Page 1 of 2 12 LastLast
Results 1 to 15 of 18

Math Help - Problem with a homogeneous equation

  1. #1
    Member
    Joined
    Aug 2010
    Posts
    138

    Problem with a homogeneous equation

    I don't know why these simple homogeneous equations are causing me so many problems, but they're kicking the teddy bear stuffing out of me.

    This is the problem I'm looking at:

    \frac{dy}{dx} = - \frac{4x + 3y}{2x + y}


    Book Answer:

    \left\vert{} y + x \right\vert{} \left\vert{} y + 4x \right\vert{}^2 = c


    Here's what I'm doing:

    v + x \frac{dv}{dx} = - \frac{4x + 3xv}{2x + xv} = - \frac{x(4+3v)}{x(2 + v)} = - \frac{4 + 3v}{2 + v}

    x \frac{dv}{dx} = - \frac{4 + 3v}{2 + v} - \frac{v(2 + v)}{2 + v} = \frac{-4 -3v}{2 + v} + \frac{-2v - v^2}{2 + v}

    x \frac{dv}{dx} = \frac{- v^2 - 5v - 4}{2 + v} = - \frac{v^2 + 5v + 4}{2 + v}

    x \frac{dv}{dx} = - \frac{(v + 1)(v + 4)}{2 + v}

    \frac{1}{x} dx = - \frac{2 + v}{(v + 1)(v + 4)}

    \frac{2 + v}{(v + 1)(v + 4)} = \frac{a}{v + 1} + \frac{b}{v + 4}

    a = \frac{1}{3}

    b = \frac{2}{3}

    \frac{1}{x} dx = - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv

    \int \frac{1}{x} dx = \int - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv

    ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v \right\vert{} - \frac{2}{3} ln \left\vert{} v \right\vert{} + c = -ln \left\vert{} v \right\vert{} + c

    e^{ln \left\vert{} x \right\vert{}} = e^{-ln \left\vert{} v \right\vert{} + c} = e^{-ln \left\vert{} v \right\vert{}} e^c = v^{-1} c

    x = \frac{1}{v} c = c \frac{x}{y}

    ...which is clearly wrong. Where am I making a mistake?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Quote Originally Posted by Lancet View Post
    I don't know why these simple homogeneous equations are causing me so many problems, but they're kicking the teddy bear stuffing out of me.

    This is the problem I'm looking at:

    \frac{dy}{dx} = - \frac{4x + 3y}{2x + y}


    Book Answer:

    \left\vert{} y + x \right\vert{} \left\vert{} y + 4x \right\vert{}^2 = c


    Here's what I'm doing:

    v + x \frac{dv}{dx} = - \frac{4x + 3xv}{2x + xv} = - \frac{x(4+3v)}{x(2 + v)} = - \frac{4 + 3v}{2 + v}

    x \frac{dv}{dx} = - \frac{4 + 3v}{2 + v} - \frac{v(2 + v)}{2 + v} = \frac{-4 -3v}{2 + v} + \frac{-2v - v^2}{2 + v}

    x \frac{dv}{dx} = \frac{- v^2 - 5v - 4}{2 + v} = - \frac{v^2 + 5v + 4}{2 + v}

    x \frac{dv}{dx} = - \frac{(v + 1)(v + 4)}{2 + v}

    \frac{1}{x} dx = - \frac{2 + v}{(v + 1)(v + 4)}

    \frac{2 + v}{(v + 1)(v + 4)} = \frac{a}{v + 1} + \frac{b}{v + 4}

    a = \frac{1}{3}

    b = \frac{2}{3}

    \frac{1}{x} dx = - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv

    \int \frac{1}{x} dx = \int - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv

    ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v \right\vert{} - \frac{2}{3} ln \left\vert{} v \right\vert{} + c = -ln \left\vert{} v \right\vert{} + c
    One mistake is in the immediate line above this sentence. You should have -ln|v+1|/3 - 2 ln|v+4|/3 + c.

    e^{ln \left\vert{} x \right\vert{}} = e^{-ln \left\vert{} v \right\vert{} + c} = e^{-ln \left\vert{} v \right\vert{}} e^c = v^{-1} c

    x = \frac{1}{v} c = c \frac{x}{y}

    ...which is clearly wrong. Where am I making a mistake?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by Lancet View Post
    I don't know why these simple homogeneous equations are causing me so many problems, but they're kicking the teddy bear stuffing out of me.

