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

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- Nov 15th 2009, 02:30 AMdavefultonDifferential constants
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 - Nov 15th 2009, 03:12 AMmr fantastic
- Nov 15th 2009, 03:34 AMdavefulton
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