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Math Help - DiffyQ

  1. #1
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    DiffyQ

    Find the general solution to the differential equation. Primes denote derivatives with respect to x.

    (3y^2)y' + y^3 = e^-x
    I beleive this is a Bernoulli Equation but I am still having trouble.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by bearej50 View Post
    Find the general solution to the differential equation. Primes denote derivatives with respect to x.

    (3y^2)y' + y^3 = e^-x


    I beleive this is a Bernoulli Equation but I am still having trouble.

    Note that the derivative of y^3 is
    \frac{d}{dx}y^3=3y^2y'

    try the substitution u=y^3 \implies \frac{du}{dx}=3y^2\frac{dy}{dx}

    So we get the equation \frac{du}{dx}+u=e^{-x}

    This can be solved with an integrating factor.

    Happy Hunting

    TES
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  3. #3
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    I got that far as well. I cannot figure out what the integrating factor would be because there is no x value attached to the u
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    \frac{du}{dx}+1\cdot u =e^{-x}

    p(x)=1 \implies e^{\int 1 dx}=e^x

     <br />
e^x\frac{du}{dx}+e^xu=1 \iff \frac{d}{dx}\left[ e^x \cdot u\right]=1<br />

    Now integrating both sides you get

    e^x\cdot u=x+c \implies u=xe^{-x}+ce^{-x} \implies y^3=xe^{-x}+ce^{-x}
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