS) H — 8 ‘ i _ 1 (4 ‘1 + b («.0 {30m 2%‘- M — 8m+lbco (m — 4')?” > M‘ r w; 4 ’Zécz/l at 8664/1 4 4x Vt 5 C45 {- CLC Problem 3: Solve the higher order Diff. Eq. (5 pts) 1 N L’] “Low” +H ' -—(g :0 Avx 53 Mg‘loml+ltl\m-b‘1=0 Problem 4: Solve the Diff Eq with Ungetermined Coefﬁcients: 1 5 t A . - — :0 (PS) Wang-.6 => Auxéb. m ML V <M+{\ (M'L\;O M;”‘\ M: Z ‘36.; QIC'X'+C’ZCLX W, mt log \vx «Corw Ac,“ (g’=3f—\e«3x H :D (M w x ‘4"10” 3? - Eiﬁl—J (10 pts) _ ' ’x ‘LX :5 ﬂbgcp +cle r 4;} {89 U32“ ‘92 (0» gram A11+Bx + (- ‘8’? ZAX+ % a": ZA €09,9an W” 5‘? ZA'(ZP~=X 4— EA ~ 2 CAXﬁ-Ex “VCB : 4Y2” Q (ﬂy-Jr 62A ~ mm + (m — a I ac) Ax; »2A:4 «1A‘1&=o ZA'B’zch) Azi, éleczj ‘6‘): ’ZX'LVZK’B , 2x (39 (ac/«:39: Q6 )6 +026 , ZXL+ZX’3 ,. 1'0 We) : c‘e Kale ~ 2(o\l+1(o\~—Z> A <0r> as”; =4 1(0) r: “(L 3’0 2 CD0 ‘ a ( 4' (17’ <or>‘c. 4-ch22" SOLUlNé 3mm); TAHCOOBLya C‘aZ I C221 . ~X 2x 3' ‘aT-Ze +18 ' ZKZ+2¢X”3 W ...
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Solutions Hmwk 3 - ME 391 Homework 3 Assigned Feb 15 2008...

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