# 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?

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

$$x(t) = t \log{t} + Ct$$

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.

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.