P19m- P59m- (a1 + a3 + a5) P+ (a2 + a4) Q (B1 + B3 + B5) P + (82 + B4) Q ' (a1 + a3 + a5) Q+ (a2 + a4) P (B1 + B3 + B5) Q + (82 + B4) P ` (4.14) P = AQ+ Q = AP+ Consequently, we get A (B1 + B3 + Bs) PQ + A (32 + B4) Q² + (a1 + a3 + a5) P+(a2 + a4) Q, (B1 + B3 + B5) P² + (82 + B4) PQ (4.15) and A (B1 + 3 + B5) PQ + A (B2 + B4) P? + (a1 + a3 + a5) Q + (a2 + a4) P. (B1 + B3 + B5) Q² + (B2 + B4) PQ %3D (4.16) By subtracting (4.15) from (4.16), we obtain [(B1 + B3 + B5) + A (32 + B4)] (P² – Q²) = [(a1 + a3 + a5) – (a2 + a4)] (P –- Q).
P19m- P59m- (a1 + a3 + a5) P+ (a2 + a4) Q (B1 + B3 + B5) P + (82 + B4) Q ' (a1 + a3 + a5) Q+ (a2 + a4) P (B1 + B3 + B5) Q + (82 + B4) P ` (4.14) P = AQ+ Q = AP+ Consequently, we get A (B1 + B3 + Bs) PQ + A (32 + B4) Q² + (a1 + a3 + a5) P+(a2 + a4) Q, (B1 + B3 + B5) P² + (82 + B4) PQ (4.15) and A (B1 + 3 + B5) PQ + A (B2 + B4) P? + (a1 + a3 + a5) Q + (a2 + a4) P. (B1 + B3 + B5) Q² + (B2 + B4) PQ %3D (4.16) By subtracting (4.15) from (4.16), we obtain [(B1 + B3 + B5) + A (32 + B4)] (P² – Q²) = [(a1 + a3 + a5) – (a2 + a4)] (P –- Q).
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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