Solve the partial differential equation $x\frac{\partial^2f}{\partial x\partial y}-y\frac{\partial^2f}{\partial y^2}-\frac{\partial f}{\partial y}=0$. Solve the partial differential equation
$$
x\frac{\partial^2f}{\partial x\partial y}-y\frac{\partial^2f}{\partial y^2}-\frac{\partial f}{\partial y}=0
$$ where $x>0$, through the the variable change $u=x$ and $v=xy$.
This is from my calc 3 exam and I don't think it should be hard because we only had a few pages on partial differential equations. But I don't know how to do this one at all.
Edit: from $f(u,v)$ I get the following.
Since $u_y=0$ and $v_y=x$ we have
$$
f_y=f_uu_y+f_vv_y=f_vx.
$$ I'm stuck att calculating $f_{yy}$ since I don't know how I should handle the $x$ in $f_vx$ when deriving with respect to $v$.
$$
f_{yy}=(f_y)_uu_y+(f_y)_vv_y=(f_vx)_vx 
$$
I'm thinking since $v=xy$ then $x=v/y$ so
$$
(f_vx)_vx=f_{vv}x^2+f_v(x/y).
$$
Now
$$
f_{xy}=f_{yx}=(f_y)_uu_x+(f_y)_vv_x=(f_vx)_u+(f_vx)_uy=f_{vu}x+f_{v}+f_{vu}xy+f_vy.
$$
Putting everything in the equation gives
$$
(f_{vu}x+f_{v}+f_{vu}xy+f_vy)x-y(f_{vv}x^2+f_v(x/y))-f_vx=f_{vu}x^2+f_{v}x+f_{vu}x^2y+f_vyx-f_{vv}yx^2-f_vx-f_vx=0.
$$
Looks horrible.
 A: When changing variables, it can be helpful to write the function of the new variables with a different name and leave the dependence of $u$ and $v$ on $x$ and $y$ explicit. Then you substitute in $u$ and $v$ only after all the derivatives are calculated.
Calling our function $g(u,v)$, we would have
$$
f(x, y) = g(x, xy)
$$
Now applying derivatives to this and using the chain rule gives
\begin{eqnarray}
\frac{\partial f}{\partial y} &=& \left[\frac{\partial g}{\partial v}\right](x,xy) =\frac{\partial (xy)}{\partial y} = x\frac{\partial g}{\partial v}(x,xy) = u\frac{\partial g}{\partial v}\\
\frac{\partial^2 f}{\partial y^2} &=& \frac{\partial}{\partial y}\left[x\frac{\partial g}{\partial v}(x,xy)\right] = x^2\frac{\partial^2 g}{\partial v^2}(x,xy) = u^2\frac{\partial^2 g}{\partial v^2} \\
\frac{\partial^2 f}{\partial x\partial y} &=& \frac{\partial}{\partial x}\left[x\frac{\partial g}{\partial v}(x,xy)\right] = \frac{\partial g}{\partial v}(x,xy) + x\frac{\partial^2 g}{\partial u\partial v}(x,xy) + x y \frac{\partial^2 g}{\partial v^2}(x,xy) \\&=& \frac{\partial g}{\partial v}+u\frac{\partial^2 g}{\partial u\partial v}+v\frac{\partial^2 g}{\partial v^2}
\end{eqnarray}
Putting these into the equation and using $x = u, y = v/u$ gives
\begin{multline}
u\left[\frac{\partial g}{\partial v} + u\frac{\partial^2 g}{\partial u\partial v} + v \frac{\partial^2 g}{\partial v^2}\right] - \frac{v}{u}\left[u^2\frac{\partial^2 g}{\partial v^2}\right]-u\frac{\partial g}{\partial v} \\= u\left[\frac{\partial g}{\partial v} + u\frac{\partial^2 g}{\partial u\partial v} + v \frac{\partial^2 g}{\partial v^2} - v \frac{\partial^2 g}{\partial v^2} - \frac{\partial g}{\partial v}\right] = u^2\frac{\partial^2 g}{\partial u\partial v} = 0.
\end{multline}
This easily solves to $g(u,v) = A(u) + B(v)$ for some functions $A,B$ determined by the boundary conditions, which in turn gives $f(x, y) = g(x,xy) = A(x) + B(xy)$.
A: Let $g = f_y$. Then you can rewrite your equation as $xg_x-yg_y=g$. Now consider, in abuse of notation, $g=g(u,v)=g(x,y)$, with $u=x, v=xy$. Applying chain rule
$$
g_x=g_uu_x+g_vv_x=g_u + yg_v=g_u+\frac{v}{u}g_v \\
g_y = g_uu_y+g_vv_y= 0 + xg_v=ug_v
$$
Substituting, one gets
$$
u(g_u+\frac{v}{u}g_v)-\frac{v}{u}ug_v =ug_u=g \Rightarrow g= uh(v)
$$
Finally $f_y =xh(xy) \Rightarrow f = H(xy) + j(x)$
