# Homework Help: Variable Separation and Partial Fractions

1. Sep 7, 2010

### erok81

1. The problem statement, all variables and given/known data

dx/dt=3x(x-5); x(0)=2

2. Relevant equations

dx/dt=F(x)G(t)

dx/dt=x'

3. The attempt at a solution

First I separated the variables using F=3x(x-5) and G=3

x'/(3x(x-5))=3

$$\int\frac{x'}{(3x(x-5)}dt=\int3 dt$$

On the right hand side I end up with 3t. For the left hand side I get an integral I need apply partial fractions.

After solving the partial fraction I have:

$$\int\frac{1}{5x-25}-\frac{1}{5x}$$

Integrating that leaves me with:

1/5ln|5x-25|-1/5ln|5x|=3t+c

Multiplying both sides by 5:

ln|5x-25|-ln|5x|=15t+c

Applying log property where ln|a|-ln|b|=ln|a/b|

$$ln\frac{5x-25}{5x}$$

Reducing that by 5:

$$ln\frac{x-5}{x}$$

So finally I have...after taking e of both sides:

$$\frac{x-5}{x}=e^{15t}+c$$

I am kind of stuck here solving for c...assuming I've done the above correctly.

The initial condition shows x(0)=2.

I've tried sub'ing in x=2 and t=0 to no avail. I've either done something wrong at the top or messed up this last part.

Any help will be appreciated.

2. Sep 7, 2010

### Samuelb88

All you need to do is solve for x! Try multiplying both sides of the equation by x, then isolating all 'x-terms' on one side. From there, you it will become clear what other steps need to be taken to determine your solution.

3. Sep 7, 2010

### Dick

You don't need to solve for x. What makes you think you've done anything wrong so far? It looks ok to me. Just put x=2 and t=0 and solve for c. What answer are you expecting to get?

4. Sep 7, 2010

### erok81

The answer in the back of the book is:

$$x(t)=\frac{10}{2+3e^{15t}}$$

Besides the fact that I can't seem to get to that answer, I've tried about six of these using the same method and end up with the same result - a similar answer but not correct.

If I plug in the values x=2 and t=0 I get this...

$$x(t)=e^{15t}-\frac{5}{2}$$

5. Sep 7, 2010

### Dick

Ok, I see what you are doing wrong. When you exponentiate log((x-5)/x)=15*t+c you get (x-5)/x=e^(15t+c)=C*e^(15t). The c doesn't remain an additive constant. I missed that, sorry. Now find C and solve for x.

Last edited: Sep 7, 2010
6. Sep 8, 2010

### erok81

Ooooh, that is exactly what I was doing wrong in every problem I tried.

For some reason I'd do e^15t+c which is e^15t*e^c. e^c is still a constant but rather than keep it as a multiplier, I'd throw in as a plus c. Now I see what I was doing wrong.

Perfect. Thank you!

7. Sep 8, 2010

### erok81

Actually, I think I am still missing something.

Using (x-5)/x=ce^15t and x(0)=2, I come up with x(t)=-3/2e^15t

Which isn't even close.

8. Sep 8, 2010

### Dick

How did you manage that? I get (x(t)-5)/x(t)=(-3/2)*e^(15t) to start out with ok, now you still have to solve for x(t).

9. Sep 8, 2010

### erok81

Maybe that is my problem. I used x(0)=2 to get x(t)=-3/2e^15t

(x-5)/x=ce^15t -> (2-5)/2=ce^15(0) which got me x(t)=-3/2e^15t...if that makes sense. To me it seems that is the same thing as solving (x(t)-5)/x(t)=(-3/2)*e^(15t) except I wouldn't have had the -3/2 like you have..since I had to use x(0)=2 to get that.

10. Sep 8, 2010

### Dick

I don't know what's going through your head. Putting t=0 and x=2 as you said gives c=(-3/2). That's fine. That gives you (x(t)-5)/x(t)=(-3/2)*e^(15t). That's NOT the same as x(t)=(-3/2)*e^(15t). !!??? If the solution were x(t)=(-3/2)*e^(15t) then if I put t=0 into that I get x(0)=(-3/2).

11. Sep 8, 2010

### erok81

Trust me, I don't know what's going through my head either. My professor skipped these and the section of the book doesn't go over them either so I have zero idea what I am doing. Sorry for the thick-headedness.

Let me give it another try.

12. Sep 8, 2010

### erok81

Ok...I think this is where I am confused on this step.

I don't see the difference in this problem between (x(t)-5)/x(t) and (x-5)/x when I am solving for x. I feel like I've solved one and I'm just doing it again.

Wait...for the x(t) I am not solving for an actual value like I was before when I solved for c before. I am trying solve for x(t)=some equation. Right?

Being this stupid is painful. :rofl:

13. Sep 8, 2010

### Dick

There is no difference between x and x(t). You just don't always write the '(t)' part to save space. You have (x-5)/x=(-3/2)e^(15t). Now just use algebra to solve for x. Try it. That is x(t). Yes. You are to get a form where x(t) is expressed as a function of t.

14. Sep 8, 2010

### erok81

Ok, I definitely get it now and see where I was confused. I was doing some of it out of order compared to how I normally do them.

Thanks again for putting up with me. I really appreciate it. I wouldn't have understood these without the help.

15. Sep 8, 2010

### erok81

Just tried it again and success! I have the exact answer that appears in the back of the book.