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Say we have a differential equation:

$$df(\mu) = g(f(\mu))dv(\mu)$$

I was wondering under what conditions we get something like this (integrating from $\mu_1$ to $\mu_2$):

$$\int df(\mu) \approx g(f(\mu_1))\int dv(\mu)$$

I will give you an example I encountered, (lets call this main equation):

$$\frac{d\alpha(\mu)}{d\mu}\mu = -\alpha^2(\mu) $$

It has exact solution, integrating from $\mu_1$ to $\mu_2$ and solving for $\alpha(\mu_2)=\alpha_2$:

$$\alpha_2 = \alpha_1(1 + \alpha_1ln\frac{\mu_2}{\mu_1})^{-1}$$

Now is the interesting part: assuming that $\alpha_1ln\frac{\mu_2}{\mu_1}$ is small and taking Taylor expansion we get:

$$\alpha_2 \approx \alpha_1 -\alpha_1^2ln\frac{\mu_2}{\mu_1}$$

However this answer could have been obtained from the main equation treating $\alpha^2 = \alpha_1^2$ as constant. From main equation:

$$d\alpha(\mu) = -\alpha_1^2(\mu)\frac{d\mu}{\mu} $$

$$\alpha_2 = \alpha_1 -\alpha_1^2ln\frac{\mu_2}{\mu_1}$$

So two very different approximations were made, but by some magic results were the same. Could somebody give me insight on the second equation I wrote?

Thank you!

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