Is the Integration Factor Correct in This Differential Equation Solution?

[mr_coffee]

ello ello!
Here is the problem:
Let g(t) be the solution of the initial value problem:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/da/9fc0e62bc2df5df4721f38c5634c1f1.png
with g(1) = 1 .
Find g(t).
g(t) = ?


Heres what i did:
2ty' + y = 0;
y' + y/(2t) = 0;

I(t) = e^(2t) dt
integrate:
I = t^2;

t^2*y = C;
y = C/t^2;

apply intial condition: g(1) = 1;
1 = C/1
C = 1;
Appply constant:
y = 1/t^2;
which si wrong!

Anyone know what I did?

 

[d_leet]

I can't figure out what you did, but the equation is seperable so you don't need an integrating factor.

 

[HallsofIvy]

what i did:
2ty' + y = 0;
y' + y/(2t) = 0;

I(t) = e^(2t) dt

NO, the coefficient of y is $\frac{1}{2t}$ not 2t:
$$I(t)= \int e^{\frac{1}{2t}}dt$$
not
[tex]\int e^{2t}dt[/itex]<br /> <br /> I think you will find that impossible to integrate so do this as a separable equation.[/tex]