bondary value problem

**charikaar** · December 6th 2009, 09:21 AM

u=x and v= $- 1/3x^2$ are two solutions of the differential equation
$x^2y'' + 2xy' - 2y = 0$
such that u(0)=0 and 2v(1)+v'(1)=0
solve the boundary value problem
$x^2y'' + 2xy' - 2y = x^2$ y(0)=0 and $2y(1) + y'(1) = 0$
if u and v are solutions does that mean $y_1$ =u and $y_2$ =v
whats the difference between u(0)=0 and 2v(1)+v'(1)=0 and y(0)=0 and 2y(1) + y'(1) = 0
they have just replaced y=u=v. I will be able to solve this using green's function once I know what are $y_1$ and $y_2$ equal to?
Thanks for any help

**shawsend** · December 7th 2009, 01:39 PM

Originally Posted by charikaar

u=x and v= $- 1/3x^2$ are two solutions of the differential equation
$x^2y'' + 2xy' - 2y = 0$
such that u(0)=0 and 2v(1)+v'(1)=0
solve the boundary value problem
$x^2y'' + 2xy' - 2y = x^2$ y(0)=0 and $2y(1) + y'(1) = 0$
if u and v are solutions does that mean $y_1$ =u and $y_2$ =v
whats the difference between u(0)=0 and 2v(1)+v'(1)=0 and y(0)=0 and 2y(1) + y'(1) = 0
they have just replaced y=u=v. I will be able to solve this using green's function once I know what are $y_1$ and $y_2$ equal to?
Thanks for any help

I think you should have written the two particular solutions as:

$y_1=x$

$y_2=-1/3 \frac{1}{x^2}$

It's not hard to use reduction of order with $y_1$ to obtain the general solution of the non-homogeneous equation:

$y(x)=c_1 \frac{1}{x^2}+c_2 x+1/4 x^2$

Not sure though what you wish to do with the boundary information you gave.

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