Differential constants

**davefulton** · November 15th 2009, 02:30 AM

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

**mr fantastic** · November 15th 2009, 03:12 AM

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

Does your answer work when you substitute it into the DE?

(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).

**davefulton** · November 15th 2009, 03:34 AM

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

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

sorry about the sloppiness

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