PDE separation of variables, weird textbook example Getting a good grasp of these type of questions now but the end of this one is tripping me up a bit.
Question: Use separation of variables to find a product solution for the partial differential equation
$$y*∂_{xy}u + u = 0,$$
where u is a function of x and y.

My attempt:
Substitute $u(x,y)=X(x)Y(y)$ into the PDE to obtain
$$yX'Y'+XY=0$$
Then using separation of variables and the separation constant $-\lambda,$
$$ \frac{X'}{X}=\frac{-Y}{yY'}=-\lambda$$
So we have
$$X'=-\lambda X,\ \ \ \ \ \ \ Y'=\frac{1}{\lambda y}Y$$
which according to the textbook has these solutions:
$$X=c_1e^{-\lambda x},\ \ \ \ \ \ \ Y=c_2y^{1/\lambda}.$$
with the product solution
$$u=XY=c_3e^{-\lambda x}y^{1/\lambda}.$$
Now I understand the solution for $X$, but I cannot figure out how to obtain $c_2y^{1/\lambda}$ for $Y$. I think the starting point is  $$Y=c_2e^{(1/\lambda y)*y} $$
but that doesn't seem to get me there unless I've missed something really obvious...
 A: To derive 
$Y = c_2 y^{1/\lambda} \tag 1$
from
$Y' = \dfrac{dY}{dy} = \dfrac{1}{\lambda y} Y, \tag 2$
write it as
$(\ln Y)' = \dfrac{Y'}{Y} = \dfrac{1}{\lambda y} = \dfrac{1}{\lambda} \dfrac{1}{y} = \dfrac{1}{\lambda} (\ln y)'; \tag 3$
we integrate from $y_0$ to $y$:
$\ln Y(y) - \ln Y(y_0) = \displaystyle \int_{y_0}^y (\ln Y(z))' \; dz =  \dfrac{1}{\lambda} \int_{y_0}^y (\ln z)' \; dz =  \dfrac{1}{\lambda} (\ln y - \ln y_0);  \tag 4$
or
$\ln \left ( \dfrac{Y(y)}{Y(y_0)} \right ) = \dfrac{1}{\lambda} \ln \left ( \dfrac{y}{y_0} \right ) = \ln \left ( \dfrac{y^{1/\lambda}}{y_0^{1/\lambda}} \right ), \tag 5$
whence
$Y(y) = \dfrac{Y(y_0)}{y_0^{1/\lambda}} y^{1/\lambda} = c_2 y^{1/\lambda}, \tag 6$
where
$c_2 = \dfrac{Y(y_0)}{y_0^{1/\lambda}}. \tag 7$
A: $$
\frac{X'}{X} = -\frac 1y \frac{Y}{Y'} = \lambda
$$
so
$$
X(x) = C_1 e^{\lambda x}\\
Y(y) = C_2 y^{-\lambda }
$$
hence
$$
u(x,y) = C_0 \left(\frac{e^x}{y}\right)^{\lambda}
$$
or also
$$
u(x,y) = C_0' \left(y e^{-x}\right)^{\lambda}
$$
considering 
$$
\frac{X'}{X} = -\frac 1y \frac{Y}{Y'} = -\lambda
$$
A: $$ Y'=\frac{1}{\lambda y}Y$$
$$ \frac   {Y'}Y=\frac{1}{\lambda y}$$
Integrate
$$ \int \frac   {dY}Y=\int \frac{dy}{\lambda y}$$
$$ \ln(Y)=\frac 1 {\lambda} \ln (y) +K$$
$$Y(y)=e^{1/\lambda(\ln(y)+K)}$$
$$Y(y)=Cy^{1/\lambda}$$
