the derivate of y + y = 2 / (1+4e^2x)
y(-ln2)=pi/2
y=e^-x*tan^-1(2e^x)
i need to find out the intial value plz someone help me work it out
greatly appreciated
so nice of you to find the answer for us.
anyway, here is how to continue (you forgot the arbitrary constant by the way)
we have:
$\displaystyle y = e^{-x} \arctan (2 e^x) + Ce^{-x}$
we are told that:
$\displaystyle y(- \ln 2) = \frac {\pi}{2}$
this means:
$\displaystyle y(- \ln 2) = e^{ \ln 2} \arctan \left( \frac {2}{e^{\ln 2}} \right) + Ce^{\ln 2} = \frac {\pi}{2}$
$\displaystyle \Rightarrow 2 \arctan(1) + 2C = \frac {\pi}{2}$
$\displaystyle \Rightarrow 2 \cdot \frac {\pi}{4} + 2C = \frac {\pi}{2}$
$\displaystyle \Rightarrow C = 0$
$\displaystyle \Rightarrow y = e^{-x} \arctan (2 e^x)$