I have to find the general solution of the following:
$$x^2u_x+yu_y=1$$
in the quarter plane $x,y > 0$
I tried solving by simple ODE with an integrating factor of $x^{2y}.$
my general solution came to
$$u(x,y)=\frac{(x^{2y+1})}{(2y+1)}+f(y)$$.
My question is does my general solution make sense, also how does the "in the quarter plane $x,y>0$ affect my solution?
Also,
Find a solution which satisfies $$u(x,\frac{x}{2})=\frac{1}{x}+1$$.
Thanks!
On
$$x^2u_x+yu_y=1 \tag 1$$ Lagrange-Charpit equations : $$\frac{dx}{x^2}=\frac{dy}{y}=\frac{du}{1}$$ A first family of characteristic curves comes from $\frac{dx}{x^2}=\frac{dy}{y}$ : $$\ln|y|+\frac{1}{x}=c_1$$ A second family of characteristic curves comes from $\frac{dx}{x^2}=\frac{du}{1}$ : $$u+\frac{1}{x}=c_2$$ The general solution of the PDE $(1)$ is : $$u+\frac{1}{x}=F\left(\ln|y|+\frac{1}{x}\right)$$ $$u(x,y)=-\frac{1}{x}+F\left(\ln|y|+\frac{1}{x}\right) \tag 2$$ $F$ is an arbitrary function, to be determined according to boundary condition.
Condition : $u(x,\frac{x}{2})=\frac{1}{x}+1$ $$\frac{1}{x}+1=-\frac{1}{x}+F\left(\ln|\frac{x}{2}|+\frac{1}{x}\right)$$ $$F\left(\ln|\frac{x}{2}|+\frac{1}{x}\right)=\frac{2}{x}+1$$
Then finding $F$ involves a special function (Lambert W). This draw to think that there is a mistake in the condition specified by the OP.
If the condition is definitively $u(x,\frac{x}{2})=\frac{1}{x}+1$ the calculus is:
Let $X=\ln|\frac{x}{2}|+\frac{1}{x}$
$x=-\frac{1}{W\left(\frac12e^{-X} \right)}$
$W$ is the Lambert W function. $$F(X)=-2W\left(\frac12 e^{-X} \right)+1$$
So, the function $F$ is determined. We put it into the general solution Eq.$(2)$ where $X=\ln|y|+\frac{1}{x}$ $$u(x,y)=-\frac{1}{x}-2W\left(\frac12 e^{-\ln|y|-\frac{1}{x}} \right)+1$$ $$u(x,y)=-\frac{1}{x}-2W\left(\frac{1}{2|y|} e^{-\frac{1}{x}} \right)+1$$
EDIT: The solution presented here was derived for the equation $$x^2 u_x + y u = 1,$$ which was the original object of the question. This derivation does not work for the equation presented now in the question.
You probably made some mistake while solving the equation, because the formula for $u(x,y)$ you provided isn't a solution to the equation. However, you don't even need to use the integrating factor, since the equation is separable, $$ \frac{u_x}{1-2yu}=\frac{1}{x}. $$ You can now readily integrate both sides, but I usually do it in the following "fancy" way: see that $$ \frac{\partial}{\partial x} (1-2yu) = -2yu_x. $$ Therefore, multiplying both sides of the separated equation by $-2y$ and changing $-2yu_x$ for $(1-2yu)_x$, $$ \frac{(1-2yu)_x}{1-2yu}=-\frac{2y}{x}. $$ Treating $y$ as a constant and integrating both sides (you can define $v=1-2yu$, and the LHS will be $v'/v$, whose integral is $\log |v|$), $$ \log |1-2yu|=f(y)-2y\log |x|. $$ Since $x>0$, $\log |x|= \log x$, therefore $$ |1-2yu| = x^{-2y} \exp (f(y)). $$ The RHS is always positive. Then, solving for $u$, $$ u= \frac{1}{2y} + \frac{g(y)}{x^{2y}}, $$ in which $g(y) = \exp(f(y))/2y$. Just as Martin pointed out in his comment, the requirement of $x,y>0$ allows us to do such manipulations. If you needed the solution for $x<0$, for example, some steps would require modifications.
For the condition $$ u(x,x/2) = 1+\frac{1}{x}, $$ we can simply substitute in the solution, leading to $$ g(x/2) = x^x. $$ Now, $g$ is simply a function, regardless of on which variable we are applying it; in the equality $g(x/2)=x^x$, $x$ is a free variable, because we can change $x$ for any other letter and the equality would be the same. Therefore, using $z=x/2$, we have $g(z)=(2z)^{2z}$, and the solution is $$ u= \frac{1}{2y} + \left(\frac{2y}{x}\right)^{2y}, $$ which satisfy $u=1+1/x$ for $y=x/2$.