Separable Differential Equation (what's wrong?)

[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 :\

 

[slider142]

That's the correct approach.

 

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

 

[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.

 

[vwishndaetr]

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.

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

 

[HallsofIvy]

I have editted my post to make that clearer.