Results 1 to 7 of 7

Math Help - Separating these equations

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    77

    Separating these equations

    1) (x^2+30)y' = xy

    If I separated it like this:

    \frac{y'}{y} = \frac{x}{x^2+30}

    ...can I integrate anyway and get lnY = something?

    2) y' = 38y^2sin(x)

    This is the same above. Getting the y and x variables, but since there's not dx can I then Integrate anyway?

    3) \frac{du}{dt} = 1+u+t+tu

    This one I'm just not sure how to separate.

    Any help on these is appreciated. I know they're simple, but well, so am I.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,340
    Thanks
    21
    Quote Originally Posted by Open that Hampster! View Post
    1) (x^2+30)y' = xy

    If I separated it like this:

    \frac{y'}{y} = \frac{x}{x^2+30}

    ...can I integrate anyway and get lnY = something?

    2) y' = 38y^2sin(x)

    This is the same above. Getting the y and x variables, but since there's not dx can I then Integrate anyway?

    3) \frac{du}{dt} = 1+u+t+tu

    This one I'm just not sure how to separate.

    Any help on these is appreciated. I know they're simple, but well, so am I.
    Factor 1+u+t+tu = (u+1)(t+1)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    You separated fine.

    Then integrate and get:

    \int\frac{1}{y}dy=\int\frac{x}{x^{2}+30}dx

    ln(y)=\frac{1}{2}ln(x^{2}+30)+C

    e to both sides isolates y and gives us a constant C_{1}

    y=C_{1}\sqrt{x^{2}+30}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2009
    Posts
    77
    Thanks for the help. Just two more quick questions:

    For the second one:

    y' = 38y^2sin(x)

    \int y^{-2}dy = \int 38sin(x)

    -\frac{1}{y} = -38cos(x) + C

    y = \frac{1}{38cos(x)+C}

    Is this wrong? Is there more than one solution?

    And then for the third one, I'm solving for u, so:

    \int (u+1)du = \int (t+1)dt
    \frac{u^2}{2}+u = \frac{t^2}{2}+t+C

    Not sure where to go to isolate it from here to isolate u.

    Happy Thanksgiving everyone.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,326
    Thanks
    1298
    Quote Originally Posted by Open that Hampster! View Post
    1) (x^2+30)y' = xy

    If I separated it like this:

    \frac{y'}{y} = \frac{x}{x^2+30}

    ...can I integrate anyway and get lnY = something?
    I don't know what you mean by "integrate anyway". Yes, of course, you can integrate \frac{dy}{y}= \frac{ xdx}{x^2+ 30}.

    2) y' = 38y^2sin(x)

    This is the same above. Getting the y and x variables, but since there's not dx can I then Integrate anyway?
    Oh, I see. There is a "dx"! y'= dy/dx and you can separate them: y^{-2}dy= 38 sin(x)dx.

    3) \frac{du}{dt} = 1+u+t+tu

    This one I'm just not sure how to separate.

    Any help on these is appreciated. I know they're simple, but well, so am I.
    Danny told you how to do the last one.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2009
    Posts
    77
    Quote Originally Posted by HallsofIvy View Post
    I don't know what you mean by "integrate anyway". Yes, of course, you can integrate \frac{dy}{y}= \frac{ xdx}{x^2+ 30}.


    Oh, I see. There is a "dx"! y'= dy/dx and you can separate them: y^{-2}dy= 38 sin(x)dx.


    Danny told you how to do the last one.
    My post right above yours is my progress after the above two posters.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Nov 2009
    Posts
    263
    Quote Originally Posted by Danny View Post
    Factor 1+u+t+tu = (u+1)(t+1)
    using this, we get

    \frac{du}{(u+1)} = (t+1)dt<br />

    integrate it and we're done.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Separating Variables
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 1st 2014, 09:07 PM
  2. Separating system of equations
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: November 30th 2011, 10:28 PM
  3. Differential equations, separating variables!
    Posted in the Differential Equations Forum
    Replies: 7
    Last Post: February 13th 2010, 11:42 AM
  4. separating variables! help!
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: October 16th 2009, 06:29 AM
  5. Separating constants
    Posted in the Algebra Forum
    Replies: 7
    Last Post: September 4th 2009, 04:31 AM

Search Tags


/mathhelpforum @mathhelpforum