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Math Help - Initial Value Problems

  1. #1
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    Unhappy Initial Value Problems

    1. y(dy/dx) = (y^3 - 1) (x/y) e^(ax^2) ; y(0) = -1
    a > 0 is fixed constant


    I solved to split up the x and y components and got:
    (y^2)/(y^3 -1) dy = (xe^ax^2) dx
    n(y) = m(x) (using equation above i made this one)
    so then i made it
    m(x) - n(x) = 0

    then dm/dx - dn/dy = C where C=constant
    soo...
    3Ln(y^3 -1) - (e^(ax^2))(1- 1/4a^2) = C

    .... solve for y....

    y = cube rt (e^(1/3)((e^(ax^2))(x-1/4a^2)+C)

    sorry for all the "( )" i know it makes it hard to read.

    then when i plugged in y(0) = -1 i couldnt get it to work because i ended with...

    -1 = e^-(1/12a^2) + C

    and the e function cant be negative right? help!
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  2. #2
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     \frac{y^2}{y^3-1} ~dy = xe^{ax^2}~dx

    At this point could you not try to integrate both sides, left hand side by substitution and right hand side by parts?
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  3. #3
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    i am not quite sure how to do it without differentiating...
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  4. #4
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    We need to consider the following

     \int \frac{y^2}{y^3-1} ~dy = \int xe^{ax^2}~dx

    To solve this we need to do the opposite to differeniating.

    Do you know of this  = \int x^{n}~dx= \frac{x^{n+1}}{n+1}+c ?

    This is a simple idea on how to solve such an equation, but in your case it's not quite this simple. Have you heard of integration by parts or by substitution?
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  5. #5
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    that is what i did. but i was just verifying that the rest was good as well. that i set everything up right. its just hard to tell lol. i used substitution on the 'y' side and by parts on the 'x' side. but i got -1 = e to a power, and the e exponential can never be negative, right?
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  6. #6
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    You can have e such that e^{-a} = \frac{1}{e^{a}}
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