1. Integration of differential equations

I am trying to integrate a differential equation

$\frac{-0.1}{-0.3-0.1u}du = -0.1dx$

I would like to show the integral in the form of u(x) = ........

Any ideas of where i can start and what methods i can use to solve this differential equation?

2. On the left let z = -.3 -.1 u

Then dz = -.1 du

you have int (dz/z) = ln(z) = ln ( -.3 -.1u) = -.1 x + c

exponentiate -.3 -.1u = e^(-.1x+c) = e^(-.1x)e^(c) =C e^(-.1x)

since e^c is just a constant

-.3 - C e^(-.1x) = .1u

-3 - C e^(-.1x) = u(x)

you may have wondered why I dropped the abs value from the ln integral.

based on the initial condition C will take care of any sign.