I would like to find out what substitutions I should use to solve the following PDE:
$${u_t+(x+t)u_x+u=x}$$
My professor advised that I try the substitution ${v=\ln(u)}$. However, I think that the substitution is not suitable because we will lose the solutions where ${u<0}$.
In fact you no need to try the substitution $v=\ln u$ , since this PDE can already solvable by using method of characteristics.
$u_t+(x+t)u_x+u=x$
$u_t+(x+t)u_x=x-u$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dx}{ds}=x+t=x+s$ , letting $x(0)=x_0$ , we have $x=(x_0+1)e^s-s-1=(x_0+1)e^t-t-1$
$\dfrac{du}{ds}=x-u=(x_0+1)e^s-s-1-u$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)e^{-s}+\dfrac{(x_0+1)e^s}{2}-\dfrac{(x_0+1)e^{-s}}{2}-s=f((x+t+1)e^{-t}-1)e^{-t}+\dfrac{x+t+1}{2}-\dfrac{(x+t+1)e^{-2t}}{2}-t=f((x+t+1)e^{-t}-1)e^{-t}+\dfrac{x-t+1}{2}-\dfrac{(x+t+1)e^{-2t}}{2}$