    This is the problem I'm looking at:

    \frac{dy}{dx} = - \frac{4x + 3y}{2x + y}


    Book Answer:

    \left\vert{} y + x \right\vert{} \left\vert{} y + 4x \right\vert{}^2 = c


    Here's what I'm doing:

    v + x \frac{dv}{dx} = - \frac{4x + 3xv}{2x + xv} = - \frac{x(4+3v)}{x(2 + v)} = - \frac{4 + 3v}{2 + v}

    x \frac{dv}{dx} = - \frac{4 + 3v}{2 + v} - \frac{v(2 + v)}{2 + v} = \frac{-4 -3v}{2 + v} + \frac{-2v - v^2}{2 + v}

    x \frac{dv}{dx} = \frac{- v^2 - 5v - 4}{2 + v} = - \frac{v^2 + 5v + 4}{2 + v}

    x \frac{dv}{dx} = - \frac{(v + 1)(v + 4)}{2 + v}

    \frac{1}{x} dx = - \frac{2 + v}{(v + 1)(v + 4)}

    \frac{2 + v}{(v + 1)(v + 4)} = \frac{a}{v + 1} + \frac{b}{v + 4}

    a = \frac{1}{3}

    b = \frac{2}{3}

    \frac{1}{x} dx = - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv

    \int \frac{1}{x} dx = \int - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv

    ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v \right\vert{} - \frac{2}{3} ln \left\vert{} v \right\vert{} + c = -ln \left\vert{} v \right\vert{} + c

    e^{ln \left\vert{} x \right\vert{}} = e^{-ln \left\vert{} v \right\vert{} + c} = e^{-ln \left\vert{} v \right\vert{}} e^c = v^{-1} c

    x = \frac{1}{v} c = c \frac{x}{y}

    ...which is clearly wrong. Where am I making a mistake?
    int(1/(v+1))=ln(v+1)+C

    EDIT: Haven't seen Acbeet post.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2010
    Posts
    138
    Quote Originally Posted by Ackbeet View Post
    One mistake is in the immediate line above this sentence. You should have -ln|v+1|/3 - 2 ln|v+4|/3 + c.
    Thank you! Okay, so then I have this:

    ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v + 1 \right\vert{} - \frac{2}{3} ln \left\vert{} v + 4 \right\vert{} + c

    e^{ln\left\vert{} x \right\vert{}} = \frac{e^{-\frac{1}{3} ln \left\vert{} v + 1 \right\vert{}}}{e^{\frac{2}{3} ln \left\vert{} v + 4 \right\vert{}}} e^c

    x = \frac{(v + 1)^{-\frac{1}{3}}}{(v + 4)^{\frac{2}{3}}} c

    x = \frac{1}{(v + 1)^{-\frac{1}{3}}} * \frac{1}{(v + 4)^{\frac{2}{3}}} * c


    Is this correct so far? If so, I'm not sure how we get from here to the final answer.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by Lancet View Post
    Thank you! Okay, so then I have this:

    ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v + 1 \right\vert{} - \frac{2}{3} ln \left\vert{} v + 4 \right\vert{} + c

    e^{ln\left\vert{} x \right\vert{}} = \frac{e^{-\frac{1}{3} ln \left\vert{} v + 1 \right\vert{}}}{e^{\frac{2}{3} ln \left\vert{} v + 4 \right\vert{}}} e^c

    x = \frac{(v + 1)^{-\frac{1}{3}}}{(v + 4)^{\frac{2}{3}}} c

    x = \frac{1}{(v + 1)^{-\frac{1}{3}}} * \frac{1}{(v + 4)^{\frac{2}{3}}} * c


    Is this correct so far? If so, I'm not sure how we get from here to the final answer.
    Last line is wrong!

    Remember:

     ln(a)+ln(b)=ln(ab)


    Now you just need to return to the original variables. What is v?
    Last edited by Also sprach Zarathustra; June 11th 2011 at 06:29 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Quote Originally Posted by Lancet View Post
    Thank you! Okay, so then I have this:

    ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v + 1 \right\vert{} - \frac{2}{3} ln \left\vert{} v + 4 \right\vert{} + c

    e^{ln\left\vert{} x \right\vert{}} = \frac{e^{-\frac{1}{3} ln \left\vert{} v + 1 \right\vert{}}}{e^{\frac{2}{3} ln \left\vert{} v + 4 \right\vert{}}} e^c

    x = \frac{(v + 1)^{-\frac{1}{3}}}{(v + 4)^{\frac{2}{3}}} c

    x = \frac{1}{(v + 1)^{-\frac{1}{3}}} * \frac{1}{(v + 4)^{\frac{2}{3}}} * c
    I think there's a mistake here. I think it should be

    x = \frac{1}{(v + 1)^{\frac{1}{3}}} * \frac{1}{(v + 4)^{\frac{2}{3}}} * c

    As AsZ said, sub back in your v=y/x and see what happens.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Aug 2010
    Posts
    138
    Quote Originally Posted by Ackbeet View Post
    I think there's a mistake here. I think it should be

    x = \frac{1}{(v + 1)^{\frac{1}{3}}} * \frac{1}{(v + 4)^{\frac{2}{3}}} * c

    As AsZ said, sub back in your v=y/x and see what happens.

