# Method of characteristics and first order ODE

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:

$$\begin{cases} xu_x+(x+y)u_y=1\\ u(1,y)=y\\ \end{cases}$$

• Are you sure you are not confusing $x$ and $t$? I cannot understand how you have solved the first ODE. Commented Nov 28, 2019 at 12:03
$$\begin{cases} xu_x+(x+y)u_y=1\\ u(1,y)=y\\ \end{cases}$$ $$\frac {dx}{x}=\dfrac {dy}{x+y}=\dfrac {dz}{1}$$ 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 $$y_t-y=x=e^tx_0\implies (e^{-t}y(t))_t=x_0\implies y(t)=e^ty_0+te^tx_0.$$
• Where did the $y_t,y$ from $y_t-y=x=e^tx_0$ came from? Commented Dec 1, 2019 at 10:00
• This is the equation that you wanted to solve, I just inserted the solution for $x$. Commented Dec 1, 2019 at 10:29