# Use chain rule to express $f_x$ and $f_y$ in terms of $f_\xi$ and $f_\eta$?

Consider partial differential equation: $f_x + 2 f_y = 1$

Trick is to introduce new variables $\xi = x$ and $\eta = 2x - y$

Using chain rule to express $f_x$ and $f_y$ in terms of $f_\xi$ and $f_\eta$ and rewrite equation 1 in terms of new variable. You should find that you can then antidifferentiate, after which you can return to the original variables. Your answer should contain an arbitrary function, and you should easily be able to check it satisfies equation 1.

• Do you have any ideas of how to begin? Commented Mar 24, 2013 at 3:03
• Don't know if this helps: $df=d\xi-f_yd\eta$. Also the function $f(x,y)=3x-y+c$ satisfies the partial d.e. Commented Mar 24, 2013 at 3:46

Since $\xi = x$
\begin{align} \eta &= 2x - y \\ \implies y &= 2x - \eta \\ &= 2 \xi - \eta \end{align}
Now, \begin{align} f_{\eta} = \dfrac{\partial f}{\partial \eta} &= \dfrac{\partial f}{\partial x} \cdot \dfrac{\partial x}{\partial \eta} + \dfrac{\partial f}{\partial y} \cdot \dfrac{\partial y}{\partial \eta} \\ &= f_x \cdot (0) + f_y \cdot (-1) \\ &= - f_y \end{align} and for $f_{\xi}$ \begin{align} f_{\xi} = \dfrac{\partial f}{\partial \xi} &= \dfrac{\partial f}{\partial x} \cdot \dfrac{\partial x}{\partial \xi} + \dfrac{\partial f}{\partial y} \cdot \dfrac{\partial y}{\partial \xi} \\ &= f_x + f_y \cdot (2) \\ &= 2 f_y + f_x \end{align}
$$\therefore f_x = f_{\xi} + 2 f_{\eta}$$
• So $f_{\xi}=1$? How do you solve it then? Commented Mar 24, 2013 at 19:00