Multiplying the denominator and numerator of (4.24) by 4 [(B1 + B3 + B5) + A (B2 + B4)]? we get 4[A(B1+B3+95)-(82+B4)][S])[(8)+B3+Bs)(a2+a4)+A(B2+B4) (a,+az+as)] (B2+B4)-(B1+ß3+B3))(A+1)]| Y1 - P = S A (B2 + Ba) (S1 - 8)² – (B1 + B3 + B5) (Sı + 8)² S 2[(81 + B3 + B5) + A (B2 + B4)] (a1+a3 + a5) (S1 + 8) + S 2[(B1 + B3 + B5)+ A (B2 +Ba)] (a2 + a4) (S1 - 8) S 4[A(81+B3+B5)-(82+B4)][S1][(81+B3+Bs)(az+a4)+A(82+B4) (a1+ag+as)] (B2+B4)-(B1+ß3+Bs))(A+1)] S A (32 + B4) [S1]² – (B1 + B3 + Bs) [S1]² + S [A(B2 + B4) - (B1 + B3 + B5)]8² | +
Multiplying the denominator and numerator of (4.24) by 4 [(B1 + B3 + B5) + A (B2 + B4)]? we get 4[A(B1+B3+95)-(82+B4)][S])[(8)+B3+Bs)(a2+a4)+A(B2+B4) (a,+az+as)] (B2+B4)-(B1+ß3+B3))(A+1)]| Y1 - P = S A (B2 + Ba) (S1 - 8)² – (B1 + B3 + B5) (Sı + 8)² S 2[(81 + B3 + B5) + A (B2 + B4)] (a1+a3 + a5) (S1 + 8) + S 2[(B1 + B3 + B5)+ A (B2 +Ba)] (a2 + a4) (S1 - 8) S 4[A(81+B3+B5)-(82+B4)][S1][(81+B3+Bs)(az+a4)+A(82+B4) (a1+ag+as)] (B2+B4)-(B1+ß3+Bs))(A+1)] S A (32 + B4) [S1]² – (B1 + B3 + Bs) [S1]² + S [A(B2 + B4) - (B1 + B3 + B5)]8² | +
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Show me the steps of determine blue and inf is here. Explain step by step and i need every details.
![(a1 + a3 + a5) P+(@2+a4) Q
(B1 + B3 + B5) P + (82 + B4) Q
[A (B1 + B3 + B3) – (B2 + Ba)]PQ + A (B2 + B4) Q² – (B1 + B3 + Bs) P2
Y1 – P = AQ +
- P
(B1 + Вз + B5) Р+ (B2 + Ba) Q
(а1 + аз + о5) Р+ (02 + ад) Q
(B1 + B3 + B5) P+(82+ B4) Q
[A(B1+B3+B5)-(B2+B4)][S1][(B1+B3+B5)(a2+a4)+A(B2+B4) (@1+a3+as)]
[(81+33+B5)+A(B2+34)]²[((82+B4)-(81+B3+B5))(A+1)]
(B1 + B3 + B5)
[(a1+a3+a5)-(a2+a4)]+8`
2[(B1+B3+B5)+A(82+B4)]
[(a1+a3+a5)-(a2+a4)]-8
+ (B2 + B4) (2(31+33+B5)+A(B2+Ba)]
A (B2 + Ba) (ataz+a5)-(a2ta4)]-8'
2[(B1+33+B5)+A(B2+B4)]
2
[(@1+a3+a5)-(ag+a4)]+8`
2[(B1+33+B5)+A(ß2+B4)]
[(ai+a3+a5)-(a2+a4)]-8
(B1 + B3 + B5) (
(B1 + B3 + B5)
[(a1+a3+a5)-(a2+a4)]+8`
2[(B1+33+ß5)+A(B2+34)]
)+ (B2 + B4) ( 2(3+33+B5)+A(B2+B4)]
[(@1+a3+a5)-(a2+a4)]+8
2[(B1+B3+B5)+A(B2+B4)]
[(a1+a3+a5)-(a2+a4)]+ô`
2[(81+33+B5)+A(ß2+B4)]
([(@1+a3+a5)-(az+a4)]-8`
2[(B1+B3+B3)+A(B2+B4)]
[(a1+a3+a5)-(a2+a4)]-8
2[(B1+B3+B5)+A(B2+B4)]
(a1 + az + a5)
+ (a2 + a4)
(B1 + B3 + B5)
+ (B2 + B4)
(4.24)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2Fb8732753-eaba-47c5-bee4-b474ea333ffc%2F7o5nltf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(a1 + a3 + a5) P+(@2+a4) Q
(B1 + B3 + B5) P + (82 + B4) Q
