I have to solve these PDE using change of variables.
1)$p(x) $$\partial f \over \partial x $$+q(y) $$\partial f \over \partial y $=0
A hint is given as to use $u=\int {1\over p}dx$, $v=\int {1\over q}dy$
2)$f_x+f_{yx}=0$
I did the problem $f_x+f_{xy}=0$ with the substitution $u=f_x $.Then the equation simplifies to $U+U_y=0$
$\int {1 \over u}du$=$\int -dy$
But the problem is in $f_x+f_{yx}=0$ if I take $u=f_x $ I have a problem substituting for $f_{yx}$ part.
Any help on how I can do these problems please
Hint: If you let $u = f_x$, then we have ( since $f_{xy} = f_{yx} = \partial_y ( f_x)$ )
$$ f_x + f_{xy} =0 \iff u + u_y = 0 \iff -\int dy = \int \frac{du}{u} \iff u = C(x)\exp(-y)$$
Then all you have to solve is
$$ f_x = C(x) \exp(-y) $$
Where $C(x)$ is any function that depends on $x$. If you have some conditions on the function, you can then get an explicit form.
Edit: Otherwise all you have is
$$f(x,y) = \exp(-y) \int C(x) dx = F(x) \exp(-y) $$
where $F(x)$ is any function $F:\mathbb{R} \to \mathbb{R}$ that is differentiable
Also, for your first question consider this:
$$ p(x) f_x + q(y) f_y = 0 \iff \frac{p(x)f_x}{f_y} = - q(y) $$
What does this tell us? This should lead you to the conclusion that $f = u-v$ where $u$ and $v$ are as you defined.