Q:

find the particular solution of the semi linear equation given the data u(1,y)= ln(y^(-1/2))

semi linear equation
the general solution to that equation
(general solution)
Given the data that u(1,y)=ln(y^(-1/2)) what would the right method be to go from the general solution to the particular solution given the data?
particular solution

A:

This is the continuation of Find the general solution of $x^3u_x-u_y=e^{2u}$ to which a boundary condition is now added.
The general solution is 
$$y-\frac12 e^{-2u}=f(y-\frac{1}{2x^2}) \tag 1$$
and the condition is :
$$u(1,y)=\ln(y^{-1/2})=-\frac12\ln(y)$$
$e^{-2u}=e^{\ln(y)}=y\quad$ that we put into $(1)$ with $x=1$ :
$$y-\frac12 y=f(y-\frac{1}{2}) $$
$$\frac12 y=f(y-\frac{1}{2})$$
Let $\quad X=y-\frac{1}{2} \quad\implies\quad y=X+\frac12$
$$\frac12 (X+\frac12)=f(X)$$
Now the function $f$ is determined :
$$f(X)=\frac12X +\frac14$$
We put it into $(1)$ where $X=y-\frac{1}{2x^2}$
$$y-\frac12 e^{-2u}= \frac12(y-\frac{1}{2x^2}) +\frac14$$
$$e^{-2u}= y+\frac{1}{2x^2} -\frac12$$
$$-2u= \ln\left(y+\frac{1}{2x^2} -\frac12\right)$$
The particular solution satisfying the specified condition is :
$$u=-\frac12\ln\left(y+\frac{1}{2x^2} -\frac12\right)$$

