-4 [(b+ f) – (c+ r)] [(c+r) (d + g) + a (e + s) (b+ f)] K 2 [(b+ f) – (c+ r)] (2a (e+ s) (b+ f)+2(c+r) (d +g)) K 2 [a (e + s) + (d+ g)] [(b+ f) – (c + r)] 8 K 2 (a (e + s) + (d + g)) [(b+ f) – (c + r)] & K -4 [(b+ f) – (c + r)[c +r) (d + g) + a (e + s) (b+ f)] K 4 [(b+ f) – (c +r)[c+r) (d+ g) + a (e + s) (b+ f)] K 2 [a (e + s) + (d + g)] [(5+ f) – (c+ r)] & K 2 [a (e + s) + (d + g)] [(b+ f) – (c+ r)] ô K = 0. where K = 2 [a (e + s) + (d + g)] [(d + g) ([(b+ f) – (c + r)] - 8) + (e +s) ([(b+ f) – (c+r)] + 6)] Similarly, we can show that bro + cx-1 + fæ-2+ ræ_3 dro + ex–1+ gx_2+ sx_3 (b+ f) P+ (c+r)Q az, + aQ + = P. %3D (d+g) P+ (e+ s) Q By using the mathematical induction, we have In = Q and rn+1 = P, n > -4.
-4 [(b+ f) – (c+ r)] [(c+r) (d + g) + a (e + s) (b+ f)] K 2 [(b+ f) – (c+ r)] (2a (e+ s) (b+ f)+2(c+r) (d +g)) K 2 [a (e + s) + (d+ g)] [(b+ f) – (c + r)] 8 K 2 (a (e + s) + (d + g)) [(b+ f) – (c + r)] & K -4 [(b+ f) – (c + r)[c +r) (d + g) + a (e + s) (b+ f)] K 4 [(b+ f) – (c +r)[c+r) (d+ g) + a (e + s) (b+ f)] K 2 [a (e + s) + (d + g)] [(5+ f) – (c+ r)] & K 2 [a (e + s) + (d + g)] [(b+ f) – (c+ r)] ô K = 0. where K = 2 [a (e + s) + (d + g)] [(d + g) ([(b+ f) – (c + r)] - 8) + (e +s) ([(b+ f) – (c+r)] + 6)] Similarly, we can show that bro + cx-1 + fæ-2+ ræ_3 dro + ex–1+ gx_2+ sx_3 (b+ f) P+ (c+r)Q az, + aQ + = P. %3D (d+g) P+ (e+ s) Q By using the mathematical induction, we have In = Q and rn+1 = P, n > -4.
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
Explain the determine blue
![(b+f)Q+(c+r) P
aP+
(d+g) Q+ (e + s) P
[a (d + g) – (e + s)] PQ + a (e + s) P² – (d +g) Q² + (b+ f)Q+ (c+r) P
(d+ g) Q+ (e + s) P
[a(d+g)-(e+s)|[(b+f)-(c+r)|[(c+r)(d+g)+a(e+s) (b+f)] tale+ s) (6+f)-(c+r)]+6
Ta(e+s)+(d+g)]²(e+s)-(d+g))(a+1)]
(6+S)-(c+r)]-6
2a(e+a)+(d+g)l,
2a(e+s)+(d+g)]
(d +g)
+ (e + s)
[(6+f)-(c+r)]+5
2a(e+s)+(d+g)]
- (d+g) (ae )+ (b+ f) (t)-(c+r))-6
2(a(e+a)+(d+g)],
[(b+S)-(c+r)]+6
[(6+S)-(c+r)-6
[(b+/)-(e+r)]-6
(d+g) (ae+)+(d+g)\ /
+ (e + s)
(6+p)+(+a)D
[(b+f)-(c+r)]+6
(c+r) (jate)+(d+g)l ,
(d +g)
[(6+f)-(c+r)]-8
(2(a(e+s)+(d+g)] ,
[(b+f)-(c+r)]+ố
(23)
+ (e+ s)
[(6+p)+(*+a)D]z
Multiplying the denominator and numerator of (23) by 4 [a (e + s) + (d+g)]
we get
x1 - Q
4[(b+S)-(c+r)][a (d+g)-(e+s)][(c+r)(d+g)+a(e+s)(b+S)]
(e+s) – (d+g))(a+1)|
K
ae ([(b+ f) – c) + 6)² – (d + g) ([(b+ f) –c] – 6)²
K
4 [a (e + s) + (d +g)]´ (b+ f) (ae+s) +(d+9)] /
[(6+f)-(c+r)]-8
K
4 [a (e + s) + (d + g)]² (c + r)
(++s)-(c+r)]+6
2|a(e+s) +(d+g),
K
4[(b+S)-(c+r)][a (d+g)-(e+s)][(c+r)(d+g)+a(e+s)(b+OI
((e+s) – (d+g))(a+1)]
K
la (e + s) + (d+ g9)] [(b+ f) – (c+r)]²
K
la (e + s) + (d + g)] &² + 2 [a (e + s) + (d+g)] [(b + f) – (c+r)] &
+
K
2 (b+ f) [a (e + s) + (d+g)] ([(b+ f) - (c+r)] – 6)
+
K
2 (c+r) [a (e+ s) + (d+g)] ([(b+f) - (c+r)]+6)
K
-4 [(b+ f) – (c+r)] [(c+r) (d+g) + a (e + s) (b+ f)]
K
2 [a (e + s) – (d+ g)] [(b + f) – (c + r))?
