I have $xu_x+2xy^2u_y=u$ with the initial condition $u(x,1)=sinx$ for $x>0$ so a parametrization is $(x(0,s),y(0,s),u(0,s))=(s,1,\sin(s))$ for $s>0$
what I have done so far: $a=x'(t)=x, b=y'(t)=2xy^2, c=u'(t)=u$ by solving these ODE and substituting the initial condition I've got $$x(t,s)=se^t, y(t,x)=\frac{1}{1-2xt}, u(t,s)=\sin(s)e^t$$ by extracting from $y(t,s)$ I've got $t=\frac{y-1}{2xy}$ and substituting in $x(t,s)$ gives $s=xe^{\frac{1-y}{2xy}}$ and lastly substituting $t$ and $s$ in $u(t,s)$ gives $$u(x,y)=\sin(xe^{\frac{1-y}{2xy}})e^{\frac{y-1}{2xy}},$$ but it seems to be wrong because I've been trying to differentiate and substitute back in the PDE and it's not giving me what I want.
$$xu_x+2xy^2u_y=u$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{x}=\frac{dy}{2xy^2}=\frac{du}{u}$$ A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{2xy^2}$ $$x+\frac{1}{2y}=c_1$$ A second characteristic equation comes from solving $\frac{dx}{x}=\frac{du}{u}$ $$\frac{u}{x}=c_2$$ General solution of the PDE on the form of implicit equation $c_2=F(c_1)$ : $$\boxed{u=xF\left(x+\frac{1}{2y}\right)}$$ $F$ is an arbitrary function (to be determined according to the specified condition).
Condition :
$u(x,1)=\sin(x)=xF\left(x+\frac{1}{2}\right)$
Let $X=x+\frac{1}{2}\quad\implies\quad x=X-\frac12$
$\sin(X-\frac12)=(X-\frac12)F(X)$ $$F(X)=\frac{\sin(X-\frac12)}{X-\frac12}$$ The function $F(X)$ is determined. We put it into the above general solution where $X=x+\frac{1}{2y}\quad\implies\quad F(x+\frac{1}{2y})=\frac{\sin(x+\frac{1}{2y}-\frac12)}{x+\frac{1}{2y}-\frac12}$
$$u=x\:\frac{\sin(x+\frac{1}{2y}-\frac12)}{x+\frac{1}{2y}-\frac12}$$ This is the particular solution which satisfies both the PDE and the condition.
If the boundary condition $u(x,1)=\sin(x)$ is limited to the range $x>0$ the above particular solution is valid in the range $\quad x+\frac{1}{2y}>\frac12$