Solve $xu_x+(x+y)u_y=1$ when $u(1,y)=y$ 
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

$$\begin{cases}
a=x\\
b=x+y\\
c=1
\end{cases}\Rightarrow \begin{cases}
\frac{dx}{dt}=x\iff \frac{dx}{x}=dt\iff ln(x)=t+c_1\iff x=ke^t\\
\frac{dy}{dt}=x+y\iff \frac{dy}{dt}=ke^t+y\iff y=(kt+m)e^t\\
\frac{du}{dt}=1\iff du=dt\iff u=t+c_2
\end{cases}$$
How do I solve $?_1$ and what should be my next step?
 A: $$xu_x+(x+y)u_y=1$$
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 $u-\ln|x|=F\left(\frac{y}{x}-\ln|x|\right)$
$$u(x,y)=\ln|x|+F\left(\frac{y}{x}-\ln|x|\right)$$
where $F$ is an arbitrary function.
This function is determined according to the boundary condition :
$u(1,y)=y=\ln|1|+F\left(\frac{y}{1}-\ln|1|\right)=F(y)$
Thus $F(y)=y$ and as a consequence $F\left(\frac{y}{x}-\ln|x|\right)=\frac{y}{x}-\ln|x|$ .
$u(x,y)=\ln|x|+\left(\frac{y}{x}-\ln|x|\right)$
The final solution is :
$$u(x,y)=\frac{y}{x}$$
A: Let us follow the method of characteristics.


*

*$\frac{\text d}{\text d t}x = x$, letting $x(0) = x_0$ we know $x = x_0e^t$;

*$\frac{\text d}{\text d t}y = x+y$, letting $t(0) = t_0$ we know $y = (x_0 t + y_0)e^t$;

*$\frac{\text d}{\text d t}u = 1$, letting $u(0) = u_0$ we know $u = t + u_0$.


Since the prescribed data is $u(1,y) = y$, we set $x_0 = 1$ and $u_0 = y_0$ -- in other words, when $t=0$ we impose $x=1$ and $u=y$. Using the equation of characteristics, we have
$$
\begin{aligned}
x &= e^t\\
y &= (t+y_0) e^t\\
u &= t+ y_0
\end{aligned}
\qquad\text{so that}\qquad
u(x,y) = \frac{y}{x} .
$$
The characteristics are the curves $y = x\, (\ln x + y_0)$ for $(x,y)$ in $\Bbb {R_+^*}\times \Bbb{R}$, which are represented below:

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