I have ADE $$\frac{\partial c}{\partial t}+u_o\frac{\partial c}{\partial x}=D_o \frac{\partial^2 c}{\partial x^2}$$ where $u_o$ and $D_o$ are constants and initial and B.C. are $c(x,0)=0$ and $c(x_o,t)=c_o, \ \ c(\infty, t) =0$ where $x_o, c_o$ are also constants.
I tried to convert ADE into ODE using transformation $\eta = x - ct \implies C(\eta) = c(x,t)$ where $c$ is transformation parameter, then ODE is $$(-c+u_o) \frac{\partial C}{\partial \eta}=D_o \frac{\partial^2 C}{\partial \eta^2} \ \ \ \ \ \to \ \ \ (1)$$
But one same transformation I found on another thread is pde - Advection diffusion equation. The same transformation applied but converted to Diffusion equation not in ODE. Difference is that their chain rule is used but I proceed as follow $$\frac{\partial c(x,t)}{\partial t}=\frac{\partial C}{\partial \eta} \frac{\partial \eta}{\partial t}=-c \frac{\partial C}{\partial \eta}$$
Now question is that, either my result into ODE eq $(1)$ is correct or I should use chain rule and it should be converted into heat equation as in other thread?
Added: Also, please help me to apply these transformation on initial and B.C. Transformation on condition $c(\infty, t)=0$ is not problem but how I apply these conditions on $c(x_o,t)=c_o$ as for $x=x_o \implies \eta=x_o-ct$ so condition becomes $c(x_o,t)=C(x_o-ct)=c_o$