    Actually, I had that correct on paper. That was a typo when I posted it - sorry.

    Now, from there, if I substitute back in, the farthest I can take it is this:

    x = \frac{x^{1/3}}{(y + x)^{1/3}}  * \frac{x^{2/3}}{(y + 4x)^{2/3}} * c


    I'm not quite sure how to proceed from here.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by Lancet View Post
    Actually, I had that correct on paper. That was a typo when I posted it - sorry.

    Now, from there, if I substitute back in, the farthest I can take it is this:

    x = \frac{x^{1/3}}{(y + x)^{1/3}}  * \frac{x^{2/3}}{(y + 4x)^{2/3}} * c


    I'm not quite sure how to proceed from here.
    a^m \cdot a^n=a^{m+n}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Aug 2010
    Posts
    138
    Ah! Okay, now I've got it. One thing I'm not sure of is when I take the final step and cube everything, why do the terms end up in absolute value form? Wouldn't cubing preserve whatever signs were there previously?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by Lancet View Post
    Ah! Okay, now I've got it. One thing I'm not sure of is when I take the final step and cube everything, why do the terms end up in absolute value form? Wouldn't cubing preserve whatever signs were there previously?
    You loose the absolute value in post No:4(after integration).
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Aug 2010
    Posts
    138
    Quote Originally Posted by Also sprach Zarathustra View Post
    You loose the absolute value in post No:4(after integration).
    The book answer is:

    \left\vert{} y + x \right\vert{} \left\vert{} y + 4x \right\vert{}^2 = c


    The answer I got to is:

    (y + x)(y + 4x)^2 = c


    Does that mean the book is incorrect, and no absolute values are needed in the final answer?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by Lancet View Post
    The book answer is:

    \left\vert{} y + x \right\vert{} \left\vert{} y + 4x \right\vert{}^2 = c


    The answer I got to is:

    (y + x)(y + 4x)^2 = c


    Does that mean the book is incorrect, and no absolute values are needed in the final answer?

    Here, in your post #4, you lost your absolute values.

    Quote Originally Posted by Lancet View Post
    Thank you! Okay, so then I have this:

    ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v + 1 \right\vert{} - \frac{2}{3} ln \left\vert{} v + 4 \right\vert{} + c

    e^{ln\left\vert{} x \right\vert{}} = \frac{e^{-\frac{1}{3} ln \left\vert{} v + 1 \right\vert{}}}{e^{\frac{2}{3} ln \left\vert{} v + 4 \right\vert{}}} e^c

    x = \frac{(v + 1)^{-\frac{1}{3}}}{(v + 4)^{\frac{2}{3}}} c


    Are you seeing that?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member
    Joined
    Aug 2010
    Posts
    138
    Quote Originally Posted by Also sprach Zarathustra View Post
    Here, in your post #4, you lost your absolute values.
    Are you seeing that?

    Oh! Okay, now I understand. Thanks! And thanks to everyone else who helped me!
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    May I ask, how do you know what sustitution to use for v?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  15. #15
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Quote Originally Posted by bugatti79 View Post
    May I ask, how do you know what sustitution to use for v?

    Thanks
    Once you've recognized the equation as homogeneous, then either v = y/x or v = x/y will render the equation separable. How do I know this? From reading my textbook. There is a proof for this fact in my textbook.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. using the homogeneous equation...
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: February 16th 2011, 03:15 PM
  2. homogeneous equation
    Posted in the Differential Equations Forum
    Replies: 7
    Last Post: February 4th 2011, 05:19 PM
  3. [SOLVED] Is x^3 + y^3 -xy^2 dy/dx = 0 a homogeneous equation?
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: October 19th 2010, 09:03 PM
  4. Replies: 0
    Last Post: March 6th 2009, 10:26 PM
  5. non-homogeneous equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 3rd 2007, 12:28 AM

Search Tags


/mathhelpforum @mathhelpforum