Solve the initial value problem: $$\begin{cases}\frac{\partial u}{\partial t}&=(2x+y)\frac{\partial u}{\partial x}+(x+2y)\frac{\partial u}{\partial y},\\u(x,y,0)&=e^{x}\end{cases}$$ for $u=u(x,y,t)$.
The coefficients are $a(x,y)=(2x+y,x+2y)$. The characteristic curves have the equation $\frac{d}{dt}\gamma_{x}(t)=-a(\gamma_{x}(t))$. Let $\gamma_{x}(t)=(x(t),y(t))$. We obtain the system of equations, $\frac{dx}{dt}=2x+y$ and $\frac{dy}{dt}=x+2y$ $\Rightarrow \frac{dx}{2x+y}=\frac{dy}{x+2y}\Rightarrow\int(x+2y)dx=\int(2x+y)dy=\frac{x^{2}}{2}+2xy=2xy+\frac{y^{2}}{2}+c\Rightarrow x^{2}-y^{2}=c' $
Here, I'm not quite sure how to proceed. Any help is appreciated.
$$\frac{\partial u}{\partial t}=(2x+y)\frac{\partial u}{\partial x}+(x+2y)\frac{\partial u}{\partial y}-\frac{\partial u}{\partial t}=0$$ Sorry I have not enough time to edit more details and explanations.
Chappit-Lagrange system : $$\frac{dx}{2x+y}=\frac{dy}{x+2y}=\frac{dt}{-1}=\frac{du}{0}$$ First characteristic equation : $$u=c_1$$ Second characteristic equation, from : $\frac{dx+dy}{x+y}=\frac{dt}{-1}$ : $$(x+y)e^{3t}=c_2$$ Third characteristic equation, from : $\frac{dx-dy}{3x+y}=\frac{dt}{-1}$ : $$(x-y)e^{t}=c_3$$ Solution on the form $c_1=F(c_2,c_3)$ : $$u=F\big((x+y)e^{3t}\:,\:(x-y)e^{t} \big)$$ $F$ : arbitrary function of two variables, to be determined according to he condition $u(x,y,0)=e^x$ : $$u(x,y,t)=\exp\left(\frac{(x+y)e^{3t}+(x-y)e^{t}}{2} \right)$$