Un-1 n=3,4,... & a1 =a0.a2=(1+ag)is theIf an=recurrence relation of y'-x -y = 0, then the power series solution isy = ao + a, x + a2x +....= ao + aox + (1+ ao)M%3Dn=0 n!y = 90-a1 x- azx²+.°0 - 9g* - (1+ ao) En=on!....1o + ag* + (1 + ac) En=o n!x ++%3D....V=-0+a x+ azx².-9o + 9o* + (1 + a,)En=on!+....

Let y=∑n=0∞anxn be the series solution Given: a1=a0, a2=1+a02, an=an-1n , n=3, 4,....

(1+a) Un-1.n=3,4,.... & a1 =a0.a2=* If an = is the recurrence relation of y'-x-y= 0, then the power series solution is y = ao + a, x+ azx? +....= ao + ax+ (1 + a) E- = ao + acx + (1 + a») Ex" %3D y= a0-a1 x-azx²+....= a0 - aox- (1+ a) E n=on! 1 - °o + °o* + (1 + a,) %3D Σ n=on! ... 1 o + °g* + (1 + ao)_ n=on! x + ....

(1+a) Un-1.n=3,4,.... & a1 =a0.a2=* If an = is the recurrence relation of y'-x-y= 0, then the power series solution is y = ao + a, x+ azx? +....= ao + ax+ (1 + a) E- = ao + acx + (1 + a») Ex" %3D y= a0-a1 x-azx²+....= a0 - aox- (1+ a) E n=on! 1 - °o + °o* + (1 + a,) %3D Σ n=on! ... 1 o + °g* + (1 + ao)_ n=on! x + ....

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Similar questions

7. Find the sum of t1 + t2 + t3 + .. tn + …. where tn = 8 (5)"
Using the power series method about x = 0, the recurrence relation of the differential !! equation: (x2 + 1)y"-4xy' +6y 0 is: (k+2)Cx+Cz-a k 2,3, . Ck+2 (k-2) (k-3)Ck (k+2) (k+1) ,k = 0,1, . (k+2)(k+1) ... This option This option O This option O This option (k+1)Ck k = 0,1, ... Ck+2 (1-2k)Ck (k +2)(k+1) k = 1,2, .. C42 = - %3D 4(k+2) This option This option
If a bank pays interest at a rate of i compounded m times a year, then the amount of money Pk at the end of k time periods (where one time period recurrence relation Pk with initial condition P₁ = the initial amount deposited. Find an explicit formula for P = Pk-1 = 1/mth of a year) satisfies the The given recurrence relation defines a geometric sequence with constant multiplier ' which is PO for every integer n ≥ 0. Therefore, Pn 1+ = --P-(A)} n m
1/2, 1/3, 1/4, 1/5, ... recursively. Include initial condition. аn — аn-1 + n+1' ª1 =1/2 n+1 аn — аn-1 + a1 = 1/2 n n An = -аn-1, aj — 1/2 n+1 n+1 -An-1, a1 =1/2 An
I know the answer is x+(1/2)(x^2)+(1/2)(x^3)+(13/24)(x^4)+(11/15)(x^5) but can you help me on verifying the answer? Somehow it matches the original equation is what I'm looking for.
Find the general solutions in powers of x and state the recurrence relation: (2 - x2)y" - xy' + 16y = 0
The roots of fourth order non linear homogeneous recurrence relation are 1, —2,3 аnd — 4 with F (n) —Зп — 2" + 3". %3D What is the form of particular solution ? a P = An + B +C2" + Dn3n a P = An + B + Cn2" + Dn3" a P = An + Bn² + C2" + Dn3" a P = Al" + B(- 2)" + C3" + D(-4)n
Directions: Find the sum or difference of the following _1. (* 2) + (* 8) _2. (* 4) + (' 8) 3. ( 1) + (* 10) _4. (* 11) + (* 11) _5. 0 + (* 9) _6. (* 5) – (* 10) _7. ( 12) – (* 7) _8. (* 4) – (' 4)
mard / My courses / ECON1003 | S2 21/22/ General / Graded Online Quia Gon 5 5x + 1 If y = then y' will be et 2x-1 – In 7' ered ked out of 5(2x-1 – In 7) – (5x + 1) (+;) (2x-1 – In 7)2 A. Flag question 5(2x"' – In 7) – (5x + 1)(-2x² – ÷) (2x 1- In 7)2 5(2x¬! – In 7) – (5x + 1) (-) C. (2x In 7)2 5(2x¬ – In 7) – (5x +1) (;-=) (2x¬ – In 7) Select one: OB larrece Roherts FCON1005 Assaccmant # 1 Type here to search S1080 B. D.