Could somebody please check I have got the value of the constant correct for this D.E
$\displaystyle \frac{dy}{dt}= - \frac{8t}{1+t^2}$
I get
$\displaystyle y = \frac{1}{(1+t^2)^4}$
where C = c2 - c1 = 0
Thank you, sorry I missed off the y term
So if
$\displaystyle \frac{dy}{dt}=-\frac{8ty}{(1+t^2}$
then
$\displaystyle \frac{1}{y}dy=-\frac{8t}{(1+t^2}dt$
integrating
$\displaystyle \ln(y)+c1=-4\ln(1+t^2) + c2$
therefore
for y(0)=1
$\displaystyle y=\frac{1}{(1+t^2)^4}$
sorry about the sloppiness