Find the General solution $u(t,x)$ to the partial differential equation $u_x+tu=0$.

Question

Find the General solution $u(t,x)$ to the partial differential equation $u_x+tu=0$.
Here is what I've tried.  I feel like this should be easy, but I'm not catching on to something here.
$$u_x+tu=0implies -u_x=tu$$
Ansatz...$u=frac{1}{t}e^{-xt}$ and $u_x=-e^{-xt}$
Thus: $int_{0}^x -e^{-st}ds=int_{0}^xfrac{1}{t}e^{-st}ds$
$implies te^{-xt}-(-1)=frac{1}{t^2}e^{-xt}-frac{1}{t^2}$
Solving for $frac{1}{t}e^{-xt}$
$implies u(t,x)=frac{1}{t}e^{-xt}=t^2e^{-xt}+t+frac{1}{t}$
As I type all of this it occurs to me that my ansatz worked without all the other work.  (that appears to not work.)
Is it ok to just stick with the ansatz and say $u(t,x)=frac{1}{t}e^{-xt}=f(t)e^{-xt}$?
Please help me understand where my thinking goes astray.
Any Guidance would be greatly apppreciated.

JJacquelin · Answer

$$frac{du}{dx}+tu=0$$
$$frac{du}{u}=-t:dx$$
Since $t$ is supposed to be not function of $x$ one can integrate wrt $x$.
$$ln|u|=-t:x+f(t)quad text{because}quad frac{df(t)}{dx}=0$$
$f$ is any function.
$$u=e^{-t:x+f(t)}$$
Let $quad e^{f(t)}=F(t)quad ;quad F$ is any function.
$$u(x,t)=F(t)e^{-t:x}$$

C-Web · Answer

Here is my attempt at a solution.  I think I need clearer explanation for the part in bold.
$u_x+tu=0implies u_x=tu$
Since the derivative contains the function assume exponential.
$u=e^{-xt}$; $frac{partial u}{partial x}=-te^{-xt}$
Thus $u(t,x)=e^{-xt}$
since $t$ is treated as a constant with regard to the differentiation, then any function of $t$ essentially acts as a constant thus...
$u(t,x)=f(t)e^{-xt}$

Robert Israel · Answer

This "partial differential equation" is really a parametric family of ordinary differential equations, since there's no derivative with respect to $t$.  You know how to solve that
ordinary differential equation...

Find the General solution $u(t,x)$ to the partial differential equation $u_x+tu=0$.

3 Answers

Add your own answers!

Ask a Question