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Math Help - Differential eqn

  1. #1
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    Differential eqn

    y' = 2(y^2)x - (e^ -2x)(y^2) , y(0) = 1/2
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  2. #2
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    Quote Originally Posted by DavidWagner View Post
    y' = 2(y^2)x - (e^ -2x)(y^2) , y(0) = 1/2
    \frac{dy}{dx} = y^2 \left( 2x - e^{-2x}\right) is seperable.
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    ∫(dy/y^2) = ∫2x e^(-2x)dx
    -1/y = x^2 + (e^(-2x))/2 + C
    y(x)= -1/(x^2 + ((e^(-2x)/2))/2 +C
    y(0) implies C= -2/5
    Is this correct?
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  4. #4
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    Quote Originally Posted by DavidWagner View Post
    ∫(dy/y^2) = ∫2x e^(-2x)dx
    -1/y = x^2 + (e^(-2x))/2 + C
    y(x)= -1/(x^2 + ((e^(-2x)/2))/2 +C
    y(0) implies C= -2/5
    Is this correct?
    I don't see anything obviously wrong so it's probably OK.
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