Results 1 to 4 of 4

Math Help - Differential Equations, what are they? a few ?'s

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    32

    Differential Equations, what are they? a few ?'s

    I'm having a tough time figuring out exactly what differential equations are. after reading the chapter on it 3 times over, I do not understand how one equation is a solution of the diff.eq. of some other equation. Can somebody explain to this to me? I've got 2 problems that I found that would help me learn what it is if somebody can explain.

    first one:
    show that y = x - x^-1 is a solution of the differential equation xy' + y = 2x

    there's only 1 example in my book and after following its same steps, i get y' = (1/x^2) + 1, how is that a solution of xy' + y = 2x?

    second:
    1) what can be concluded about a solution of the equation y' = -y^2 just by looking at it?
    2) verify all members of family y = 1/(x+C) are solutions of y' =-y^2
    3) think of a solution NOT a meber of the family in part 2
    4) find a solution of the initial-value problem: y' = -y^2 y(0)=.5

    the second one really has me wondering of how to approach y' = -y^2 and especially in 2), how is that function a solution?

    All help appreciated
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,832
    Thanks
    1602
    Quote Originally Posted by deltemis View Post
    I'm having a tough time figuring out exactly what differential equations are. after reading the chapter on it 3 times over, I do not understand how one equation is a solution of the diff.eq. of some other equation. Can somebody explain to this to me? I've got 2 problems that I found that would help me learn what it is if somebody can explain.

    first one:
    show that y = x - x^-1 is a solution of the differential equation xy' + y = 2x

    there's only 1 example in my book and after following its same steps, i get y' = (1/x^2) + 1, how is that a solution of xy' + y = 2x?

    second:
    1) what can be concluded about a solution of the equation y' = -y^2 just by looking at it?
    2) verify all members of family y = 1/(x+C) are solutions of y' =-y^2
    3) think of a solution NOT a meber of the family in part 2
    4) find a solution of the initial-value problem: y' = -y^2 y(0)=.5

    the second one really has me wondering of how to approach y' = -y^2 and especially in 2), how is that function a solution?

    All help appreciated
    x\,\frac{dy}{dx} + y = 2x

    Can you see that the LHS is a product rule expansion of \frac{d}{dx}\,(x\,y)?


    So \frac{d}{dx}\,(x\,y) = 2x

    x\,y = \int{2x\,dx}

    x\,y = x^2 + C

    y = x + Cx^{-1}.


    If C = -1 then x - x^{-1} is a solution.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2009
    Posts
    226
    The derivative of y = x - x^{-1} with respect to x is y' = 1 + \frac{1}{x^2}.

    Substitute y = x - x^{-1} and y' = 1 + \frac{1}{x^2} in xy' + y = 2x. Simplify to an obviously true statement, such as 0 = 0.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2009
    Posts
    32
    wow thank you both for fast and simplified answers, I hadn't noticed that product expansion and I wouldn't have even thought of substitution

    I get it now, thank you both very much
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 'Differential' in differential equations
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: October 5th 2010, 11:20 AM
  2. Replies: 2
    Last Post: May 18th 2009, 04:49 AM
  3. Differential Equations
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 24th 2007, 05:52 AM
  4. Replies: 5
    Last Post: July 16th 2007, 05:55 AM
  5. Replies: 3
    Last Post: July 9th 2007, 06:30 PM

Search Tags


/mathhelpforum @mathhelpforum