# Problem manipulating solution of a differential equation!

1. May 1, 2012

### the0

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

$Y'(u) = A(u)Y(u)$

$V(u)$ is the general solution

The question asks to show that if $A(u)$ is antisymetric for all $u$
i.e. $^{t}A(u) = -A(u)$ for all $u$
Then $^{t}V(u).V(u) = I$

2. Relevant equations

A hint says to use the fact that $V(0) = I$

3. The attempt at a solution

Using the fact given in the hint I have a solution

(differentiate $^{t}V(u).V(u)$ and show that this is zero, therefore $^{t}V(u).V(u)$ is constant, and since the hint implies $^{t}V(0).V(0) = I$ the problem is solved)

HOWEVER, I do not understand why $V(0) = I$!
Maybe it's something obvious which I just cannot spot!
Please could somebody explain this to me

2. May 2, 2012

### dikmikkel

V(0)=I because $V(x) = K_1e^{\int A(x)\text{d}x}$
i.e. every exp function with argument 0 is 1.
So you have to prove that it is the solution V(x) maybe?