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    • Thread Starter

    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

    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!!!

    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?
    • Thread Starter

    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!!!

    If you look at 2t/(1-t^2), you'll notice that the top is the derivative of the bottom (sort of!).
    • Thread Starter

    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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