# Math Help - Differential constants

1. ## Differential constants

Could somebody please check I have got the value of the constant correct for this D.E

$\frac{dy}{dt}= - \frac{8t}{1+t^2}$

I get

$y = \frac{1}{(1+t^2)^4}$

where C = c2 - c1 = 0

2. Originally Posted by davefulton
Could somebody please check I have got the value of the constant correct for this D.E

$\frac{dy}{dt}= - \frac{8t}{1+t^2}$

I get

$y = \frac{1}{(1+t^2)^4}$

where C = c2 - c1 = 0

(You will need to show all your work if you want your error(s) corrected. And you will need to include whatever boundary condition came with the question).

3. Thank you, sorry I missed off the y term

So if

$\frac{dy}{dt}=-\frac{8ty}{(1+t^2}$

then

$\frac{1}{y}dy=-\frac{8t}{(1+t^2}dt$

integrating

$\ln(y)+c1=-4\ln(1+t^2) + c2$

therefore

for y(0)=1

$y=\frac{1}{(1+t^2)^4}$