# Thread: Y''/y = (-λ^2 - 1)

1. ## Y''/y = (-λ^2 - 1)

Hello! I was trying to solve a Laplace's Equation and I got stuck on this step.

Here's the original question:

$\displaystyle u_{xx}+u_{yy}+u=0$
$\displaystyle 0<x<1$
$\displaystyle 0<y<1$
$\displaystyle u(0,y) = 0$
$\displaystyle u(1,y) = 0$
$\displaystyle u(x,0) = 0$
$\displaystyle u_y(x,1) = sin(3\pi x)$

So I assumed $\displaystyle u(x,y) = X(x)Y(y)$

Used separation of variables:

$\displaystyle \frac{Y''}{Y} + 1 = \frac{X''}{X} = -\lambda^2$

And I got:

$\displaystyle X(x) = Acos\lambda x + Bsin\lambda x$

And I don't know how to solve for Y

Here's my equation:

$\displaystyle Y'' = (-\lambda^2 - 1)Y$

2. Notice that $\displaystyle \lambda$ is just an eigenvalue, so this is really just a second order linear ODE. So, treating $\displaystyle (- \lambda^2 -1)$ as a single coefficient, we get that

$\displaystyle Y(y) = C*e^{y \sqrt{-\lambda^2 -1}} + D*e^{-y\sqrt{-\lambda^2 -1}}$

3. I thought of that too but it looked really messy to me so I didn't take it as an answer.

I think I could also do this:

$\displaystyle Y''(y) = ( -\lambda^2 - 1)Y(y)$
$\displaystyle Y''(y) = -( \lambda^2 + 1)Y(y)$
$\displaystyle Y(y) = Ccos(y*\sqrt{\lambda^2+1}) + Dsin(y*\sqrt{\lambda^2+1})$

Because I think this form works better with the initial conditions. =P