K
2 [(c+r) + (b+ f] [a (e+ s) + (d+ g)] [(b+ f) – (c+r)]
K
2 [a (e + s) + (d+g)] [(b+f) - (c+r)] &
K
2 [a (e + s) + (d+ g)] [(b+ f) – (c+ r)] 6
K
-4 [(b+ f) – (c+ r)] [(c+ r) (d + g) + a (e + s) (b+ f)]
K
2 [(b+ f)- (c+r)) ([a (e + s) – (d+ g)] [(b+ f) – (c+r)])
K
2 [(b+ f) – (c+r)] ([(b+ f) + (c+ r)] [a (e + s) +(d+ g)])
K
2 [a (e + s) + (d+ g)] [(b+ f) - (c+r)] 8
K
2 [a (e + s) + (d+ g)] [(b + f) – (c+r)] &
K
-4 [(b+ f) – (c+ r)] [(c+r) (d + g) +a (e+ s) (b+ f)]
K
2 [(b+ f) – (c+r)] (2a (e+ s) (b+ f) +2 (c+r) (d+g))
+
K
2 (a (e + s) + (d +g)] [(b+ f) – (c +r)] &
K
2 [a (e + s) + (d+ g)] [(b+ f) - (c+r)] 8
K
-4 [(b+ f) – (c+r)] [(c+r) (d+ g) + a (e + s) (b+ f)]
K
4 [(b+f) - (c+r)[(c+r) (d+g) + a (e+ s) (b+ f)]
K
2 [a (e+ s) + (d + 9)] [(b+ f) – (c+ r)] ô
+
K
2 [a (e + s) + (d + g)] [(b+ f) – (c+r)] &
K
= 0.
where
2 [a (e + s) + (d + g) [(d+ g) ([(b+ f) - (c+r)] – 6)
+(e + s) ([(b+ f) - (c+r)] + 8)]
%3D
Similarly, we can show that
bro + cx-1 + fx-2 + rx-3
(b+f) P+ (c+r) Q
x2 = ax, +
= aQ +
= P.
