Confusion With A Separable Differential Equation

**Spudwad** · March 31st 2010, 11:06 AM

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

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

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

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

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

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

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

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

So for the initial starting conditions:

$70 = 8 - Ce^5$

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

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

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

**Lafexlos** · March 31st 2010, 12:22 PM

Hi Spudwad,

There should be minus sign here because of $\frac{d\ln(8-B)}{dB}=-\frac{1}{8-B}$ ,
So yu will get,

$(-1) \ln{(8-B)} = 5t + C$

$\ln{(8-B)^{-1}} = 5t + C$

Try with this. If you still get problems, write again

Thread: Confusion With A Separable Differential Equation

LinkBack

Thread Tools

Display

Confusion With A Separable Differential Equation

Similar Math Help Forum Discussions

Separable Differential Equation

Separable Differential Equation

separable differential equation dy/dx = y/x

Another separable differential equation

separable differential equation

Search Tags