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Differential Equations - Initial Value Problem

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose that the initial value problem
    y' = 7(x^2) + (5y^2) − 6, y(0)=−2
    has a solution in an interval about x=0.

    Find y'(0) =
    Find y''(0) =
    Find y'''(0) =

    2. Relevant equations

    get it into standard form: dy/dt + p(t)y = g(t)
    find integrating factor = e^([tex]\int[/tex]p(t)dt + k

    multiply everything by integrating factor, simplify left-hand-side and then integrate both sides

    using initial condition, solve for C
    solve for y

    3. The attempt at a solution

    i don't seem to be able to get it into standard form

    i tried doing y' - 5y^2 = 7x^2 -6
    which gave me an integrating factor of e^-5xy

    i tried following the rest of the steps with that integrating factor but its not working
  2. jcsd
  3. Sep 11, 2009 #2


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    Science Advisor
    Homework Helper

    You don't actually have to solve the equation to find the derivatives of y(x) at x=0. To find y'(0) just plug x=0, y=(-2) into the equation. To find the higher derivatives, just differentiate the equation with respect to x a couple of times.
  4. Sep 11, 2009 #3
    oh wow i feel like such a moron
    thank you so much!!

    edit: wait, for y''(0)

    i differentiated it, and i got 14x?

    edit(2): nvm i solved it
    Last edited: Sep 12, 2009
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