Intial Value problem, Diff EQ, my steps look right but answer = wrong wee

[mr_coffee]

Hello everyone I'm stuck on this problem:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/18/0c06a4e72ea888ea65443a46afd42f1.png
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/90/dff750d2ab4fefb41eccca8ce709a01.png
Looks simple enough but the answer they said is wrong. Here is my work:
(ln(y))^4/y dy = x^4 dx;
after integrating both sides:
1/5 * ln(y)^5 = (x^5/5) + C;

I raised both sides to e^
y^5 = [e^(x^5)*e^(5C)]
y = [e^(x^5)*e^(5C)]^(1/5);

now i subed in the I.C.
e^2 = (e^(5C))^(1/5);
e^2 = e^C;
ln(e^2) = C;
2 = C;

y = (e^(x^5)*e^10)^(1/5) which was wrong!~ wee!

Any help would be great! Linux isn't liking my scanner at the moment so I'm going to have to type out my problems for now :(
I think i screwd pu here but not sure:
ln(e^2) = C;
is that
2 = C or
ln(2*e) = C

 

[qtp]

i get that
$$( \ln y )^5 = x^5 + C$$
and
$$y = e^{\frac{x}{5}} e^C$$
then when $$y(1) = e^2$$
we have
$$e^2 = e^\frac{1}{5} e^C$$
solving for C
$$C = \frac{9}{5}$$
then
$$y = e^{\frac{x}{5} + \frac{9}{5}}$$

 

[mr_coffee]

Thanks for the responce but
i submitted it and is also wrong!
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/15/e1bf20f7f42406e76279e9bc8fe45a1.png

 

[TD]

So we have after integrating, then solving for y:

$$\frac{{\left( {\ln y} \right)^5 }}{5} = \frac{{x^5 }}{5} + c \Leftrightarrow y = e^{\left( {x^5 + 5c} \right)^{1/5} }$$

Filling in the initial condition:

$$e^2 = e^{\left( {1 + 5c} \right)^{1/5} } \Leftrightarrow 2 = \left( {1 + 5c} \right)^{1/5} \Leftrightarrow 32 = 1 + 5c \Leftrightarrow c = \frac{{31}}{5}$$

So we conclude:

$$y = e^{\left( {x^5 + 31} \right)^{1/5} }$$

 

[mr_coffee]

TD you are the man, I'm e-mailing u some money hah. Thanks !

 

[qtp]

yes i redid the problem and got the same answer... i made a mistake with the exponent :( sorry about that