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

2. Originally Posted by jam333
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:

$y = e^{-x} \arctan (2 e^x) + Ce^{-x}$

we are told that:

$y(- \ln 2) = \frac {\pi}{2}$

this means:

$y(- \ln 2) = e^{ \ln 2} \arctan \left( \frac {2}{e^{\ln 2}} \right) + Ce^{\ln 2} = \frac {\pi}{2}$

$\Rightarrow 2 \arctan(1) + 2C = \frac {\pi}{2}$

$\Rightarrow 2 \cdot \frac {\pi}{4} + 2C = \frac {\pi}{2}$

$\Rightarrow C = 0$

$\Rightarrow y = e^{-x} \arctan (2 e^x)$

3. thanks helped me outalot