1. ## bondary value problem

u=x and v=$\displaystyle - 1/3x^2$ are two solutions of the differential equation
$\displaystyle x^2y'' + 2xy' - 2y = 0$
such that u(0)=0 and 2v(1)+v'(1)=0
solve the boundary value problem
$\displaystyle x^2y'' + 2xy' - 2y = x^2$ y(0)=0 and $\displaystyle 2y(1) + y'(1) = 0$
if u and v are solutions does that mean $\displaystyle y_1$=u and $\displaystyle y_2$=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 $\displaystyle y_1$ and $\displaystyle y_2$ equal to?
Thanks for any help

2. Originally Posted by charikaar
u=x and v=$\displaystyle - 1/3x^2$ are two solutions of the differential equation
$\displaystyle x^2y'' + 2xy' - 2y = 0$
such that u(0)=0 and 2v(1)+v'(1)=0
solve the boundary value problem
$\displaystyle x^2y'' + 2xy' - 2y = x^2$ y(0)=0 and $\displaystyle 2y(1) + y'(1) = 0$
if u and v are solutions does that mean $\displaystyle y_1$=u and $\displaystyle y_2$=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 $\displaystyle y_1$ and $\displaystyle y_2$ equal to?
Thanks for any help
I think you should have written the two particular solutions as:

$\displaystyle y_1=x$

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

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

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

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