While solving an PDE using method of characteristics on of the equations I got is:
$$y_t=x+y\Rightarrow y_t-y=x$$
This is a linear first order ODE that can be solve as follow:
$$y_h=c_2e^t$$
Then using variation of parameters should $y_p=c_s(s)e^t$ or $y_p=c_2(t,s)e^t$?
The solution is still $y=y_h+y_p$?
The PDE is:
$$ \begin{cases} xu_x+(x+y)u_y=1\\ u(1,y)=y\\ \end{cases} $$
Which I once asked about
On
$$\begin{cases} xu_x+(x+y)u_y=1\\ u(1,y)=y\\ \end{cases} $$ $$\frac {dx}{x}=\dfrac {dy}{x+y}=\dfrac {dz}{1}$$ So you need to solve this system of DE: $$\begin{cases}\dfrac {dx}{x}=\dfrac {dy}{x+y} \\ \dfrac {dz}{1}=\dfrac {dx}{x} \tag{2}\end{cases}$$ $$\dfrac {dx}{x}=\dfrac {dy}{x+y} $$ $$(x+y)dx=xdy $$ $$ydx-xdy=-xdx$$ $$\frac {ydx-xdy}{x^2}=-\frac 1 x dx$$ $$\frac {xdy-ydx}{x^2}=\frac 1 x dx$$ $$d\left (\frac y x \right )= \frac 1 x dx$$ Integrate: $$ y (x) = x\ln x+C_2x$$ The second one is easy to integrate: $$\dfrac {dz}{1}=\dfrac {dx}{x} \tag{2}$$ $$z+C_1=\ln x$$ You can surely take from there.
You need to treat this equation together with the equation $$ x_t=x $$ for $x$ or the coefficient of $u_x$, as alone there is no control on what $x(t)$ is. Now that $x(t)=e^tx_0$ you can solve the original equation $$ y_t-y=x=e^tx_0\implies (e^{-t}y(t))_t=x_0\implies y(t)=e^ty_0+te^tx_0. $$