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Math Help - Homogeneous Equation Help

  1. #1
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    Homogeneous Equation Help

    LaTex appears to not be working so ill type normally.

    Should and can the equation below be solved using the y'=v+xv', y=vx substitution method? I try to do so and i get stuck halfway through.

    -2y'+(4y/x)=y^3/x

    -2(v+xv')+(4vx/x)=(v^3x^3)/x

    -2v-2xv'+4v=v^3x^2

    -2xv'+2v=v^3x^2

    -2xv'=v^3x^2-2v

    v'=-(v^3x^2-2v)/2x


    Thats where i get stuck. Ive tried other ways but i still get stuck.
    Thanks in advance for the help.
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  2. #2
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    Re: Homogeneous Equation Help

    It's a Bernoulli Equation.
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  3. #3
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    Re: Homogeneous Equation Help

    Ok thank you. I managed to figure that one out. Took me a while but thanks for your input. But now im stuck at another question.

    y'=\frac{x^4y^4+4x^4}{y^3}

    So I tried to check if it was homogenous using f(tx,ty)=f(x,y) and I concluded its not homogenous. Is that right?

    I attempted to rearrange it and work it out as a Bernoulli equation but still I didnt get to the correct answer. Any guidance to which method I should be trying?
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  4. #4
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    Re: Homogeneous Equation Help

    If you divide each term by y^3 you'll find this one is also a Bernoulli equation.
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  5. #5
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    Re: Homogeneous Equation Help

    Hello, petenice!

    y'\:=\:\frac{x^4y^4+4x^4}{y^3}

    The variables are Separable!

    \frac{dy}{dx} \;=\; \frac{x^4(y^4+4)}{y^3} \quad\Rightarrow\quad \frac{y^3}{y^4+4}\,dy \;=\;x^4\,dx
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  6. #6
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    Re: Homogeneous Equation Help

    I was able to figure it out once you guys set me in the right direction. I was making simple mistakes in the algebra. Thanks for the help everyone.
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