Please use BESSEL FUNCTIONS, if you do not know this method, do not try to answer the question with different methods. I attached the note for the question.Q: Find the general solutions using BESSEL FUNCTIONS 9. Problem 5.3.1.19:ху" + 3у' + х3у %3D0 Generalized form of Bessel's equationGiven the following ordinary differential equation:d²y+ x(a + 2bx")dx2dy+ [c + dx2s – b(1 – a – r)x" + b²x2r]y = 0dx-|The general solution isVīdi-xs ) + A2Z-pVidly(x) = x(1-a)/2e-bx"/rSSwhere:1/21-2

Answered: 9. Problem 5.3.1.19: ху" + Зу' +…

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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Please use BESSEL FUNCTIONS, if you do not know this method, do not try to answer the question with different methods. I attached the note for the question.

Q: Find the general solutions using BESSEL FUNCTIONS

9. Problem 5.3.1.19:
ху" + 3у' + х3у %3D0

Generalized form of Bessel's equation
Given the following ordinary differential equation:
d²y
+ x(a + 2bx")
dx2
dy
+ [c + dx2s – b(1 – a – r)x" + b²x2r]y = 0
dx
-
|
The general solution is
Vīdi
-xs ) + A2Z-p
Vidl
y(x) = x(1-a)/2e-bx"/r
S
S
where:
1/2
1
-
2

Expert Solution

Step 1

Step:- 1

Given generalized differential equation (Bessel form ) is

$x^{2} \frac{d^{2} y}{d x^{2}} + x (a + 2 b x^{r}) \frac{d y}{d x} + [c + d x^{2 s} - b (1 - a - r) x^{r} + b^{2} x^{2 r}] y = 0 d i v i d e i t b y x^{2} S o, I t c a n b e w r i i t e n a s \frac{d^{2} y}{d x^{2}} + \frac{(a + 2 b x^{r})}{x} \frac{d y}{d x} + \frac{[c + d x^{2 s} - b (1 - a - r) x^{r} + b^{2} x^{2 r}]}{x^{2}} y = 0 \frac{d^{2} y}{d x^{2}} + \frac{(a + 2 b x^{r})}{x} \frac{d y}{d x} + [c x^{- 2} + d x^{2 s - 2} - b (1 - a - r) x^{r - 2} + b^{2} x^{2 r - 2}] = 0 - - - (1)$

step:- 2

Now given differential equation is

$x y'' + 3 y' + x^{3} y = 0 d i v i d e t h e e q u a t i o n b y x, w e g e t y'' + \frac{3}{x} y' + x^{2} y = 0 - - - (2)$

Comparing equation (2) with equation (1), we get

$3 = a + 2 b x^{r}$

on comparing constant term with constant and coefficient of variable, we get

$3 = a + 2 b x^{r} \Rightarrow 3 + 0 x^{r} = a + 2 b x^{r}$

$\Rightarrow a = 3 a n d b = 0$ ---(3)

$c x^{- 2} + d x^{2 s - 2} - b (1 - a - r) x^{r - 2} + b^{2} x^{2 r - 2} = x^{2}$

Since, by (3) we have b=0, so

$c x^{- 2} + d x^{2 s - 2} - 0 (1 - a - r) x^{r - 2} + 0 x^{2 r - 2} = x^{2} \Rightarrow c x^{- 2} + d x^{2 s - 2} = x^{2}$

on comparing constant term with constant and coefficient of variable, we get

$c x^{- 2} + d x^{2 s - 2} = 0 x^{- 2} + x^{2} \Rightarrow c = 0, d = 1, 2 s - 2 = 2 \Rightarrow s = 2 \Rightarrow c = 0, d = 1, s = 2 - - - - (4)$

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