# Urnt? easy integration factor D.E. but its != correct, ahh!

1. Jan 22, 2006

### mr_coffee

ello ello!
Here is the problem:
Let g(t) be the solution of the initial value problem:

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! :surprised

Anyone know what I did?

2. Jan 22, 2006

### d_leet

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

3. Jan 22, 2006

### HallsofIvy

Staff Emeritus
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]

I think you will find that impossible to integrate so do this as a separable equation.