dro + ex-1+ gx-2+ sx-3
(d+g) P+(e+ s) Q
By using the mathematical induction, we have an = Q and xn+1 = P, n 2
-4.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff32990f0-b218-411f-bb33-f8cb71f1fa43%2Fc517faaf-0b8d-47bc-a961-d0997bf507e9%2Fge4eg4q_processed.png&w=3840&q=75)
Transcribed Image Text:(b+f)Q+(c+r) P
aP+
(d+g) Q+ (e + s) P
[a (d + g) – (e + s)] PQ + a (e + s) P² – (d +g) Q² + (b+ f)Q+ (c+r) P
(d+ g) Q+ (e + s) P
[a(d+g)-(e+s)|[(b+f)-(c+r)|[(c+r)(d+g)+a(e+s) (b+f)] tale+ s) (6+f)-(c+r)]+6
Ta(e+s)+(d+g)]²(e+s)-(d+g))(a+1)]
(6+S)-(c+r)]-6
2a(e+a)+(d+g)l,
2a(e+s)+(d+g)]
(d +g)
+ (e + s)
[(6+f)-(c+r)]+5
2a(e+s)+(d+g)]
- (d+g) (ae )+ (b+ f) (t)-(c+r))-6
2(a(e+a)+(d+g)],
[(b+S)-(c+r)]+6
[(6+S)-(c+r)-6
[(b+/)-(e+r)]-6
(d+g) (ae+)+(d+g)\ /
+ (e + s)
(6+p)+(+a)D
[(b+f)-(c+r)]+6
(c+r) (jate)+(d+g)l ,
(d +g)
[(6+f)-(c+r)]-8
(2(a(e+s)+(d+g)] ,
[(b+f)-(c+r)]+ố
(23)
+ (e+ s)
[(6+p)+(*+a)D]z
Multiplying the denominator and numerator of (23) by 4 [a (e + s) + (d+g)]
we get
x1 - Q
4[(b+S)-(c+r)][a (d+g)-(e+s)][(c+r)(d+g)+a(e+s)(b+S)]
(e+s) – (d+g))(a+1)|
K
ae ([(b+ f) – c) + 6)² – (d + g) ([(b+ f) –c] – 6)²
K
4 [a (e + s) + (d +g)]´ (b+ f) (ae+s) +(d+9)] /
[(6+f)-(c+r)]-8
K
4 [a (e + s) + (d + g)]² (c + r)
(++s)-(c+r)]+6
2|a(e+s) +(d+g),
K
4[(b+S)-(c+r)][a (d+g)-(e+s)][(c+r)(d+g)+a(e+s)(b+OI
((e+s) – (d+g))(a+1)]
K
la (e + s) + (d+ g9)] [(b+ f) – (c+r)]²
K
la (e + s) + (d + g)] &² + 2 [a (e + s) + (d+g)] [(b + f) – (c+r)] &
+
K
2 (b+ f) [a (e + s) + (d+g)] ([(b+ f) - (c+r)] – 6)
+
K
2 (c+r) [a (e+ s) + (d+g)] ([(b+f) - (c+r)]+6)
K
-4 [(b+ f) – (c+r)] [(c+r) (d+g) + a (e + s) (b+ f)]
K
2 [a (e + s) – (d+ g)] [(b + f) – (c + r))?
K
2 [(c+r) + (b+ f] [a (e+ s) + (d+ g)] [(b+ f) – (c+r)]
K
2 [a (e + s) + (d+g)] [(b+f) - (c+r)] &
K
2 [a (e + s) + (d+ g)] [(b+ f) – (c+ r)] 6
K
-4 [(b+ f) – (c+ r)] [(c+ r) (d + g) + a (e + s) (b+ f)]
K
2 [(b+ f)- (c+r)) ([a (e + s) – (d+ g)] [(b+ f) – (c+r)])
K
2 [(b+ f) – (c+r)] ([(b+ f) + (c+ r)] [a (e + s) +(d+ g)])
K
2 [a (e + s) + (d+ g)] [(b+ f) - (c+r)] 8
K
2 [a (e + s) + (d+ g)] [(b + f) – (c+r)] &
K
-4 [(b+ f) – (c+ r)] [(c+r) (d + g) +a (e+ s) (b+ f)]
K
2 [(b+ f) – (c+r)] (2a (e+ s) (b+ f) +2 (c+r) (d+g))
+
K
2 (a (e + s) + (d +g)] [(b+ f) – (c +r)] &
K
2 [a (e + s) + (d+ g)] [(b+ f) - (c+r)] 8
K
-4 [(b+ f) – (c+r)] [(c+r) (d+ g) + a (e + s) (b+ f)]
K
4 [(b+f) - (c+r)[(c+r) (d+g) + a (e+ s) (b+ f)]
K
2 [a (e+ s) + (d + 9)] [(b+ f) – (c+ r)] ô
+
K
2 [a (e + s) + (d + g)] [(b+ f) – (c+r)] &
K
= 0.
where
2 [a (e + s) + (d + g) [(d+ g) ([(b+ f) - (c+r)] – 6)
+(e + s) ([(b+ f) - (c+r)] + 8)]
%3D
Similarly, we can show that
bro + cx-1 + fx-2 + rx-3
(b+f) P+ (c+r) Q
x2 = ax, +
= aQ +
= P.
dro + ex-1+ gx-2+ sx-3
(d+g) P+(e+ s) Q
By using the mathematical induction, we have an = Q and xn+1 = P, n 2
-4.
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