Question

du/dt = (u^2+4)/(t^2+4)
I got to the point where I have:
(tan(u/2))^-1=(tan(t/2))^-1 +c
but I don't know where to go from there. Please help!

Answer 1

Take the tangent of both sides:
\[\frac{u}{2}=\tan\left(\tan^{-1}\frac{t}{2}+C\right)\]
Use the angle sum identity for tangent:
\[\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}\]
This gives
\[\frac{u}{2}=\frac{\tan\left(\tan^{-1}\dfrac{t}{2}\right)+\tan C}{1-\tan\left(\tan^{-1}\dfrac{t}{2}\right)\tan C}\]
The tangent of a constant is another constant, so you reduce to
\[\frac{u}{2}=\frac{\tan\left(\tan^{-1}\dfrac{t}{2}\right)+C}{1-C\tan\left(\tan^{-1}\dfrac{t}{2}\right)}\]
Since \(\tan x\) and \(\tan^{-1}x\) are inverse functions, you have \(\tan\left(\tan^{-1}x\right)=x\), which gives you
\[\frac{u}{2}=\frac{\dfrac{t}{2}+C}{1-C\dfrac{t}{2}}\]
Some simplifying gives
\[\frac{u}{2}=\frac{t+2C}{2-Ct}\]
Twice a constant is another constant:
\[\frac{u}{2}=\frac{t+C}{2-Ct}\]
And isolate \(u\):
\[u=\frac{2t+C}{2-Ct}\]

Answer 2

The constant rule will let you simplify further, actually, to
\[u=\frac{t+C}{1-Ct}\]