Advection equation with decay

Question

I tried to solve the advection equation with decay

$$ u_t + 2u_x - u - t = 0 $$

Firstly, I solved

$$ \frac{dt}{1} = \frac{dx}{2} $$

obtaining $ u_0 (x,t)= \Phi(x-2t) $

If I treat the equation like solving

$$ u_t + 2u_x = u + t $$

Then I must solve

$$ \frac{dt}{1} = \frac{dx}{2} = \frac{du}{u + t }$$

But yields $u(x,t)= \Phi(x-2t) e^t - t $, that is wrong.

I solved treating the equation like

$$ u_t + 2u_x - u = t $$

Firstly, I obtained the homogeneous-like part

$$ u_t + 2u_x - u = 0$$

finding the solution

$$ u(x,t)= \Phi(x-2t) e^t$$

Then I estimated the "nonhomogeneous-like" part, guessing a polynomial in $t$ such as

$$ u^\star (x,t) = At+ B$$, replacing in the equation and obtaining

$$u^\star (x,t) = - t+ 1$$

$$ u(x,t)= \Phi(x-2t) e^t - t + 1$$, that is wrong again.

Can Anyone tell me how to solve it and obtaining the

$$ u(x,t)= \Phi(x-2t) e^t - t - 1$$

correct solution without using variable changes (I have already solved and checked with variable changes, but I want to solve the equation pretending this last method does not exist)? Thanks in advance

Answer 1

If the particular solution is $$ u_p(t) = At + B $$

Then we have

$$ A - (At+B) = t $$

This gives $$ - A = 1 $$ $$ A-B = 0 $$

Thus $ A = B = -1$

So this method will lead you to the correct solution $$ u(t) = \Phi(x-2t)e^t -t - 1 $$