Differentials and Integrating

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October 20th 2012, 09:41 PM
astuart

Differentials and Integrating

*EDIT: Whoops, wrong forum. This can be moved if needed. Sorry.

Hello,

I'm really having a bit of trouble here. I'm doing an online course with limited recorded lectures and a textbook which is helpful only some of the time, so I'm hoping for a little bit of guidance here. (Headbang)

I have problem in which I've managed to write a differential equation to represent the problem. In the problem, C is defined as calories per day, so it's not an 'arbitrary constant'. I'm supposed to solve the equation, but I don't really know how to go about it.

$\frac {dw}{dt} = \frac {C-17.5w}{3500}$
$= \frac {1}{C-17.5w} \, \frac {dw}{dt} = \frac {1}{3500}$
$= \int \frac {1}{C-17.5w}\,dw = \int 3500^{-1} dt$

I fall down a little bit here, as I don't know exactly how to deal with C (the constant), whether the dt on the right, is just t??

$= \frac{1}{C} \, \int \frac {1}{-17.5w} dw = \frac {1}{3500t} + M$
$= \frac {1}{C} \, -17.5\,ln|w| = \frac {1}{3500t} + M$

From here it just kind of falls to pieces, and I don't know whether I am supposed to get the 1/C to the other side, then divide by 17.5, then get rid of the natural log, or whether I messed up an earlier step.

I really wish the lectures explained this sort of question a little better...
October 20th 2012, 10:18 PM
Prove It

Re: Differentials and Integrating

Quote:

Originally Posted by astuart

*EDIT: Whoops, wrong forum. This can be moved if needed. Sorry.

Hello,

I'm really having a bit of trouble here. I'm doing an online course with limited recorded lectures and a textbook which is helpful only some of the time, so I'm hoping for a little bit of guidance here. (Headbang)

I have problem in which I've managed to write a differential equation to represent the problem. In the problem, C is defined as calories per day, so it's not an 'arbitrary constant'. I'm supposed to solve the equation, but I don't really know how to go about it.

$\frac {dw}{dt} = \frac {C-17.5w}{3500}$
$= \frac {1}{C-17.5w} \, \frac {dw}{dt} = \frac {1}{3500}$
$= \int \frac {1}{C-17.5w}\,dw = \int 3500^{-1} dt$

I fall down a little bit here, as I don't know exactly how to deal with C (the constant), whether the dt on the right, is just t??

$= \frac{1}{C} \, \int \frac {1}{-17.5w} dw = \frac {1}{3500t} + M$
$= \frac {1}{C} \, -17.5\,ln|w| = \frac {1}{3500t} + M$

From here it just kind of falls to pieces, and I don't know whether I am supposed to get the 1/C to the other side, then divide by 17.5, then get rid of the natural log, or whether I messed up an earlier step.

I really wish the lectures explained this sort of question a little better...

First of all $\displaystyle \begin{align*} \frac{1}{3500} \end{align*}$ is a constant, so its integral will be $\displaystyle \begin{align*} \frac{1}{3500}t \end{align*}$ . As for the LHS, rewrite it as $\displaystyle \begin{align*} -\frac{1}{17.5}\int{\frac{-17.5}{C - 17.5w}\,dw} \end{align*}$ so that you can use the substitution $\displaystyle \begin{align*} u = C - 17.5w \implies du = -17.5\,dw \end{align*}$ and the integral becomes $\displaystyle \begin{align*} -\frac{1}{17.5}\int{\frac{1}{u}\,du} \end{align*}$ .
October 20th 2012, 10:18 PM
MarkFL

Re: Differentials and Integrating

Your separation of the variables is good up to here:

$\int\frac{1}{C-17.5w}\,dw=\int 3500^{-1}\,dt$

although I would write:

$-\int\frac{1}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt$

Now, in your next step, you pulled a $\frac{1}{C}$ out in front of the integral as if it were a factor, which is is not. You did not have:

$\int\frac{1}{-17.5Cw}\,dw=\int 3500^{-1}\,dt$

If you did, then what you did would have been fine. So, let's go back to:

$-\int\frac{1}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt$

Let's multiply the integral on the left by $1=\frac{17.5}{17.5}$

$-\frac{17.5}{17.5}\int\frac{1}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt$

$-\frac{1}{17.5}\int\frac{17.5}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt$

$-\frac{2}{35}\int\frac{17.5\,dw}{17.5w-C}=\frac{1}{3500}\int\,dt$

Now, observe that the integral on the left is of the form $\int\frac{du}{u}=\ln|u|+c_1$ ...can you finish?
October 21st 2012, 01:39 AM
astuart

Re: Differentials and Integrating

Quote:

Originally Posted by MarkFL2

Your separation of the variables is good up to here:

$\int\frac{1}{C-17.5w}\,dw=\int 3500^{-1}\,dt$

although I would write:

$-\int\frac{1}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt$

Now, in your next step, you pulled a $\frac{1}{C}$ out in front of the integral as if it were a factor, which is is not. You did not have:

$\int\frac{1}{-17.5Cw}\,dw=\int 3500^{-1}\,dt$

If you did, then what you did would have been fine. So, let's go back to:

$-\int\frac{1}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt$

Let's multiply the integral on the left by $1=\frac{17.5}{17.5}$

$-\frac{17.5}{17.5}\int\frac{1}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt$

$-\frac{1}{17.5}\int\frac{17.5}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt$

$-\frac{2}{35}\int\frac{17.5\,dw}{17.5w-C}=\frac{1}{3500}\int\,dt$

Now, observe that the integral on the left is of the form $\int\frac{du}{u}=\ln|u|+c_1$ ...can you finish?

I think so...However, I don't understand why $\frac {1}{17.5}$ became $\frac{2}{35}$ . I realize they're equivalent, but is there a reason for doing it?

$-\frac{1}{-17.5}\int\frac{17.5}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt$

$= ln |17.5w - C| = \frac {\frac{t}{3500}}{\frac{1}{-17.5}}$

$= ln |17.5w - C| = -0.005t$

$= 17.5w - C = e^{-0.005t}$

$= w = \frac {e^{-0.005t} + c} {17.5}$

Is this right? When I ran it through Wolfram, it gave me a different answer, which I just can't see how it got to that point..
October 21st 2012, 07:31 AM
MarkFL

Re: Differentials and Integrating

You forgot to include a parameter representing the constant of integration.