Find a solution bounded near $x=0$ of the following ODE
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Fine a solution bounded near $x=0$ of the following ODE
$$x^2y''+xy'+( lambda ^2x^2-1)y=0$$
my attempt :
this is Bessel's equation so let $u=lambda x$ then $y(x)=y(frac{u}{lambda})$
Also $y'(x)=lambda Y'(u)$ and $y''(x)=lambda^2Y''(u)$
then the give equation reduced to $u^2Y''(u)+uY'(u)+(lambda^2-1)Y(u)=0$
then the general solution is $Y(u)=AJ_1(u)+BY_1(u)=AJ_1(lambda x)+BY_1(lambda x)=y(x)$
But what a solution bounded near $x=0$ ?
differential-equations boundary-value-problem
edited Nov 14 at 12:45
ablmf
2,33942251
asked Nov 14 at 12:37
Inverse Problem
923918
add a comment |
up vote
0
down vote
favorite
Fine a solution bounded near $x=0$ of the following ODE
$$x^2y''+xy'+( lambda ^2x^2-1)y=0$$
my attempt :
this is Bessel's equation so let $u=lambda x$ then $y(x)=y(frac{u}{lambda})$
Also $y'(x)=lambda Y'(u)$ and $y''(x)=lambda^2Y''(u)$
then the give equation reduced to $u^2Y''(u)+uY'(u)+(lambda^2-1)Y(u)=0$
then the general solution is $Y(u)=AJ_1(u)+BY_1(u)=AJ_1(lambda x)+BY_1(lambda x)=y(x)$
But what a solution bounded near $x=0$ ?
differential-equations boundary-value-problem
edited Nov 14 at 12:45
ablmf
2,33942251
asked Nov 14 at 12:37
Inverse Problem
923918
You should choose $B=0$.
– Jon
Nov 14 at 12:51
@Jon..how we can choose B=0
– Inverse Problem
Nov 14 at 12:52
$A$ and $B$ are arbitrary constant. If you are asked for a solution bounded in 0, $J_1$ is and $Y_1$ is not. So, you have to fix $B=0$.
– Jon
Nov 14 at 12:54
$yequiv 0$ is a bounded solution ;)
– MPW
Nov 14 at 12:54
@MPW The trivial one! :)
– Jon
Nov 14 at 12:56
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Fine a solution bounded near $x=0$ of the following ODE
$$x^2y''+xy'+( lambda ^2x^2-1)y=0$$
my attempt :
this is Bessel's equation so let $u=lambda x$ then $y(x)=y(frac{u}{lambda})$
Also $y'(x)=lambda Y'(u)$ and $y''(x)=lambda^2Y''(u)$
then the give equation reduced to $u^2Y''(u)+uY'(u)+(lambda^2-1)Y(u)=0$
then the general solution is $Y(u)=AJ_1(u)+BY_1(u)=AJ_1(lambda x)+BY_1(lambda x)=y(x)$
But what a solution bounded near $x=0$ ?
differential-equations boundary-value-problem
edited Nov 14 at 12:45
ablmf
2,33942251
asked Nov 14 at 12:37
Inverse Problem
923918
Fine a solution bounded near $x=0$ of the following ODE
$$x^2y''+xy'+( lambda ^2x^2-1)y=0$$
my attempt :
this is Bessel's equation so let $u=lambda x$ then $y(x)=y(frac{u}{lambda})$
Also $y'(x)=lambda Y'(u)$ and $y''(x)=lambda^2Y''(u)$
then the give equation reduced to $u^2Y''(u)+uY'(u)+(lambda^2-1)Y(u)=0$
then the general solution is $Y(u)=AJ_1(u)+BY_1(u)=AJ_1(lambda x)+BY_1(lambda x)=y(x)$
But what a solution bounded near $x=0$ ?
differential-equations boundary-value-problem
differential-equations boundary-value-problem
edited Nov 14 at 12:45
ablmf
2,33942251
asked Nov 14 at 12:37
Inverse Problem
923918
edited Nov 14 at 12:45
ablmf
2,33942251
asked Nov 14 at 12:37
Inverse Problem
923918
edited Nov 14 at 12:45
ablmf
2,33942251
edited Nov 14 at 12:45
ablmf
2,33942251
edited Nov 14 at 12:45
ablmf
2,33942251
2,33942251
asked Nov 14 at 12:37
Inverse Problem
923918
asked Nov 14 at 12:37
Inverse Problem
923918
asked Nov 14 at 12:37
Inverse Problem
923918
923918
You should choose $B=0$.
– Jon
Nov 14 at 12:51
@Jon..how we can choose B=0
– Inverse Problem
Nov 14 at 12:52
$A$ and $B$ are arbitrary constant. If you are asked for a solution bounded in 0, $J_1$ is and $Y_1$ is not. So, you have to fix $B=0$.
– Jon
Nov 14 at 12:54
$yequiv 0$ is a bounded solution ;)
– MPW
Nov 14 at 12:54
@MPW The trivial one! :)
– Jon
Nov 14 at 12:56
add a comment |
You should choose $B=0$.
– Jon
Nov 14 at 12:51
@Jon..how we can choose B=0
– Inverse Problem
Nov 14 at 12:52
$A$ and $B$ are arbitrary constant. If you are asked for a solution bounded in 0, $J_1$ is and $Y_1$ is not. So, you have to fix $B=0$.
– Jon
Nov 14 at 12:54
$yequiv 0$ is a bounded solution ;)
– MPW
Nov 14 at 12:54
@MPW The trivial one! :)
– Jon
Nov 14 at 12:56
You should choose $B=0$.
– Jon
Nov 14 at 12:51
You should choose $B=0$.
– Jon
Nov 14 at 12:51
@Jon..how we can choose B=0
– Inverse Problem
Nov 14 at 12:52
@Jon..how we can choose B=0
– Inverse Problem
Nov 14 at 12:52
$A$ and $B$ are arbitrary constant. If you are asked for a solution bounded in 0, $J_1$ is and $Y_1$ is not. So, you have to fix $B=0$.
– Jon
Nov 14 at 12:54
$A$ and $B$ are arbitrary constant. If you are asked for a solution bounded in 0, $J_1$ is and $Y_1$ is not. So, you have to fix $B=0$.
– Jon
Nov 14 at 12:54
$yequiv 0$ is a bounded solution ;)
– MPW
Nov 14 at 12:54
$yequiv 0$ is a bounded solution ;)
– MPW
Nov 14 at 12:54
@MPW The trivial one! :)
– Jon
Nov 14 at 12:56
@MPW The trivial one! :)
– Jon
Nov 14 at 12:56
add a comment |
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You should choose $B=0$.
– Jon
Nov 14 at 12:51
@Jon..how we can choose B=0
– Inverse Problem
Nov 14 at 12:52
$A$ and $B$ are arbitrary constant. If you are asked for a solution bounded in 0, $J_1$ is and $Y_1$ is not. So, you have to fix $B=0$.
– Jon
Nov 14 at 12:54
$yequiv 0$ is a bounded solution ;)
– MPW
Nov 14 at 12:54
@MPW The trivial one! :)
– Jon
Nov 14 at 12:56