How to think when solving $3\frac{\partial f}{\partial x}+5\frac{\partial f}{\partial y}=0$? Solve this differential equation
$$3\frac{\partial f}{\partial x}+5\frac{\partial f}{\partial y}=0$$
Usually, when we get these problems, they tell us what variable change is smart to do and we just chunk through the chain rule and end up with an answer. Now, you have to think for yourself what variables to put. Therefore, I did not know how to do.
I know the variable substitution that they made, but not why the made it. How should I think here? 
My first guess was $e^{5x-3y}+C$ which does indeed solve it, but that solution is not general enough. Any one function $g(5x-3y)$ will do it, according to the solution manual.
 A: The given equation says that the gradient $\nabla f(x,y)$ is at all points $(x,y)$ orthogonal to the vector $(3,5)$. This implies that $f$ is constant on all lines which are parallel to this vector. These are the lines of slope ${5\over3}$, which can be written as $$\ell_c: \quad  5x-3y=c\ .$$ The value of $f$ on $\ell_c$ is determined by its "number" c. It follows that there is a function $$g: \>{\mathbb R}\to{\mathbb R},\qquad c\mapsto g(c)\ ,$$ such that one has
$$f(x,y)=g(5x-3y)\qquad\bigl((x,y)\in{\mathbb R}^2\bigr)\ .\tag{1}$$
Conversely: If $f$ is given by $(1)$ with an arbitrary differentiable $g$ then by the chain rule we have
$$\nabla f(x,y)=g'(5x-3y)\>(5,-3)\ ;$$
whence $\nabla f$ is orthogonal to $(3,5)$ at all points.
A: I don't think it's so instructive to think of this in terms of substitutions, it doesn't explain what is really going on. The way to think about it is that $\displaystyle 3f_x+5f_y=\lim\limits_{h\to 0}\frac{f(x+3h,y+5h)-f(x,y)}{h}$, so by your equation we have that $f$ is constant along the lines $(x(t),y(t))=(x_0+3t,y_0+5t)$ for fixed $(x_0,y_0)$, ie. lines of the form $\displaystyle 3(y-y_0)=5(x-x_0)$, or equivalently $\displaystyle 3y-5x=3y_0-5x_0$. So the value $3y-5x$ determines the value of the function (since if $\displaystyle 3y_1-5x_1=3y_2-5x_2$ then $\displaystyle f(x_1,y_1)=f(x_2,y_2)$ by the above), which is equivalent to saying that $f(x,y)=g(3y-5x)$, and conversely any function of this form clearly solves the PDE.
A: Hint: if you use the change of variables $s = 5 x - 3 y, t = 3 x + 5 y$, what do you get?
A: This problem can be solved by using the method of characteristics which for a such a linear PDE reads: 
$$ \frac{dx}{3} = \frac{dy}{5} = \frac{df}{0}. $$ While the third fraction tells us that $f = C$, for some constant $C$, the two first gives us the characteristic curve where $f$ is constant, i.e:
$$ 5 dx - 3 dy = 0 \Rightarrow 5x-3y = D.$$ Put now $C$ as a function of $D$ to have:
$$f(x,y) = C(5x  - 3y), $$ where $C$ is now an arbitrary function of $5x-3y$.
Cheers!
A: Basically, when given a change of variables $u=u(x,y),v=v(x,y)$, you will use the Chain  rule to transform your equation with unknowns $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ (i.e. the gradient $\nabla f(x,y)$) into an equation with unknowns $\dfrac{\partial f}{\partial u}$ and $ \dfrac{\partial f}{\partial v}$ (i.e. the gradient $\nabla f(u,v)$).
So you may want to think about $0=3\dfrac{\partial f}{\partial x}+5\dfrac{\partial f}{\partial y}$ as $0=\dfrac{\partial x}{\partial u}\dfrac{\partial f}{\partial x}+\dfrac{\partial y}{\partial u}\dfrac{\partial f}{\partial y}=\dfrac{\partial f}{\partial u}$.
So $\dfrac{\partial x}{\partial u}=3$ and $\dfrac{\partial y}{\partial u}=5$, and $x(u,v)=3u+g(v)$ and $y(u,v)=5u+h(v)$. Choos $g$ and $h$ for the transformation to be invertible and computations easy.
For example, take $x=3u-2v$ and $y=5u-3v$ (so $u=-3x+2y$ and $v=-5x+3y$), we obtain that $\dfrac{\partial f}{\partial u}=0$, so $f$ is a constant function of $u$: $f(u,v)=\phi(v)$ for some function $\phi$. Back to the $(x,y)$-coordinates, we have $f(x,y)=\phi(3y-5x)$ for some function $\phi$.