[A (B1 + B3 + B3) – (B2 + Ba)]PQ + A (B2 + B4) Q² – (B1 + B3 + Bs) P2
Y1 – P = AQ +
- P
(B1 + Вз + B5) Р+ (B2 + Ba) Q
(а1 + аз + о5) Р+ (02 + ад) Q
(B1 + B3 + B5) P+(82+ B4) Q
[A(B1+B3+B5)-(B2+B4)][S1][(B1+B3+B5)(a2+a4)+A(B2+B4) (@1+a3+as)]
[(81+33+B5)+A(B2+34)]²[((82+B4)-(81+B3+B5))(A+1)]
(B1 + B3 + B5)
[(a1+a3+a5)-(a2+a4)]+8`
2[(B1+B3+B5)+A(82+B4)]
[(a1+a3+a5)-(a2+a4)]-8
+ (B2 + B4) (2(31+33+B5)+A(B2+Ba)]
A (B2 + Ba) (ataz+a5)-(a2ta4)]-8'
2[(B1+33+B5)+A(B2+B4)]
2
[(@1+a3+a5)-(ag+a4)]+8`
2[(B1+33+B5)+A(ß2+B4)]
[(ai+a3+a5)-(a2+a4)]-8
(B1 + B3 + B5) (
(B1 + B3 + B5)
[(a1+a3+a5)-(a2+a4)]+8`
2[(B1+33+ß5)+A(B2+34)]
)+ (B2 + B4) ( 2(3+33+B5)+A(B2+B4)]
[(@1+a3+a5)-(a2+a4)]+8
2[(B1+B3+B5)+A(B2+B4)]
[(a1+a3+a5)-(a2+a4)]+ô`
2[(81+33+B5)+A(ß2+B4)]
([(@1+a3+a5)-(az+a4)]-8`
2[(B1+B3+B3)+A(B2+B4)]
[(a1+a3+a5)-(a2+a4)]-8
2[(B1+B3+B5)+A(B2+B4)]
(a1 + az + a5)
+ (a2 + a4)
(B1 + B3 + B5)
+ (B2 + B4)
(4.24)
![Multiplying the denominator and numerator of (4.24) by 4 [(81 + B3 + B5) + A (B2 + Ba)]?
we get
4[A(B1+B3+Bs)-(B2+B4)][S1][(31+B3+B5)(ag+a4)+A(B2+B4) (a1ta3+as)]
T((B2+B4)-(B1+B3+Bs))(A+1)]
Y1 – P =
A (B2 + Ba) (S1 – 6)² – (B1 + B3 + B5) (Si + 6)?
+
S
2[(81 + B3 + B5)+ A (B2+ B4)] (aı + a3 + a5) (S1 + 8)
S
2[(B1 + B3 + B5) + A (B2 + B4)] (a2 + a4) (S1 - 8)
S
4[A(81+B3+35)-(82+84)][S1][(B,+83+B5)(a2+a4)+A(82+B4) (a+a3+as)]
[((B2+84)-(81+B3+Bs))(A+1)]T
S
A (32 + B4) [Si]? – (B1 + 33 + B3) [Si]?
S
[A (32 + Ba) - (B1 + B3 + Bs)|8²
+
S
14
2[(81 + B3 + B5)+A (B2+B4)] (a2 + a4) [S1]
+
S
2[(B1 + B3 + B5) + A (B2 + Ba)] (a1 +a3 +a5) [S1]
S
2A (B2 + Ba) [(a1 + a3 + a5) – (a2 +a4)] &+2 (B1 + B3 + B5) [S1] &
S
2[(B1 + B3 + B5) + A (B2 + B4)] (a1 + a3 + a5) 6 – 2[(B1 + 3 + B5) + A (B2 + Ba)] (a2 + a4) 8
S
4[A(81+B3+Bs)-(32+B4)][S]((8,+83+Bs)(a2+a4)+A(B2+B4)(a1ta3+as)]
T((B2+B4)-(B1+B3+Bs))(A+1)]
4[S1][(B1+B3+B5)(a2+a4)+A(B2+B4)(a+a3+as)][A(82+B4)-(B1+83+B3)]
[((B2+B4)-(81+B3+Bs))(A+1)]
A (B2 + Ba) [S1]? – (B1 + B3 + B5) [S1]²
S
[A (B2 + B4) – (B1 + B3 + Bs) [S1]²
S
2[(B1 + B3 + B5)+ A (B2 + B4)] (a2 +a4) [S1]
S
2[(B1 + B3 + B5) + A (B2 + Ba)] (aı +a3 +a5) [S1]
2 [S1] [(B1 + B3 + B5) + A (B2 + Ba)]S
S
2 [S1] [(B1 + B3 + Bs) + A (B2 + Ba)]8
S
4 [S1] [(B1 + B3 + B5) (a2 + a4)+A (32+ B4) (a1 + a3 + a5)]
S
4 [S1] [(B1 + B3 + Bs) (a2+ a4) + A (B2 + BA) (a1 + a3 + a5)]
S
2[(a1 + a3 + as) - (a2 + a4)|[(P1 + B3 + Bs) + A (B2 + B4)]S
S
2[(a1 + a3 + a5) - (a2+ a4)][(81 + B3 + B5) + A (B2 + B4)]8
+
S
0.
