Hello kevinlightman Originally Posted by

**kevinlightman** I'm trying to solve the system of equations:

$\displaystyle (tan 2x) g'(x) + 2g(x) = 0, f'(x) - f(x) g(x)=0

$

I rearrange for g(x) in the second equation and substitute that into the first which after a little rearrangement gives:

tan(2x)((-(f'(x))^2)/(f(x)) + f''(x)) + 2f'(x) = 0

I tried to use the integrating factor method but I can't seem to get anything from it, any help please?

If you use the integrating factor method on the first equation, the solution comes out as:$\displaystyle g(x) = A\csc(2x)$

The second equation then becomes:$\displaystyle f'(x)-A\csc(2x)f(x) = 0$

which again can be solved using the integrating factor, which turns out to be $\displaystyle (\tan x ) ^{-A/2}$. This gives:$\displaystyle f(x) = B (\tan x) ^{A/2}$

Grandad