Results 1 to 3 of 3

Thread: u and v are solutions

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    19

    u and v are solutions

    Clearly u = cos x and v = sin x are solutions of y'' + y = 0. Show that  z = u^2 and z = uv are solutions of z''' + 4z' = 0


    I dont understand how to start this off. Im not asking for the answer just trying to understand what is going on or what to do.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,035
    Thanks
    49
    Is what's meant to be clear, clear? Just in case a picture helps...



    ... differentiating downwards.

    Spoiler:

    _________________________________________
    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    18,205
    Thanks
    2428
    Quote Originally Posted by onemore View Post
    Clearly u = cos x and v = sin x are solutions of y'' + y = 0. Show that  z = u^2 and z = uv are solutions of z''' + 4z' = 0


    I dont understand how to start this off. Im not asking for the answer just trying to understand what is going on or what to do.
    z= u^2= cos^2(x). What is (cos^2(x))'''+ 4(cos^2(x))'?

    z= uv= cos(x)sin(x). What is (cos(x)sin(x))'''+ 4(cos(x)sin(x))'?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Oct 4th 2011, 08:21 PM
  2. solutions
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 11th 2010, 05:07 AM
  3. Replies: 6
    Last Post: Jul 26th 2010, 11:45 AM
  4. help with ODE. *i got no solutions
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Oct 10th 2009, 09:12 AM
  5. Solutions to ODE y'(t) = t^2 +y(t)^2
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Jan 18th 2009, 12:44 AM

Search Tags


/mathhelpforum @mathhelpforum