PDE separation of variables, weird textbook example

Question

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.

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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...

Answer 1

$$ Y'=\frac{1}{\lambda y}Y$$ $$ \frac {Y'}Y=\frac{1}{\lambda y}$$ Integrate $$ \int \frac {dY}Y=\int \frac{dy}{\lambda y}$$ $$ \ln(Y)=\frac 1 {\lambda} \ln (y) +K$$ $$Y(y)=e^{1/\lambda(\ln(y)+K)}$$ $$Y(y)=Cy^{1/\lambda}$$

Answer 2

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

$c_2 = \dfrac{Y(y_0)}{y_0^{1/\lambda}}. \tag 7$

Answer 3

$$ \frac{X'}{X} = -\frac 1y \frac{Y}{Y'} = \lambda $$

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

$$ \frac{X'}{X} = -\frac 1y \frac{Y}{Y'} = -\lambda $$