where
S
2 [(B1 + B3 + Bs) + A (B2 + Ba)] ×
[(B1 + B3 + Bs) (Si + 8) + (B2 + Ba) (Sı – 8)]
15](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2Fb8732753-eaba-47c5-bee4-b474ea333ffc%2Frpz9te9_processed.png&w=3840&q=75)
Transcribed Image Text:Multiplying the denominator and numerator of (4.24) by 4 [(81 + B3 + B5) + A (B2 + Ba)]?
we get
4[A(B1+B3+Bs)-(B2+B4)][S1][(31+B3+B5)(ag+a4)+A(B2+B4) (a1ta3+as)]
T((B2+B4)-(B1+B3+Bs))(A+1)]
Y1 – P =
A (B2 + Ba) (S1 – 6)² – (B1 + B3 + B5) (Si + 6)?
+
S
2[(81 + B3 + B5)+ A (B2+ B4)] (aı + a3 + a5) (S1 + 8)
S
2[(B1 + B3 + B5) + A (B2 + B4)] (a2 + a4) (S1 - 8)
S
4[A(81+B3+35)-(82+84)][S1][(B,+83+B5)(a2+a4)+A(82+B4) (a+a3+as)]
[((B2+84)-(81+B3+Bs))(A+1)]T
S
A (32 + B4) [Si]? – (B1 + 33 + B3) [Si]?
S
[A (32 + Ba) - (B1 + B3 + Bs)|8²
+
S
14
2[(81 + B3 + B5)+A (B2+B4)] (a2 + a4) [S1]
+
S
2[(B1 + B3 + B5) + A (B2 + Ba)] (a1 +a3 +a5) [S1]
S
2A (B2 + Ba) [(a1 + a3 + a5) – (a2 +a4)] &+2 (B1 + B3 + B5) [S1] &
S
2[(B1 + B3 + B5) + A (B2 + B4)] (a1 + a3 + a5) 6 – 2[(B1 + 3 + B5) + A (B2 + Ba)] (a2 + a4) 8
S
4[A(81+B3+Bs)-(32+B4)][S]((8,+83+Bs)(a2+a4)+A(B2+B4)(a1ta3+as)]
T((B2+B4)-(B1+B3+Bs))(A+1)]
4[S1][(B1+B3+B5)(a2+a4)+A(B2+B4)(a+a3+as)][A(82+B4)-(B1+83+B3)]
[((B2+B4)-(81+B3+Bs))(A+1)]
A (B2 + Ba) [S1]? – (B1 + B3 + B5) [S1]²
S
[A (B2 + B4) – (B1 + B3 + Bs) [S1]²
S
2[(B1 + B3 + B5)+ A (B2 + B4)] (a2 +a4) [S1]
S
2[(B1 + B3 + B5) + A (B2 + Ba)] (aı +a3 +a5) [S1]
2 [S1] [(B1 + B3 + B5) + A (B2 + Ba)]S
S
2 [S1] [(B1 + B3 + Bs) + A (B2 + Ba)]8
S
4 [S1] [(B1 + B3 + B5) (a2 + a4)+A (32+ B4) (a1 + a3 + a5)]
S
4 [S1] [(B1 + B3 + Bs) (a2+ a4) + A (B2 + BA) (a1 + a3 + a5)]
S
2[(a1 + a3 + as) - (a2 + a4)|[(P1 + B3 + Bs) + A (B2 + B4)]S
S
2[(a1 + a3 + a5) - (a2+ a4)][(81 + B3 + B5) + A (B2 + B4)]8
+
S
0.
where
S
2 [(B1 + B3 + Bs) + A (B2 + Ba)] ×
[(B1 + B3 + Bs) (Si + 8) + (B2 + Ba) (Sı – 8)]
15
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