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**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