Several times now, I have ended up with the incorrect answer for this differential equation:

$\displaystyle \frac {dB} {dt} + 5B = 40$

With starting conditions: B(1) = 70. Here is how I've approached it:

$\displaystyle \frac {dB} {dt} = 40 - 5B = 5(8-B)$

$\displaystyle \frac {dB} {8-B} = 5 dt$

$\displaystyle ln(8-B) = 5t + C$

$\displaystyle 8-B = Ce^{5t}$

$\displaystyle B = 8 - Ce^{5t}$

So for the initial starting conditions:

$\displaystyle 70 = 8 - Ce^5$

$\displaystyle \frac {62} {e^5} = -C$

And so $\displaystyle B = 8 + \frac {62} {e^5}e^{5t}$

I appreciate if anyone can guide me as to where I went wrong. Thanks for the help.