Results 1 to 7 of 7

Math Help - Implicit form question

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    28

    Question Implicit form question

    Hi, I am able to find the general solution in implicit form for single line equations but I'm really having trouble with this & would appreciate some assistance.

    \frac{dy}{dx}=\frac{6y^\frac{2}{3} sin x}{(2 - cos x)^3}

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Did you try separating variables?.

    \frac{1}{6}\int y^{\frac{-2}{3}}dy=\int\frac{sin(x)}{(2-cos(x))^{3}}dx

    Integrate both sides and don't forget the constant on the right.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2009
    Posts
    28
    Quote Originally Posted by galactus View Post
    Did you try separating variables?.

    \frac{1}{6}\int y^{\frac{-2}{3}}dy=\int\frac{sin(x)}{(2-cos(x))^{3}}dx

    Integrate both sides and don't forget the constant on the right.
    Thank you & I'll give it a go now.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Dec 2009
    Posts
    28
    Okay so integrating both sides gives me this:

    \frac{y^5/3}{10}=-\frac{1}{2(2-cos x)^2} + c

    Is this the implicit form or do I need to go further?

    I have an initial condition y(0)=1 which I presume I can solve for c & then substitute back into this equation?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Yes, use your initial condition

    \frac{1}{2}y^{\frac{1}{3}}=\frac{-1}{2(cos(x)-2)^{2}}+C

    y^{\frac{1}{3}}=\frac{-1}{(cos(x)-2)^{2}}+2C

    y=\left(\frac{-1}{(cos(x)-2)^{2}}+C_{1}\right)^{3}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Dec 2009
    Posts
    28
    Quote Originally Posted by galactus View Post
    Yes, use your initial condition

    \frac{1}{2}y^{\frac{1}{3}}=\frac{-1}{2(cos(x)-2)^{2}}+C

    y^{\frac{1}{3}}=\frac{-1}{(cos(x)-2)^{2}}+2C

    y=\left(\frac{-1}{(cos(x)-2)^{2}}+C_{1}\right)^{3}
    Right I can see how you've done that (including the initial fraction for y which I left undone!) but where does the +C_{1} come from?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Dec 2009
    Posts
    28
    Quote Originally Posted by bigroo View Post
    Right I can see how you've done that (including the initial fraction for y which I left undone!) but where does the +C_{1} come from?
    I'm still not sure about that +C_{1} so I've left it as +2C. Maybe someone will tell me otherwise. Also why did the denominator reverse?

    Substituting back into general equation for y(0)=1:

    1=\left(\frac{-1}{(2-cos(0))^{2}}+2C\right)^{3}

    C=1

    Hence y=\left(\frac{-1}{(cos(x)-2)^{2}}+2\right)^{3}

    Would this satisfy my original question now or have I made it explicit? Thanks
    Last edited by bigroo; December 9th 2009 at 08:29 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Implicit Form
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: July 16th 2009, 07:08 AM
  2. find implicit form
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: July 5th 2009, 01:08 PM
  3. Differential equation in implicit form
    Posted in the Differential Equations Forum
    Replies: 8
    Last Post: March 16th 2009, 07:28 AM
  4. General solution of eqn in implicit form
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: February 24th 2009, 07:44 AM
  5. implicit form
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 30th 2008, 02:00 AM

Search Tags


/mathhelpforum @mathhelpforum