Solving an ODE without Lambert W function

Question

I have a question regarding the possibility of solving the following ODE:

$$\left[2x(t)+t\right]x^{\prime}(t)=1$$

such that $x(0)=-1$.

If we make the substitution $w(t)=2x(t)+t$, we obtain the following equation:

$$-2\ln[w(t)+2]+w(t)=t+C$$

which can be solved for $w$ (and hence $x$), although both $w$ and $x$ will be expressed in terms of the Lambert W function. I won't be posting the full solution here - WolframAlpha shows it step-by-step. Nevertheless, after applying the initial condition that $x(0)=-1$, $x$ is simplified to a much more digestible form: $x(t)=-\frac{t}{2}-1$.

Therefore, I was wondering - is it possible to arrive at this solution without the use of Lambert W function? More specifically, is it possible to somehow apply the initial condition earlier and hence to avoid having to find the general solution for $x(t)$ (and thereby avoiding stumbling on the Lambert W function)?

I would be grateful for some advice on this.

Answer 1

Instead of looking for $x(t)$, first look for $t(x) $:

$$x' = \dfrac{\mathrm dx}{\mathrm dt} = \dfrac{1}{(\mathrm dt/\mathrm dx)} = \dfrac{1}{t'}\\\,\\(2x+t)=\dfrac{1}{x'}=t'\\\,\\ t'-t=2x$$

is a linear ODE with condition $t(1)=0$.

Answer 2

Therefore, I was wondering - is it possible to arrive at this solution without the use of Lambert W function? More specifically, is it possible to somehow apply the initial condition earlier and hence to avoid having to find the general solution for $x(t)$ (and thereby avoiding stumbling on the Lambert W function)?

Yes this is possible. Using your substitution $w(t)=2x(t)+t$, one sees that $(w(t)+2)^2\mathrm e^{-w(t)}=C\mathrm e^{-t}$, for some constant $C$.

The initial condition $x(0)=-1$ reads $w(0)=-2$ hence $C=0$. Since $\mathrm e^{-w(t)}\ne0$, one gets $w(t)+2=0$ for every $t$, that is, $x(t)=-\frac12(t+2)$.

As you noted, this is specific to the initial condition $w(0)=-2$.