I have the following PDE before me:
$xu_x+u_y=x \sinh(y)+u$ with initial condition $u(0,y)=0$
I wrote the characteristic equations as below:
$x_\tau=x, x(0,s)=0$ $y_\tau=1, y(0,s)=s$ $u_\tau=x \sinh(y)+u, u(0,s)=0$
Solving initial condition problem for $x,y$, I got:
$x=0$
$y=\tau+s$
After putting $x=0$ in equation for $u_\tau$, I got:
$u_\tau=u, u(0,s)=0$
Solving initial condition problem for $u$, I got:
$u=0$.
My question is whether I am doing all this correctly. Is the final answer to this question going to be $u=0$. Please suggest.
$$xu_x+u_y=x \sinh(y)+u$$ Charpit Lagrange characteristic ODEs :
$$\frac{dx}{x}=\frac{dy}{1}=\frac{du}{x \sinh(y)+u}$$ A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{1}$ : $$x\:e^{-y}=c_1$$ $\sinh(y)=\frac12\left(\frac{x}{c_1}-\frac{c_1}{x} \right)$
A second characteristic equation comes from solving $\frac{dx}{x}=\frac{du}{x \sinh(y)+u}$
$\frac{du}{dx}=\sinh(y)+\frac{u}{x}=\frac12\left(\frac{x}{c_1}-\frac{c_1}{x} \right)+\frac{u}{x}\quad$ which solution is : $$\frac{u}{x}-\frac{x}{2c_1}-\frac{c_1}{2x}=c_2$$ The general solution of the PDE on implicit form $c_2=F(c_1)$ is : $$\frac{u}{x}-\frac{x}{2c_1}-\frac{c_1}{2x}=F(x\:e^{-y})$$ $F$ is an arbitrary function (to be determined according to the boundary condition). $$\frac{u}{x}-\frac{e^y}{2}-\frac{e^{-y}}{2}=F(x\:e^{-y})$$ The explicit form of the general solution of the PDE is : $$\boxed{u(x,y)=x\:\cosh(y)+x\:F(x\:e^{-y})}$$ CONDITION :
$u(0,y)=0=0*\:\cosh(y)+0*F(0\:e^{-y})$
This is true any function $F$. Thus the function $F$ cannot be determined.
Conclusion : They are infinity many solutions which satisfy both the PDE and the specified condition. The equation of those solutions is the above equation of the general solution (any function $F$ ).