Could someone please explain what type of problem this is? I can't seem to find any good examples anywhere. I know it's a chain rule problem, but almost all examples i find is something like: find $ \dfrac{\partial f}{\partial t}$ of some given function.
And it doesnt look like a typical differential equation problem either. Our teacher said it should be solved using the chain rule
Find the solution to: $y\dfrac{\partial f}{\partial x} + x\dfrac{\partial f}{\partial y} = xyf(x,y)$
That satisfies $f(0,y)=y^2$
Use:
$u = ax^2+y^2$
$v = e^{-1/2x^2}$
Im hard stuck with this problem. Our teacher just mentioned the chain rule fast in the end of a lecture. My textbook doesnt explain it very thoroughly either.
All I've gotten so far is:
$\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial x} + \dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial x} = \dfrac{\partial f}{\partial u} \cdot 2ax + \dfrac{\partial f}{\partial v} \cdot -xe^{-1/2x^2}$
$\dfrac{\partial f}{\partial y} = \dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial y} + \dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial y} = \dfrac{\partial f}{\partial u} \cdot 2y + \dfrac{\partial f}{\partial v} \cdot 0$
Substitution:
$y\cdot (\dfrac{\partial f}{\partial u} \cdot 2ax + \dfrac{\partial f}{\partial v} \cdot -xe^{-1/2x^2}) + x \cdot (\dfrac{\partial f}{\partial u} \cdot 2y) = xyf(x,y)$
$2a\dfrac{\partial f}{\partial u} - e^{-1/2x^2}\dfrac{\partial f}{\partial v} + 2\dfrac{\partial f}{\partial u} = f(x,y)$
Now I have no clue where to go from here. I've looked at similar examples from the book and online but they all seem so different that i cant really apply it to my problem. I feel really stupid right now.
Any help is greatly appreciated
