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Solve the Initial Value problem

  1. Feb 6, 2012 #1
    dy/dt=t^(2)y^(3) , y(0)=-1

    I need help solving this

    I put the integral (dy/y^3)= integral (t^2)dt
    but idk what to do after that or if thats even right
     
    Last edited: Feb 6, 2012
  2. jcsd
  3. Feb 6, 2012 #2

    ehild

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    Hi Imanri,

    You dropped an integral sign...

    [tex]\int{\frac{dy}{y^3}}=\int{t^2dt}[/tex]

    Do the integration on both sides.

    ehild
     
  4. Feb 6, 2012 #3
    Thank you, but after I find that what do I do with y(0)=1
     
  5. Feb 6, 2012 #4

    ehild

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    You have an undetermined constant after integration. Substitute t=0 and y=-1 in your integral and solve for c.

    ehild
     
  6. Feb 6, 2012 #5
    Thank you
     
  7. Feb 6, 2012 #6

    ehild

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    What is your solution?

    ehild
     
  8. Feb 7, 2012 #7
    y=-1/(t^(3)+C)
     
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