How Does the Substitution x=es, y=et Transform the PDE for f(x,y)?

[CrimsnDragn]

Homework Statement

the substitution x=e^s, y=e^t converts f(x,y) into g(s,t) where g(s,t)=f(e^s, e^t). If f is known to satisfy the partial differential equation

x²(d²f/dx²) + y²(d²f/dy²) + x(df/dx) + y(df/dy) = 0

show that g satisfies the partial-differential equation

(d²g/ds²) + (d²g/dt²) = 0

The Attempt at a Solution

I feel like this is a simple problem - All I need to figure out is how to find the partial derivative of f(x,y) and the double partial derivative, but I'm not sure how to do it. Are the values of x and y relevant for the first part of the question?

what would df/dx be? D1f(x,y) = (y)f'(x,y) by the chain rule? and similarly df/dy = D2f(x,y) = (x)f'(x,y)? Someone please help!

 

[lanedance]

i assume they're meant to be partials... & I don't really understand what you've written at the end of the post but here's some ideas I hope help

now as you're given
$$x(s)=e^s$$
$$y(t)=e^t$$

invert these and then re-write as
$$s(x)=ln(s)$$
$$t(y)=ln(y)$$

so the relation function g & f is given as
$$g(s,t) = f(x(s), y(t))$$

we can re-write it and think of it as
$$f(x,y) = g(s(x), t(y))$$

now try taking a partial w.r.t x and using the chain rule on the RHS

$$\frac{\partial f(x,y)}{\pratial x} = \frac{\partial }{\partial x} g(s(x), t(y))$$
the partial w.r.t. x means the other variable (y) in this case is kept constant