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# differential equations questions watch

1. 1.solve the differential equation with the given initial conditions :
(1-t^2)dy/dt = 2t, y=0 when t=0, for -1<t<1

2.find the solution curve of the following differential equation which passes through the given points
(x+1)dy/dx =x-1 , through the origin for x>-1

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ok um.. for the first one i can get to
dy/dt = 2t/(1-t^2) then i used partial fractions to get to
dy/dt=t/1-t + t/1+t and i dont know how to integrate this or where to go with it

and for the second one i can get to
dy/dx=x-1/x+1 :P like i got anywhere with this!!!
2. You don't really need partial fractions for the first one. Try using the sub u=1-t^2.

And for the second one:
(x-1)/(x+1) = (x+1-2)/(x+1) = (x+1)/(x+1) - 2/(x+1) = 1 - 2/(x+1)

I take it you can go on from here?
3. yes i can!!! yes i can!!! ok i'm in lurve with u rite now :P
its just that i wasnt getting the whole thing and i've got 4 exercises to do in one day and i can do this!!! :P btw ok um... cud u tell me um... how u got to wat to substitute for the for first one... and i guess for the second one if i have a prob like that i'll stick to this method!!!
4. If you look at 2t/(1-t^2), you'll notice that the top is the derivative of the bottom (sort of!).
5. ok so thats the trick? thanks!!! thanks... u c u gave me hope i can get thru doing 4 exercises in one day btw i'm on the first :P and its 9 pm where i am!! hope!!!

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Updated: April 9, 2006
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