First Order Linear Diff. Eq.

amy

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

Find the General Solution:
2. Relevant equations

3. The attempt at a solution

so I used the seperation of variable method to get
Then I took the integral of both side to get
then I got rid of the ln(y) and factored out the x on the other side to get

The back of the book tells me I should get
I think what I'm having trouble with is the algebra involved in simplifying the exponent amongst other things...  So how do I get what I have, to what I'm supposed to get?  

    Try integrating ln(x)/x again. It didn't go very well. Try differentiating what you got, you won't get ln(x)/x.  

    I thought that might be wrong.  Well since it can be seperated into (lnx)*(1/x) I integrated them seperatley and multiplied them.  So the integration table in my book has the integral of ln(x) to be (xlnx-x) then times (ln(x)) gives me (ln(x))(xlnx-x)...  I can't seem to find how to integrat ln in my old calc. text.   Is that where I'm screwing up?  

    Try a simple u-substitution. The right one makes for a very simple result.  

    Whoa! There is no such integral product rule. It's a simple u substitution. Set u=ln(x) and change the variable to u.  

    Dang!  (said in best Napolean Dynamite voice)...  I really gotta go back and refresh myself on the ol' integration methods.  Thanks for your help

I used u=lnx of cours to get du=1/x so it simplifies to udu etc...

Thanks for the help!!!  

    since my text is saying integral of  ln(abs(u)) = u ln(abs(u)) - u, is where I went wrong???  but I wouldn't think abs value would make a big diff.