3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Do you have any ideas of how to begin? $\endgroup$
    – Ron Gordon
    Commented Mar 24, 2013 at 3:03
  • $\begingroup$ 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. $\endgroup$
    – kuzand
    Commented Mar 24, 2013 at 3:46

1 Answer 1

3
$\begingroup$

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} $ $$ \begin{align} f_{\xi} = \dfrac{\partial f}{\partial \xi} &= \dfrac{\partial f}{\partial x} \cdot \dfrac{\partial x}{\partial \xi} + \dfrac{\partial f}{\partial y} \cdot \dfrac{\partial y}{\partial \xi} \\ &= f_x + f_y \cdot (2) \\ &= 2 f_y + f_x \end{align} $$

$$ \therefore f_x = f_{\xi} + 2 f_{\eta} $$

$\endgroup$
1
  • $\begingroup$ So $f_{\xi}=1$? How do you solve it then? $\endgroup$
    – kuzand
    Commented Mar 24, 2013 at 19:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .