Folks,
Could anyone check my attached work. Mathematica seems to output u(x,y)=-3+2x for the particular solution...
Thanks
If we parametrize x and y we have
$\displaystyle x(s) \text{ and } x(0)=1$
and
$\displaystyle y(s) \text{ and } y(0)=y_0$
Then
$\displaystyle u(x(s),y(s))$
Now if we take the derivative with respect to s we have
$\displaystyle \frac{d}{ds}u(x(s),y(s))=\frac{dx}{ds}u_x+\frac{dy }{ds}u_y=u+3$
This gives the system of equations
$\displaystyle \frac{dx}{ds}=x, \quad x(0)=1$
$\displaystyle \frac{dy}{ds}=y, \quad y(0)=y_0$
$\displaystyle \frac{du}{ds}=u+3$
Solving this gives
$\displaystyle x=e^{s} \quad y=y_0e^{s} \quad u=Ce^{s}-3$
Now if we sub out
$\displaystyle e^{s}$
we get
$\displaystyle u(x,y)=Cx-3 \implies u(1,y)=-1=C-3 \iff C=2$
$\displaystyle u(x,y)=2x-3$
Here's your mistake
First let $\displaystyle e^{f} = g$ so $\displaystyle u = x g\left(\dfrac{y}{x}\right) - 3$. If $\displaystyle u(1,y) = -1$ then $\displaystyle -1 = g(y) - 3$ so $\displaystyle g(y) = 2$ so $\displaystyle u = x \cdot 2 - 3$ (your Mathematica and TheEmptySet's solution)