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and L_:(2)= ¿L_;(2)+1+ Lo(2) (27) Because the Y,(1) are easy to calculate [Y,(1)=1 for all p], one immediately concludes from (26) that the general solution of (26) is (1– (1 – 42)/2\o+1 (1+(1– 42)2\+1 + A2(2) | " (28) L,= A,(2) 2 2 The problem now is to determine the constants A(2) and A2(i). From the definition (25), it is clear that for any p the first term in the sum L,(2) is proportional to i?*'. This is compatible with the result (28) only if A2{2) =0 (29) The expression of A,(1) is then easy to find from the boundary condition (27) and one ends up with - (1 - 42)/2\P+3 1 L„(1)= (- (30) 2

and L_:(2)= ¿L_;(2)+1+ Lo(2) (27) Because the Y,(1) are easy to calculate [Y,(1)=1 for all p], one immediately concludes from (26) that the general solution of (26) is (1– (1 – 42)/2\o+1 (1+(1– 42)2\+1 + A2(2) | " (28) L,= A,(2) 2 2 The problem now is to determine the constants A(2) and A2(i). From the definition (25), it is clear that for any p the first term in the sum L,(2) is proportional to i?*'. This is compatible with the result (28) only if A2{2) =0 (29) The expression of A,(1) is then easy to find from the boundary condition (27) and one ends up with - (1 - 42)/2\P+3 1 L„(1)= (- (30) 2

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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Can you substitude equation 28 into equation 27 and try and achieve equation 30.

and L-1(2)= ¿L_,(2)+ 1+ Lo(2) (27) Because the Y,(1) are easy to calculate [Y,(1)=1 for all p], one immediately concludes from (26) that the general solution of (26) is 1+(1– 42)/2\P+1 (28) p+1 (1– 42)'/2) L, = 4,(4)(- A2(2) (T +. 2 The problem now is to determine the constants A,(2) and A2(i). From the definition (25), it is clear that for any p the first term in the sum L,(2) is proportional to iP+!. This is compatible with the result (28) only if A2(2) =0 (29) The expression of A,(2) is then easy to find from the boundary condition (27) and one ends up with ´1-(1– 42)/2\P+3 1 L„(1) =( | wwww (30)
Transcribed Image Text:and L-1(2)= ¿L_,(2)+ 1+ Lo(2) (27) Because the Y,(1) are easy to calculate [Y,(1)=1 for all p], one immediately concludes from (26) that the general solution of (26) is 1+(1– 42)/2\P+1 (28) p+1 (1– 42)'/2) L, = 4,(4)(- A2(2) (T +. 2 The problem now is to determine the constants A,(2) and A2(i). From the definition (25), it is clear that for any p the first term in the sum L,(2) is proportional to iP+!. This is compatible with the result (28) only if A2(2) =0 (29) The expression of A,(2) is then easy to find from the boundary condition (27) and one ends up with ´1-(1– 42)/2\P+3 1 L„(1) =( | wwww (30)
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