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
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}

[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.