# Differential constants

• Nov 15th 2009, 02:30 AM
davefulton
Differential 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 AM
mr fantastic
Quote:

Originally Posted by davefulton
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

(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).
• Nov 15th 2009, 03:34 AM
davefulton
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

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