# Homework Help: Why do we do this in Differential Equations?

1. Sep 21, 2010

I have always gone along with this, but now I am not so sure:

Take the DE:

$$y' +4y = 0$$

If you carry out the integration, we get

$$y = e^{-4x} +C\qquad(1)$$

but we always pull some stuff like "well since we could write C as ec1, since it's arbitrary", we can write the solution like

$$y = c_1e^{-4x}\qquad(2)$$.

I am not buying that they are the same. Yes, I know... the solution in (2) satisfies the original DE. Here is my problem, (1) gives a family of curves that differ by a vertical shift whereas (2) gives a family of curves that differ by a "stretch" or compression about the y-axis. How can these be the same thing?

So why do I bring this up anyway? I was doing the following initial value problem

$$y' = y^2 + y -6 \qquad(3)$$

I got a solution of

$$\ln\frac{y-2}{y+3} = 5x + e^{C}\qquad(4)$$

you can see that I used "the trick" in (4), letting co =eC so that when I exponentiate, I get a nice

$$y(x) =\frac{3Ce^{5x}+2}{1-Ce^{5x}}\qquad(5)$$

now comparing this to the book solution, this time they did not use the trick and got

$$y(x) =\frac{3e^{5x + c}+2}{1-e^{5x+c}}\qquad(6)$$

Naturally, I figured that (5) and (6) are the same. But when I go to use the initial value of y(5) = 10, if I use my version (5), I get that C is virtually zero since it is given from

$$10 - 10Ce^{25} = 3Ce^{25} +2\qquad(7)$$

However, the book plugs it into (4) of all god damn places and gets that C = -25.

What am screwing up conceptually here?

2. Sep 21, 2010

### ╔(σ_σ)╝

C =0 in the first equation.

3. Sep 21, 2010

### Dick

Start at the beginning. y=exp(-4x)+C only works as a solution in y'+4y=0 if C=0. C*exp(-4x) always works no matter what C is. Just substitute. They aren't the same. Nobody is 'pulling some stuff'.

4. Sep 21, 2010

Hi sigma face I really have no idea what you mean or how it helps. What do you mean C = 0 in the first equation? That is the general solution! Why would it equal zero?

5. Sep 21, 2010

### Staff: Mentor

No, you don't. If you substitute this into the DE you get -4e^(-4x) + 4e^(-4x) + C != 0 unless C happens to be 0.

If you separate the original DE, you get
$$\frac{dy}{y} = -4dx$$

Integrating both sides gives you
$$ln|y| = -4x + C$$

Exponentiating both sides gives you
$$e^{ln|y|} = e^{-4x + C}$$

The right side is the same as $$e^C\cdot e^{-4x}$$

$e^C$ is just a constant, so it can be renamed c1.

6. Sep 21, 2010

### ╔(σ_σ)╝

Hey :-),
If you sub in your answer into the differential equation you would arrive at the same conclusion.

Edit
Mark44 already took care of it.:-)

7. Sep 21, 2010

Ok. So it has to equal zero, but people are definitely pulling stuff. What about the rest of what I wrote? What about the initial value problem? Why would it depend on which form I use? And how would I know what form to use?

So is my (5) NOT the same as their (6) ?

Last edited: Sep 21, 2010
8. Sep 21, 2010

### Dick

If the books C is C', then your C is e^(C'). If the book got -25 for C'. Then your C is e^(-25). Sure, it's almost zero. Might be better to work with the books definition of 'C' if you want to get easy to work with numbers. But there is nothing wrong with your solution either.

9. Sep 21, 2010

### Staff: Mentor

You described your second problem as being an initial value problem. You gave the DE, but not the initial condition. You need that to be able to get the constant.

10. Sep 21, 2010

Hi Mark the IV problem is in the OP along with the IV

EDIT:

I apologize, they plug it into their version of (4) which is

$$\ln\frac{y-2}{y+3} = 5x + c_1$$

and they solve for c1 = -25

My main problem is that my book is the one who said that I should write it as a constant multiplier! Not as an added constant ... so i don't know why they would taylor an IV problem solution like this.

Last edited: Sep 21, 2010
11. Sep 21, 2010

### Staff: Mentor

Ah, I see it now. It was buried down later in what you wrote.