task is to transform the differential equation
$$x\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial x\partial y}+\frac{\partial f}{\partial x}=xe^{-2y}$$
using the variabel substitution
$$u=xe^{-y}, v=y.$$
I have started with the chain rule for the partial derivatives of the first order, I get
$$\frac{\partial f}{\partial x}=e^{-y}\cdot\frac{\partial f}{\partial u}$$
$$\frac{\partial f}{\partial y}= -xe^{-y}\cdot\frac{\partial f}{\partial u}+\frac{\partial f}{\partial v}$$
Then when trying to transform $\frac{\partial^2 f}{\partial x^2}$ I get
$$\frac{\partial^2 f}{\partial x^2}=\frac{\partial }{\partial x}(\frac{\partial f}{\partial x})=\frac{\partial }{\partial x}(e^{-y}\frac{\partial f}{\partial u})$$
But now I am stuck. I don't think I can use the product rule for derivatives here since I want to take the derivative with respect to x and $e^{-y}$ is not a function of x.
EDIT:
I have now manage to transform all the partial derivatives that are included in the equation, I get
$$\frac{\partial^2 f}{\partial x^2}=\frac{\partial^2 f}{\partial u^2}\cdot e^{-2y}$$
$$\frac{\partial^2 f}{\partial x \partial y}=\frac{\partial }{\partial x}(-xe^{-y}\cdot \frac{\partial f}{\partial u}) + \frac{\partial }{\partial x}(\frac{\partial f}{\partial v}) = -e^{-y}\cdot \frac{\partial f}{\partial u}-xe^{-2y}\cdot \frac{\partial^2 f}{\partial u^2}+\frac{\partial^2 }{\partial u \partial v}$$
If i insert that to the equation I get
$$xe^{-2y}\frac{\partial^2 f}{\partial u^2}-e^{-y}\frac{\partial f}{\partial u}-xe^{-2y}\cdot \frac{\partial^2 f }{\partial u^2}+\frac{\partial^2 f}{\partial u \partial v}+e^{-y}\frac{\partial f}{\partial u}=xe^{-2y}$$
which simplifies to
$$\frac{\partial^2 }{\partial u \partial v} = xe^{-2y}$$
My textbook has written the answer as
$$\frac{\partial^2 }{\partial u \partial v} = u$$
Did I do something wrong along the way or is my answer correct?
Best Answer
HINT $e^{-y}$ is constant with respect to $x$.