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I just read a paper in which an ODE was solved using a step I don't understand.

$${dI \over \kappa (x) dx} = I$$ Let $$\tau = \int_0^x \kappa(x)dx$$
Rewrite equation as $${dI \over d\tau} = I$$ and solve.

How does the final expression follow from the definition of $\tau$?

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2 Answers 2

up vote 4 down vote accepted

If $\tau(x) = \displaystyle \int_{0}^x \kappa(y) dy$, then $d \tau = \kappa(x) dx$

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Chain rule:

$$\frac{dI}{dx}=\frac{dI}{d\tau}\frac{d\tau}{dx}=\frac{dI}{d\tau}\kappa(x).$$

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