What is the solution to Killing equations for the given metric?

[PhyPsy]

Homework Statement

Find all Killing vector solutions of the metric
g_x{_x}=x^2, g_x{_y}=g_y{_x}=0, g_y{_y}=x
where (x^a)=(x^0, x^1)=(x, y)

Homework Equations

Killing equations:
L_Xg_a{_b} = X^e\partial_eg_a{_b}+g_a{_d}\partial_bX^d+g_b{_d}{\partial}_aX^d = 0

The Attempt at a Solution

L_Xg_x{_x} = X^a+x\partial_xX^a=0
L_Xg_x{_y}= L_Xg_y{_x}=x\partial_yX^a+\partial_xX^a=0
L_Xg_y{_y}=X^a+2x\partial_yX^a=0
In the back of the book, it says the solution is \frac{\partial}{{\partial}y}. I don't really know what they mean by that. I've always seen \frac{\partial}{{\partial}y} as an operation that takes the partial derivative of something with respect to y, not a value. I thought perhaps it meant any function that depends only on y and not on x, but if I plug that into the equations above, I get:
f(y)\neq0
x\partial_yf\neq0
f(y)+2x\partial_yf\neq0
I'm pretty sure I've got the equations correct; I just don't know what they mean when they say the solution is \frac{\partial}{{\partial}y}.

 

[PhyPsy]

The equations in the first post are wrong. The equations should be:L_Xg_x{_x}=2xX^x+2x^2\partial_xX^x=0L_Xg_x{_y}=L_Xg_y{_x}=x^2\partial_yX^x+x\partial_xX^y=0L_Xg_y{_y}=X^x+2x\partial_yX^y=0
The first equation restricts you to X^x=\frac{1}{x}f(y), and anything I do with the second and third equations, I come up with -\frac{1}{2x^3}f(y)=\partial_y\partial_yf(y). Obviously, there is no solution to this except f(y)=0, so I'm still finding the only possible solution to be X^x=0, X^y=c.

I found out that \frac{\partial}{{\partial}y} indeed means any function of y. I'm not sure why they would offer that as a solution when there are 2 vectors I am solving for. Can anyone figure out why \frac{\partial}{{\partial}y} is a solution in the back of the book?

 

[Dick]

d/dx and d/dy are a basis for the space of tangent vectors. So a general vector is written as X^x*d/dx+X^y*d/dy. Saying d/dy is a solution is just saying X^x=0 and X^y=1 is a solution. And it is, isn't it? You've already shown that. If you think of what a Killing vector means as an isometry of the metric, then d/dy corresponds to translation in the y direction. And it's pretty obvious that's an isometry, yes?