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Math Help - Ordinary differential equations

  1. #1
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    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)
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  2. #2
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    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
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  3. #3
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    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
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  4. #4
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    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|)
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  5. #5
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    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
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  6. #6
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    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
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  7. #7
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    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.
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