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Solve the Bessel differential equation

A differential equationInhomogeneous modified Bessel differential equationSecond order differential equation with multiple bessel functionsBessel's Differential Equation - textbook queries:How can I solve the differential equation $(y'/y)'=ay$?Solution of a Modified Bessel Differential Equation with Complex CoefficientSolve the Bessel Equation to show it satisfies the differential equationSecond Solution of Bessel Equation of Order $p$Alternate forms of Bessel Equationchecking the Solution of Bessel differential equation

0

$begingroup$

Show that $J_n(x) / x^n$ is a solution of
$$fracd^2 yd x^2+left(frac1+2 nxright) fracd yd x+y=0$$
and that $sqrt(x) J_n(k x)$ is a solution of
$$fracd^2 yd x^2+left(k^2-frac4 n^2-14 x^2right) y=0$$
where, in both cases, $n$ is a positive integer.

ordinary-differential-equations special-functions bessel-functions

share|cite|improve this question

edited Mar 15 at 15:04

Andrews

1,2691421

asked Mar 15 at 4:32

Hamada AlHamada Al

528318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried the obvious method of taking some derivatives and applying the known differential equation for $J_n$?
  $endgroup$
  – jmerry
  Mar 15 at 6:27
  

  
  
  
  

add a comment | 

0

$begingroup$

Show that $J_n(x) / x^n$ is a solution of
$$fracd^2 yd x^2+left(frac1+2 nxright) fracd yd x+y=0$$
and that $sqrt(x) J_n(k x)$ is a solution of
$$fracd^2 yd x^2+left(k^2-frac4 n^2-14 x^2right) y=0$$
where, in both cases, $n$ is a positive integer.

ordinary-differential-equations special-functions bessel-functions

share|cite|improve this question

edited Mar 15 at 15:04

Andrews

1,2691421

asked Mar 15 at 4:32

Hamada AlHamada Al

528318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried the obvious method of taking some derivatives and applying the known differential equation for $J_n$?
  $endgroup$
  – jmerry
  Mar 15 at 6:27
  

  
  
  
  

add a comment | 

$begingroup$

Show that $J_n(x) / x^n$ is a solution of
$$fracd^2 yd x^2+left(frac1+2 nxright) fracd yd x+y=0$$
and that $sqrt(x) J_n(k x)$ is a solution of
$$fracd^2 yd x^2+left(k^2-frac4 n^2-14 x^2right) y=0$$
where, in both cases, $n$ is a positive integer.

ordinary-differential-equations special-functions bessel-functions

share|cite|improve this question

edited Mar 15 at 15:04

Andrews

1,2691421

asked Mar 15 at 4:32

Hamada AlHamada Al

528318

$endgroup$

share|cite|improve this question

edited Mar 15 at 15:04

Andrews

1,2691421

asked Mar 15 at 4:32

Hamada AlHamada Al

528318

share|cite|improve this question

edited Mar 15 at 15:04

Andrews

1,2691421

asked Mar 15 at 4:32

Hamada AlHamada Al

528318

edited Mar 15 at 15:04

Andrews

1,2691421

edited Mar 15 at 15:04

Andrews

1,2691421

asked Mar 15 at 4:32

Hamada AlHamada Al

528318

asked Mar 15 at 4:32

Hamada AlHamada Al

528318

1 Answer
1

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oldest

votes

0

$begingroup$

$$y''+left(frac1+2 nxright)y'+y=0tag 1$$
Change of function :

$y(x)=fracu(x)x^n quad;quad y'=fracu'x^n-nfracux^n+1 quad;quad y''=fracu''x^n-2nfracu'x^n+1+n(n+1)fracux^n+2 $

Putting them into the ODE :
$$fracu''x^n-2nfracu'x^n+1+n(n+1)fracux^n+2+left(frac1+2 nxright)left(fracu'x^n-nfracux^n+1right) +fracux^n=0$$
$$u''-2nfracu'x+n(n+1)fracux^2+left(frac1+2 nxright)left(u'-nfracuxright) +u=0$$
After simplification :
$$u''+fracu'x+(1-frac1x^2)u=0$$
This is the Bessel equation on standard form, which solution is :
$$u(x)=c_1J_1(x)+c_2Y_1(x)$$
with the Bessel functions of first and second kind.
$$y(x)=c_1fracJ_1(x)x^n+c_2fracY_1(x)x^n$$
Thus $fracJ_1(x)x^n$ is a solution of Eq.$(1)$.

