Non-homogeneous differential equation

  • #1
1. Homework Statement

I am trying to find out the soulution of the following non homogeneous differential equation with variable coefficients. The differential equation is given as follows

2. Homework Equations

y''+(1/x)y'-(A0/x)y = -B0/x

where,
A0,B0 = constants

Does anyone have any idea how to solve this one???
help is greatly appriciated..
Thanks!!
 

Answers and Replies

  • #2
798
34
First consider the EDO :
y''(x) +(1/x)y'(x) -(A0/x)y(x) = 0
Let x = k t² and determine k so that the EDO be transformed into :
y''(t) +(1/t)y'(t) -y(t) = 0
which is a basic form of Bessel equation.
Then, consider the complete EDO :
y''(x) +(1/x)y'(x) -(A0/x)y(x) = -B0/x
A particular solution is obvious. Add it to the preceeding result.
 
  • #3
I dont quite get the transformation..Could you please expand it a little bit
Thanks..
 
  • #4
798
34
A little bit more expanded ... in attachment
 

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  • #6
I got the solution but if i want to perform simulations i need a general solution of bessel funtion of the first kind...There are 1000 different forms of this funtion available..I am a bit confused which to use?

Could you help?
 
  • #7
798
34
There are 1000 different forms of this funtion available..I am a bit confused which to use?
There is only one "Modified Bessel Function of the first kind" and order zero, which symbol is I0.
There is only one "Modified Bessel Function of the second kind" and order zero, which symbol is K0.
That is all what you need.
 
  • #8
thanks mate...I couldent understand that J0 ist the 0th order..
 
  • #9
First consider the EDO :
y''(x) +(1/x)y'(x) -(A0/x)y(x) = 0
Let x = k t² and determine k so that the EDO be transformed into :
y''(t) +(1/t)y'(t) -y(t) = 0
which is a basic form of Bessel equation.
Then, consider the complete EDO :
y''(x) +(1/x)y'(x) -(A0/x)y(x) = -B0/x
A particular solution is obvious. Add it to the preceeding result.

The perticular solution is not that obvious...As after taken general form of perticular solution as y=a*x^-1 i am getting pretty wierd results..Am i doing it right?
 
  • #11
798
34
Hi mate !
The perticular solution is not that obvious...
y''(x) +(1/x)y'(x) -(A0/x)y(x) = -B0/x
Obviously, y(x) = B0/A0 is a particular solution.
 
  • #12
but there is 1/x on the RHS ...what will be the general form the Particular integral...As you can see i am not so good in math..
 
  • #13
798
34
but there is 1/x on the RHS
y''(x) +(1/x)y'(x) -(A0/x)y(x) = -B0/x
Just plug y(x) = B0/A0 into the ODE
y'(x)=0 since y(x)=constant
y''(x)=0
0 + 0 -(A0/x)(B0/A0) = -B0/x
The equality in the ODE is achieved.
So, y(x) = B0/A0 is a solution of the ODE.
what will be the general form the Particular integral
It makes no sens : "general" and "particular" are contradictory and what "integral" ?
 
  • #14
212
0
y''(x) +(1/x)y'(x) -(A0/x)y(x) = -B0/x
Just plug y(x) = B0/A0 into the ODE
y'(x)=0 since y(x)=constant
y''(x)=0
0 + 0 -(A0/x)(B0/A0) = -B0/x
The equality in the ODE is achieved.
So, y(x) = B0/A0 is a solution of the ODE.

It makes no sens : "general" and "particular" are contradictory and what "integral" ?

i think he means Yp

y=Yc+Yp

Yc= complimentary solution
Yp=particular solution (user called it particular integral)
 
  • #15
Yes generally in the calculus or differential equations books it is called as perticular integral..And this definition is also not so senseless..but anyways thats a million for your help..
 

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