# Differential equation

1. Sep 28, 2010

### fatimajan

Hello every body,
actually I'm confused and I don't know what to do?
yy'-axy=bx^5-cx^3
where a,b,c are constants
thank you

2. Sep 28, 2010

### Eynstone

Find an integrating factor for the equation.

3. Sep 29, 2010

### JJacquelin

Let t=x². This leads to a linear ODE

4. Sep 29, 2010

### fatimajan

Thank you for your guidanc, jjacquelin.
do you mean yy'-at^(1/2)y=bt^(5/2)-ct^(3/2) ?
well I see no difference between this and the initial form!
maybe I'm wrong?

5. Sep 29, 2010

### JJacquelin

Don't confuse y'(x) and y'(t) . They aren't equal :
y'(x) = y'(t)*2*t^(1/2)
So the transformed form is simpler than the initial :
2yy'-ay = bt²-ct
But, it isn't a linear ODE. I admit my mistake.

6. Sep 29, 2010

### jackmell

Ok, it looks like the equation in t is just as difficult and you asked for an analytic solution. You could solve it via power series to achieve that goal although I think it's difficult to determine the radius of convergence of a Cauchy product that would enter into the power series solutions due to the yy' term.

7. Sep 30, 2010

### JJacquelin

I agree with jackmell's comments :
The series development leads to complicated formulas for the coefficients as functions of a, b and c (Attachment below).
It is doubtfull that a closed form can be derived in the general case. May be possible in particular cases, for particular values of a, b, c, or in case of particular relationship between these parameters. This would require specific studies if more information is available concerning the parametrers values or relationship beteween them.

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8. Oct 6, 2010

### fatimajan

Thank you for your help, jackmell and jjacquelin

but dear jjacquelin,
actually I don't know about power series method well to solve the equation like you again, since I think you've made a mistake just in transformed form (yy' in lieu of 2yy') in your attachment, so I want to know this mistake makes what difference in your solution that I assumed is true?
and the last question, I have two boundary conditions:
y(0)=finit
y(-infinit)=finit

do you think they're enough to derive a closed form for our solution?

Thank you again

9. Oct 7, 2010

### JJacquelin

Hello,
You are right, but it is just a typing mistake. I forgot the "2" in typing the equation, but I didn't forget the "2" in the calculs. So the series development is correct.
Since it is a first order ODE, only one boundary condition can be settled. If you set two conditions, generally a contradiction will occur.
The bondary condition y(0)=finit sets the yo value appearing in the formulas of the coefficients.
Moreover, a condition such y(-infinit)=finit isn't usable in case of limited series development. The series provides an approximate solution only for not too large values of abs(x), but not for x approaching -infinity or +infinity.
So, don't expect that the solution given in terms of a limited series will be satisfactory in case of large negative x values.

Note:
It seems that the ODE :
yy'-axy=bx^5-cx^3
associated with the boundary condition:
y(-infinit)=finit
has no solution. So, the ODE migth be not convenient to model the physical phenomena which gives y(-infinit)=finit
As a matter of fact, if y(-infinit) is finit, then y'(-infinity)=0
and cx^3 tends to be negigible compared to bx^5. So, the relationship tends to become equivalent to :
-axy=bx^5
With a finit value of y, this is impossible, because -axy isn't equivalent to bx^5, except if (a=0 and b=0), or if (y=0 and b=0).

Last edited: Oct 7, 2010