Explicit solution for this ODE I've recently come across a very simple ODE in my work:
$$x'(t) = 1 + \frac{x}{t}$$
Obviously, if the constant were not there then the solution would be easy to obtain by the usual ``separate and integrate'' trick. I was thinking that there must be a simple closed form for the solution, but I don't see what it would be.
Motivation: there will surely be others, but this parametrizes the curve of discontinuity that naturally arises from certain initial conditions for a Riemann problem for the Burgers equation. 
Is there a trick to solve something like this?
 A: You can use an integrating factor to write down a closed form expression for the solution. This allows you to solve all ODE of the form
$$ x'(t) + p(t) x(t) = q(t), $$
though in practice it may not be possible to simplify it in the way you like. Try following the steps detailed there. You will obtain the general solution
$$ x(t) = t \log{t} + Ct $$
A: Here's a quick way to do the same thing without an integrating factor.  Use an invariant Lie group, a simple one:  $t'=\lambda t$, $x'=\lambda^\beta x$. 
$$
\frac{dx'}{dt'}=1+\frac{x}{t}
$$
$$
\frac{\lambda^\beta dx}{\lambda dt}=\lambda^0 1+\frac{\lambda^\beta x}{\lambda t}
$$
For invariance, $\beta =1$.  Stabilizers for this group are 
$$\mu=\frac{x}{t^\beta}=\frac{x}{t}$$ and $$\nu=\frac{\dot{x}}{t^{\beta -1}}=\dot{x}$$ where $\dot{x}=\frac{dx}{dt}$.  Thus the equation becomes $$\nu=1+\mu$$
With a little effort you can derive 
$$
t\frac{d\mu}{dt}=\nu-\beta \mu=1+\mu-\mu=1
$$ so
$$
d\mu=\frac{dt}{t}
$$ and
$$
\mu=\frac{x}{t}=lnt+C
$$which gives
$$
x=tlnt+Ct
$$Okay, so maybe that wasn't so quick, but it is a good example of how a Lie group may be used to solve a differential equation.
A: Let $y = 1 + \frac{x}{t}$. Then
$$\frac{dy}{dt} = \frac{\frac{dx}{dt}t - x}{t^2}$$
Substituting $\frac{dx}{dt} = 1 + \frac{x}{t}$ into the expression above, you get 
$$\frac{dy}{dt} = \frac{1}{t}$$.
The rest should be clear.
