The second order equation 81a?y" + 81ry' + (81x? – 16) y = 0 has a regular singular point at a = 0, and has series solutions of the form00(1) Insert the formal power series into the differential equation, we derive an equationCn+n-2So we have the indicial equationand a recurrence relationCn_2 for n = 2, 3, .(2) From the indicial equation we can solve the indicial roots(enter your results as a comma separated list). For any one of the indicial roots, we have c =(3) Let r, be the smaller indicial root. Then rand the recurrence relation becomesCn 2 for n = 2, 3, - . .Let co =1 From the recurrence relation, we have a solutionI+1+z+2+T+3+z+4++...Y1 = z+(4) Let r2 be the larger indicial root Then T2 =and the recurrence relation becomesCn=Cn 2 for n = 2, 3, -..Let co = 1 From the recurrence relation, we have another solution.2 = 1+T+2+Ta+0 +...(5) The general solution is given by y = Ay + By, with arbitrary constants A, B.

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Answered: 1) Insert the formal power series into…

1) Insert the formal power series into the differential equation, we derive an equation 2)zr-1 - 0 So we have the indicial equation and a recurrence relation

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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Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differential equations.

Question

The second order equation 81a?y" + 81ry' + (81x? – 16) y = 0 has a regular singular point at a = 0, and has series solutions of the form
00
(1) Insert the formal power series into the differential equation, we derive an equation
Cn+
n-2
So we have the indicial equation
and a recurrence relation
Cn_2 for n = 2, 3, .
(2) From the indicial equation we can solve the indicial roots
(enter your results as a comma separated list). For any one of the indicial roots, we have c =
(3) Let r, be the smaller indicial root. Then r
and the recurrence relation becomes
Cn 2 for n = 2, 3, - . .
Let co =1 From the recurrence relation, we have a solution
I+1+
z+2+
T+3+
z+4+
+...
Y1 = z+
(4) Let r2 be the larger indicial root Then T2 =
and the recurrence relation becomes
Cn=
Cn 2 for n = 2, 3, -..
Let co = 1 From the recurrence relation, we have another solution.
2 = 1+
T+2+
Ta+0 +...
(5) The general solution is given by y = Ay + By, with arbitrary constants A, B.

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