# Separable Differential Equation (what's wrong?)

1. Oct 7, 2009

### scud0405

So, here's the problem:

$$\frac{dy}{dx} = \frac{8y}{5x}$$

To start off, I separate the integrals, which gives me:

$$\frac{dy}{8y} = \frac{dy}{5x}$$

After that, I integrate both sides, which gives me:

$$\frac{ln8y}{8} = \frac{ln5x}{5} + c$$

Now, the question says that it runs through (4, 1), so that is saying that y(4) = 1, correct?

To solve for c, I just plug the 4 in where the Xs are and the 1 is where the Ys are?

EDIT: Sorry, posted in the wrong forum! Please move this thread for me :\

2. Oct 7, 2009

### slider142

That's the correct approach.

3. Oct 8, 2009

### HallsofIvy

Notice by the way that
$$\frac{ln(8y)}{8}= \frac{ln(5x)}{5}+ c$$
is the same as
$$\frac{ln(y)+ ln(8)}{8}= \frac{ln(x)+ ln(5)}{5}+ c$$
$$\frac{ln(y)}{8}+ 1= \frac{ln(x)}{5}+ 1+ c$$
$$\frac{ln(y)}{8}= \frac{ln(x)}{5}+ c$$

And that is the same as
$$ln(y)= \frac{8}{5} ln(x)+ 8c= ln(x^{\frac{8}{5}})+ 8c[/itex] so that [tex]y= e^{8c} x^{\frac{8}{4}}$$
and that can be written simply
$$y= Cx^{\frac{8}{5}$$
where $C= e^{8c}$

Last edited by a moderator: Oct 9, 2009
4. Oct 8, 2009

### scud0405

Oh, okay! How did the 8c become only c in:

$$\frac{8}{5} ln(x)+ 8c= ln(x^{\frac{8}{5}})+ c$$

Thanks for the help.

5. Oct 8, 2009

### vwishndaetr

8*c = c because 8, a constant, times a constant c, is still a constant.

6. Oct 9, 2009

### HallsofIvy

I have editted my post to make that clearer.