Proceed on the same manner for Eq.$(2)$ , with the change of function $y(x)=sqrtx:u(kx)$ and the change of variable $X=kx$.

share|cite|improve this answer

edited Mar 15 at 7:04

answered Mar 15 at 6:39

JJacquelinJJacquelin

45.1k21855

$endgroup$

add a comment | 

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0

$begingroup$

$$y''+left(frac1+2 nxright)y'+y=0tag 1$$
Change of function :

$y(x)=fracu(x)x^n quad;quad y'=fracu'x^n-nfracux^n+1 quad;quad y''=fracu''x^n-2nfracu'x^n+1+n(n+1)fracux^n+2 $

Putting them into the ODE :
$$fracu''x^n-2nfracu'x^n+1+n(n+1)fracux^n+2+left(frac1+2 nxright)left(fracu'x^n-nfracux^n+1right) +fracux^n=0$$
$$u''-2nfracu'x+n(n+1)fracux^2+left(frac1+2 nxright)left(u'-nfracuxright) +u=0$$
After simplification :
$$u''+fracu'x+(1-frac1x^2)u=0$$
This is the Bessel equation on standard form, which solution is :
$$u(x)=c_1J_1(x)+c_2Y_1(x)$$
with the Bessel functions of first and second kind.
$$y(x)=c_1fracJ_1(x)x^n+c_2fracY_1(x)x^n$$
Thus $fracJ_1(x)x^n$ is a solution of Eq.$(1)$.

Proceed on the same manner for Eq.$(2)$ , with the change of function $y(x)=sqrtx:u(kx)$ and the change of variable $X=kx$.

share|cite|improve this answer

edited Mar 15 at 7:04

answered Mar 15 at 6:39

JJacquelinJJacquelin

45.1k21855

$endgroup$

add a comment | 

0

$begingroup$

$$y''+left(frac1+2 nxright)y'+y=0tag 1$$
Change of function :

$y(x)=fracu(x)x^n quad;quad y'=fracu'x^n-nfracux^n+1 quad;quad y''=fracu''x^n-2nfracu'x^n+1+n(n+1)fracux^n+2 $

Putting them into the ODE :
$$fracu''x^n-2nfracu'x^n+1+n(n+1)fracux^n+2+left(frac1+2 nxright)left(fracu'x^n-nfracux^n+1right) +fracux^n=0$$
$$u''-2nfracu'x+n(n+1)fracux^2+left(frac1+2 nxright)left(u'-nfracuxright) +u=0$$
After simplification :
$$u''+fracu'x+(1-frac1x^2)u=0$$
This is the Bessel equation on standard form, which solution is :
$$u(x)=c_1J_1(x)+c_2Y_1(x)$$
with the Bessel functions of first and second kind.
$$y(x)=c_1fracJ_1(x)x^n+c_2fracY_1(x)x^n$$
Thus $fracJ_1(x)x^n$ is a solution of Eq.$(1)$.

Proceed on the same manner for Eq.$(2)$ , with the change of function $y(x)=sqrtx:u(kx)$ and the change of variable $X=kx$.

share|cite|improve this answer

edited Mar 15 at 7:04

answered Mar 15 at 6:39

JJacquelinJJacquelin

45.1k21855

$endgroup$

add a comment | 

$begingroup$

$$y''+left(frac1+2 nxright)y'+y=0tag 1$$
Change of function :

$y(x)=fracu(x)x^n quad;quad y'=fracu'x^n-nfracux^n+1 quad;quad y''=fracu''x^n-2nfracu'x^n+1+n(n+1)fracux^n+2 $

Putting them into the ODE :
$$fracu''x^n-2nfracu'x^n+1+n(n+1)fracux^n+2+left(frac1+2 nxright)left(fracu'x^n-nfracux^n+1right) +fracux^n=0$$
$$u''-2nfracu'x+n(n+1)fracux^2+left(frac1+2 nxright)left(u'-nfracuxright) +u=0$$
After simplification :
$$u''+fracu'x+(1-frac1x^2)u=0$$
This is the Bessel equation on standard form, which solution is :
$$u(x)=c_1J_1(x)+c_2Y_1(x)$$
with the Bessel functions of first and second kind.
$$y(x)=c_1fracJ_1(x)x^n+c_2fracY_1(x)x^n$$
Thus $fracJ_1(x)x^n$ is a solution of Eq.$(1)$.

Proceed on the same manner for Eq.$(2)$ , with the change of function $y(x)=sqrtx:u(kx)$ and the change of variable $X=kx$.

share|cite|improve this answer

edited Mar 15 at 7:04

answered Mar 15 at 6:39

JJacquelinJJacquelin

45.1k21855

$endgroup$

share|cite|improve this answer

edited Mar 15 at 7:04

answered Mar 15 at 6:39

JJacquelinJJacquelin

45.1k21855

answered Mar 15 at 6:39

JJacquelinJJacquelin

45.1k21855

answered Mar 15 at 6:39

JJacquelinJJacquelin

45.1k21855