unique solution for a homogeneous differential equation

**sterps** · August 26th 2009, 08:23 PM

I'm having trouble getting this question started. Im not sure how to go about solving it nor what method to use. Some of my friends say to divide by $x^2$ , while others tell me other ways.

$x^2 d^2 y/ dx^2 + 2x (x-1) dy/dx - 2(x-1)y=0$

Why doesn't the general theory guarantee a unique solution to the equation satisfying the initial conditions y(0) = 0, y'(0) = 1.

If someone could explain to me what i need to do, i would much appreciate it

thanks

**chisigma** · August 26th 2009, 11:12 PM

In a linear DE of the type...

$c_{2} (x) y^{''} + c_{1} (x) y^{'} + c_{0} (x) y=0$ (1)

... where $x \in [a,b]$ the condition of existence and unicity of the solution given the 'initial conditions' $y(x_{0}) = y_{0}, y'(x_{0}) = y'_{0}$ requires that $c_{2} (x_{0}) \ne 0$ . In your example is $c_{2} (x)= x^{2} = 0_{x=0}$ so that the requirement isn't satisfied...

Kind regards

$\chi$ $\sigma$

**sterps** · August 27th 2009, 05:42 AM

is that all?! thanks man. i appreciate it, i dont know what the others are talking about dividing by $x^2$

**sterps** · August 27th 2009, 07:40 AM

Ok i need more help with a related problem to this question

$<br /> x^2 d^2 y/ dx^2 + 2x (x-1) dy/dx - 2(x-1)y=0<br />$

Using the method of reduction of order, find a second linearly independent
solution y2(x) of the equation.

In the previous question i was able to show y1(x) = x is a solution of this equation, by taking the derivatives of y1, and then substituting back into the equation. and finding that x is indeed a solution.

Though for this question i tried using that guide at the top of the forum for reduction order, and i get an answer that doesnt look right.

$u(x) = |[(e^ (1/3x^22x-3)) / (x^2)]dx$

cant get the maths coding right :S

anyway help would be great

**chisigma** · August 27th 2009, 11:24 AM

The procedure to obain, once you have a solution of this type of equation, another solution independent from the first is illustated here...

http://www.mathhelpforum.com/math-he...tion-case.html

The second solution has a singularity in $x=0$ and that's why you aren't free to impose arbitrary 'initial conditions' in $x=0$ ...

Kind regards

$\chi$ $\sigma$

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