Results 1 to 7 of 7

Math Help - Ordinary differential equations

  1. #1
    Newbie
    Joined
    Sep 2006
    Posts
    13

    Ordinary differential equations

    Ok, can someone check my qorking is sound for these please? ta

    question is to find the solution of:

    dy/dx - (y^2)/(x+1)=0

    For which y=1 at x=0

    i've got y^-2 dy = (x+1)^-1 dx

    Then integrated both sides to get

    -y^-1 = ln|x+1| + constant

    which ends with the equation

    y = -(ln|x+1|-1)^-1.


    number 2

    general solution of:

    dx/dt + x =e^t

    ive got it as:

    Sdx + S(x)dt=S(e^t) dt

    To give an answer of

    x(t+1) = e^t +c

    which doesnt seem right or is it?

    Final one which i could do with a few pointers on is

    (dw/dx)sin x -wcosx = xsin^2 x

    find an integrating factor u(x)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by macabre View Post
    number 2

    general solution of:

    dx/dt + x =e^t

    ive got it as:

    Sdx + S(x)dt=S(e^t) dt

    To give an answer of

    x(t+1) = e^t +c

    which doesnt seem right or is it?
    x=(1/2)exp(t) is a particular integral of dx/dt+x=exp(t), as can be
    seen by differentiating it and substituting it and its derivative into
    the equation.

    The general solution is a particular integral plus a solution to the
    homogeneous equation dx/dt+x=0. Which is Aexp(-t).

    So the general solution is: x(t)=Aexp(-t)+(1/2)exp(t).

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2006
    Posts
    13

    could you clarify this please?

    I'm afraid i dont understand your reasoning, would i tbe possible to provide a step by step working? also, what do you mean by Aexp? thanks a lot
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by macabre View Post
    Ok, can someone check my qorking is sound for these please? ta

    question is to find the solution of:

    dy/dx - (y^2)/(x+1)=0

    For which y=1 at x=0
    Okay,

    y'=y^2/(x+1)
    Thus,
    (1/y^2)y'=1/(x+1)
    This is a seperable equation,
    INTEGRAL (1/y^2)y' dx= INTEGRAL 1/(x+1) dx
    Thus,
    -1/y=ln|x+1|+C
    Thus,
    y=1/(C-ln|x+1|)
    When x=0 we have,
    1=1/(C-ln 1)
    Thus,
    1=1/C
    Thus,
    C=1
    Thus,
    y=1/(1-ln|x+1|)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by macabre View Post

    number 2

    general solution of:

    dx/dt + x =e^t
    The equation, first order linear diophantine equation,
    x'+x=0
    Has a solution, we the entire number line.
    Use the formula for solving 'first order linear',
    x=1/u(x) INTEGRAL u(x) 0 dx
    Thus,
    x=C/u(x)
    Where u(x) is integrating factor, which is,
    u(x)=exp ( INTEGRAL 1 dx)=exp (x)=e^x
    Thus,
    x=Ce^(-t)
    Now,
    To solve the non-homogenous diophantine equation,
    x'+x=e^t
    We need a particular solution,
    We search for solutions of the form,
    x=Ae^t
    Thus,
    x'=Ae^t and x=Ae^t
    Thus,
    Ae^t+Ae^t=e^t
    Thus,
    2Ae^t=e^t
    Thus,
    2A=1
    Thus,
    A=1/2
    Thus, all solutions to the equation,
    x'+x=e^t are,
    Ce^(-t)+1/2e^t
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker View Post
    The equation, first order linear diophantine equation,
    x'+x=0

    eherem.. differential

    Has a solution, we the entire number line.
    Use the formula for solving 'first order linear',
    x=1/u(x) INTEGRAL u(x) 0 dx
    Thus,
    x=C/u(x)
    Where u(x) is integrating factor, which is,
    u(x)=exp ( INTEGRAL 1 dx)=exp (x)=e^x
    Thus,
    x=Ce^(-t)
    At this stage you should know that a solution to:

    dx/dt=-x

    is of the form x(t)=A exp(-t)

    Now,
    To solve the non-homogenous diophantine equation,
    x'+x=e^t
    We need a particular solution,
    We search for solutions of the form,
    x=Ae^t
    It is fairly clear that if we have a linear differential equation:

    A(x)dx/dt+B(x)=G(t) .. (1)

    That if x1 is some solution to this, and x2 is a solution to
    the homogeneous equation:

    A(x)dx/dt+B(x)=0 .. (2),

    then x1+x2 is also a solution to (1). Hence the general solution to
    (1) can be written in the form of the sum of a particular solution to
    (1) amd the general solution to (2).

    Thus,
    x'=Ae^t and x=Ae^t
    Thus,
    Ae^t+Ae^t=e^t
    Thus,
    2Ae^t=e^t
    Thus,
    2A=1
    Thus,
    A=1/2
    Thus, all solutions to the equation,
    x'+x=e^t are,
    Ce^(-t)+1/2e^t
    RonL
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by CaptainBlank View Post
    eherem.. differential
    Too much number theory
    ---
    Was that post too correct me? Or to explain me? It seems to explain me, because I think I did everyting correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Ordinary Differential Equations
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: November 17th 2010, 05:41 AM
  2. Ordinary differential equations
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: October 8th 2010, 07:31 PM
  3. Ordinary differential equations....
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: July 22nd 2009, 11:13 AM
  4. Ordinary differential equations
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: January 16th 2009, 06:13 AM
  5. Ordinary Differential Equations
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 1st 2008, 08:34 AM

Search Tags


/mathhelpforum @mathhelpforum