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Math Help - Differential equations help!

  1. #1
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    Differential equations help!

    Please help me with the following problems. These i tried solving with the substitution y=ux but ended up stuck somewhere along the way.

    1) y' = (y/x) - exp(y/x) y(1)=1

    I got as far as an integral involving -e^-u but couldn't recall how to integrate such a function...

    2) y' = (-x + 2y)/y y(1)=2

    Similarly got stuck at the integral stage.

    The following are problems involving integrating factors.

    3) y' + ln(x)y = 1 y(1)=1

    Whats the integrating factor and how does one proceed?

    4) y' + sin(x)y = sin(x) y(0)=2

    I got stuck doing the integration by parts necessary to find the general solution.

    5) y' + 3y = exp(x) y(0)=1

    6) y' + e*y = x y(1)=2



    Also there are a couple of problems involving the undetermined coefficient method. I'm completely at a loss as to what that is. If someone could explain it I'd be very grateful and could attempt those questions by myself.


    Thanks very much for any help, apologies for not posting in latex format, i tried to make everything clear and legible.
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  2. #2
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    Quote Originally Posted by Obstacle1 View Post
    Please help me with the following problems. These i tried solving with the substitution y=ux but ended up stuck somewhere along the way.

    1) y' = (y/x) - exp(y/x) y(1)=1
    Say we want to solve this on the interval x>0.

    And define a new function u=y/x
    Meaning, y=ux
    Differenciate both sides (use the product rule!)
    y'=u+u'x
    Substitue that into the differencial equation:
    u+u'x = u - exp(u)
    Thus,
    u'x=-exp(u)
    Thus,
    u'*exp(-u) = -1/x
    Integrate both sides,
    INT u'*exp(-u) dx = INT -1/x dx
    Thus, (substitution rule)
    -exp(-u) = -ln x +C
    Multiply by (-1),
    exp(-u) = ln x +C
    Then,
    -u = ln ( ln x +C) for C>=0
    Thus,
    -y/x = ln (ln x +C)
    Thus,
    y= - x * ln (ln x +C) for C>=0
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  3. #3
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    Quote Originally Posted by Obstacle1 View Post
    4) y' + sin(x)y = sin(x) y(0)=2
    A linear ODE has the form y' + P(x)y = Q(x), then an suitable integral factor will be given by u(x) = exp(int P(x)dx).

    When you get this factor, amplifies all the equation by this one, and you must modify the left member of the equation, in one derivative from a product of functions. Finally integrate each side, and the problem is killed (do not forget the constant in the right member).

    The forgotten thing, later when you find the solution to the equation, you impose the initial condition, which is y(0) = 2.
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  4. #4
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    I'm working on the integrating factor problem now and although I have the teacher's worked solutions, I'm completely at a loss as to what he's done here...

    from: d/dx [ y*exp(-cos(x)) ] = sin(x)*exp(-cos(x))

    to: y(x)*exp(-cos(x)) - y(0)*exp(-cos(0)) = (integral between x,0) of sin(s)*exp(-cos(s)) ds


    i have no idea how he makes that jump. any suggestions?
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  5. #5
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    Quote Originally Posted by Obstacle1 View Post
    I'm working on the integrating factor problem now and although I have the teacher's worked solutions, I'm completely at a loss as to what he's done here...

    from: d/dx [ y*exp(-cos(x)) ] = sin(x)*exp(-cos(x))

    to: y(x)*exp(-cos(x)) - y(0)*exp(-cos(0)) = (integral between x,0) of sin(s)*exp(-cos(s)) ds


    i have no idea how he makes that jump. any suggestions?
    You asked the same question here.
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