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.