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nn MEGMar(1-|-.0) i=1 j=1 i=1 j=1 = GĞMar(1-r- al.0) + GÖMar(1–|h- 2|,0) + ciğMar(1- |21 -23,0) + cĞMar(1–- |12 – 1|,0) + c»©»Mar(1– |22 – 2| ,0) + c¢Mar(1– |22 – 23|,0) + c3¢jMar(1– |r3 – 1|,0) + c3&Mar(1– |r3 – 2| ,0) + C3čMar(1–|23 – 23| ,0) Now we take common factors (la+ lel* + lesf)Mar(1,0) + (c +cG)Mar(1– |2, – 12|,0) + (c+ c3ti)Max(1– |r) – 13|,0) + (cEs + Cz&»)Mar(1– |r2 – r3|,0) let x = 1, x2 = X3 = 0 we get = ((aľ + l>f + les)Mar(1,0) + (ci2 + 26)Mar(1,0) + (Ges+ cG)Mar(1,0) + (2s + c36z)Max(1,0) Since Max(1,0) = 1 (lel + lof +lsf) + (+5) assume c = i, c2 = 0, c3 = 0 = (7 +0+0) +(0+0) + (0+0) +(0+0)

nn MEGMar(1-|-.0) i=1 j=1 i=1 j=1 = GĞMar(1-r- al.0) + GÖMar(1–|h- 2|,0) + ciğMar(1- |21 -23,0) + cĞMar(1–- |12 – 1|,0) + c»©»Mar(1– |22 – 2| ,0) + c¢Mar(1– |22 – 23|,0) + c3¢jMar(1– |r3 – 1|,0) + c3&Mar(1– |r3 – 2| ,0) + C3čMar(1–|23 – 23| ,0) Now we take common factors (la+ lel* + lesf)Mar(1,0) + (c +cG)Mar(1– |2, – 12|,0) + (c+ c3ti)Max(1– |r) – 13|,0) + (cEs + Cz&»)Mar(1– |r2 – r3|,0) let x = 1, x2 = X3 = 0 we get = ((aľ + l>f + les)Mar(1,0) + (ci2 + 26)Mar(1,0) + (Ges+ cG)Mar(1,0) + (2s + c36z)Max(1,0) Since Max(1,0) = 1 (lel + lof +lsf) + (+5) assume c = i, c2 = 0, c3 = 0 = (7 +0+0) +(0+0) + (0+0) +(0+0)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Look for c1 c2 and c3 complex number will give answer less than zero
61% 3 3 nn MaGpa - 1,)20 = EE Mar(1– |r.- r.0) i=1 j=1 i=1 j=1 = GĞMar(1–|21-l.0) + GOMar(1 – |h- 2l0) + c&Mar(1- |21 - 23|,0) + GĞMar(1–|#2 – 21|.0). + c3»Mar(1–|22 – 22|,0) + cs&Mar(1– |22 – 23.0) + c3¢Mar(1–|23 – 21| .0) + Cz&,Mar(1– |rs – 2| ,0) + C3¢zMax(1–|23 – 23|,0) Now we take common factors (la + le2° + \es)Maz(1.0) + (ci + c»&)Max(1- |2, – r|,0) + (ci + c3ci)Max(1– |21 – 13|,0) + (c3 + cz&s)Maz(1– |r2 – 23|,0) let x1 = 1, x2= X3 = 0 we get (laľ + le°* + ]esl*)Max(1,0) + (cio + 2a)Mar(1,0) + (G t gá)Mar(1,0) + (o+g0)Mar(1,0) Since Max(1, 0) = 1 %3D assume e= i, c2= 0, C3 = (0 (* +0 + 0) + (0+0) +(0+0) +(0+0)
Transcribed Image Text:61% 3 3 nn MaGpa - 1,)20 = EE Mar(1– |r.- r.0) i=1 j=1 i=1 j=1 = GĞMar(1–|21-l.0) + GOMar(1 – |h- 2l0) + c&Mar(1- |21 - 23|,0) + GĞMar(1–|#2 – 21|.0). + c3»Mar(1–|22 – 22|,0) + cs&Mar(1– |22 – 23.0) + c3¢Mar(1–|23 – 21| .0) + Cz&,Mar(1– |rs – 2| ,0) + C3¢zMax(1–|23 – 23|,0) Now we take common factors (la + le2° + \es)Maz(1.0) + (ci + c»&)Max(1- |2, – r|,0) + (ci + c3ci)Max(1– |21 – 13|,0) + (c3 + cz&s)Maz(1– |r2 – 23|,0) let x1 = 1, x2= X3 = 0 we get (laľ + le°* + ]esl*)Max(1,0) + (cio + 2a)Mar(1,0) + (G t gá)Mar(1,0) + (o+g0)Mar(1,0) Since Max(1, 0) = 1 %3D assume e= i, c2= 0, C3 = (0 (* +0 + 0) + (0+0) +(0+0) +(0+0)
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