= Solution
Set
$$
y=\frac{4\pi}{\alpha_\lambda}=\frac{(4\pi)^2}{\lambda^2},\qquad t=\log\mu^2.
$$
The chain rule gives
$$
\frac{dy}{dt}=-(4\pi)^2\lambda^{-3}\beta(\lambda)=\beta_0+\frac{\beta_1}{y}+O(y^{-2}).
$$
For the <ultraviolet> expansion take $\beta_0>0$ and $L=\log(\mu^2/\Lambda^2)\gg1$. The leading solution is $y\sim\beta_0L$. Substituting it in the next term gives $dy/dL=\beta_0+\beta_1/(\beta_0L)+O(\log L/L^2)$, whose integration yields
$$
\boxed{\frac{4\pi}{\alpha_\lambda}=\beta_0L+\frac{\beta_1}{\beta_0}\log L+O\!\left(\frac{\log L}{L}\right).}
$$
Choosing $\Lambda$ removes the additive constant. This proves the two displayed leading terms of the <two-loop asymptotic running coupling>.
\b[The printed remainder is too small in general.] This can be checked even for an exactly two-loop <renormalization-group beta function>, so it does not depend on unknown higher-loop effects. Put $c=\beta_1/\beta_0$. Separating variables and fixing the integration constant gives
$$
y-c\log\!\left(\frac{y+c}{\beta_0}\right)=\beta_0L.
$$
Expanding this identity one order further gives
$$
y=\beta_0L+c\log L+\frac{c^2}{\beta_0}\frac{\log L+1}{L}+O\!\left(\frac{(\log L)^2}{L^2}\right).
$$
The $\log L/L$ term cannot be absorbed into a fixed change of $\Lambda$. For example, with $\beta_0=\beta_1=1$, the exact implicit relation is $y-\log(y+1)=L$, and the next term is $(\log L+1)/L$, which is not $O(1/L)$. The unspecified $O(y^{-2})$ contribution to the <differential equation> changes the constant coefficient of $1/L$, while the coefficient $\beta_1^2/\beta_0^3$ of $\log L/L$ is already fixed. Only special cases such as $\beta_1=0$ avoid this particular obstruction.
The asymptotic formula also presupposes an asymptotically free <ultraviolet> regime. If $\beta_0=0$, division by it is invalid and the leading scaling changes. If $\beta_0<0$, the weak-coupling asymptotic direction is generally toward the <infrared> and the logarithmic variable must be chosen accordingly. These qualifications distinguish the valid perturbative solution from an unconditional formula.
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