Jtdylktuy

Solve the following PDE

$$xu_x+(x+y)u_y=1$$ when $$u(1,y)=y$$ using method of characteristics and find the projections of the characteristics on the xy plane

$$begin{cases}
a=x\
b=x+y\
c=1
end{cases}Rightarrow begin{cases}
frac{dx}{dt}=xiff frac{dx}{x}=dtiff ln(x)=t+c_1iff x=ke^t\
frac{dy}{dt}=x+yiff frac{dy}{dt}=ke^t+yiff y=(kt+m)e^t\
frac{du}{dt}=1iff du=dtiff u=t+c_2
end{cases}$$

How do I solve $?_1$ and what should be my next step?

$$xu_x+(x+y)u_y=1$$
You have got the correct system of equation which can be written on this form :
$$frac{dx}{x}=frac{dy}{x+y}=frac{du}{1}=dt$$
Solving $frac{dx}{x}=frac{dy}{x+y}$ gives a first characteristic
$$frac{y}{x}-ln|x|=c_1$$
Solving $frac{dx}{x}=frac{du}{1}$ gives a second characteristic
$$u-ln|x|=c_2$$
The general solution of the PDE is $u-ln|x|=Fleft(frac{y}{x}-ln|x|right)$
$$u(x,y)=ln|x|+Fleft(frac{y}{x}-ln|x|right)$$
where $F$ is an arbitrary function.

This function is determined according to the boundary condition :

$u(1,y)=y=ln|1|+Fleft(frac{y}{1}-ln|1|right)=F(y)$

Thus $F(y)=y$ and as a consequence $Fleft(frac{y}{x}-ln|x|right)=frac{y}{x}-ln|x|$ .

$u(x,y)=ln|x|+left(frac{y}{x}-ln|x|right)$

The final solution is :
$$u(x,y)=frac{y}{x}$$

The second equation is a linear inhomogeneous ODE of first order. The standard solution methods apply. For instance, $e^{-t}$ is an integrating factor,
$$
frac{d}{dt}(e^{-t}y(t))=e^{-t}(y'(t)-y(t))=kimplies y(t)=(kt+m)e^t
$$