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Math Help - Mostly Differential Equations and 3 others

  1. #1
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    Mostly Differential Equations and 3 others

    Hi, I am having trouble with differential equations . I studied these a long time ago and I'm having difficulty remembering them now. Also there are a coupla other problems that are not differential equations but I didnt want to open a seperate thread for them so I am including them here. I hope that's not a problem. Thank you so much







    Last edited by askmemath; April 21st 2006 at 08:20 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Umm...BIG problems!

    Anyway:

    \frac{dy}{dx}-y=3e^x

    First solve the homogeneous equation:

    \frac{dy_h}{dx}-y_h=0

    We have that m-1=0 gives m=1 so

    y_h=Ae^x

    Now we need a particular solution. Try
    y_p=Bxe^{Cx}
    (Normally we'd perhaps try y_p=Be^{Cx} first, but a quick check will show this doesn't work.)
    So
    \frac{dy_p}{dx}=Be^{Cx}+BCxe^{Cx}

    Thus \frac{dy_p}{dx}-y_p=Be^{Cx}+BCxe^{Cx}-Bxe^{Cx}=3e^x

    Thus BC-B=0 leaving B=3 and C=1, which checks in the initial condition.

    Thus: y_p=3xe^x.

    So y=y_h+y_p = Ae^x+3xe^x.

    As you haven't listed any initial conditions, the problem is done.

    -Dan
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  3. #3
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    Quote Originally Posted by askmemath
    Hi, I am having trouble with differential equations . I studied these a long time ago and I'm having difficulty remembering them now. Also there are a coupla other problems that are not differential equations but I didnt want to open a seperate thread for them so I am including them here. I hope that's not a problem. Thank you so much
    Too many.

    Why not post them separately. One per posting if possible.

    I'd do the first and last problems here.

    1.)Solve the Initial Value Problem.
    (x^2 +1)(dy/dx) +y^2 +1 = 0; y(0) = 1.

    (x^2 +1)(dy/dx) = -y^2 -1
    (x^2 +1)dy = -(y^2 +1)dx
    (x^2 +1)/dx = -(y^2 +1)/dy
    Take the reciprocals of both sides,
    dx/(x^2 +1) = -dy/(y^2 +1)
    Integrate bot sides,
    arctan(x) +C1 = -arctan(y) +C2
    arctan(y) = -arctan(x) +C
    Get the tan of both sides,
    y = tan[C -arctan(x)] -----------(i)

    Plug y(0) = 1 into (i),
    Meaning, y=1 when x=0,
    1 = tan[C -arctan(0)]
    1 = (tanC -0)/(1 +(tanC)(0))
    1 = tanC /1
    tanC = 1
    C = arctan(1) = pi/4 ----***

    Substitute that into (i),
    y = tan[pi/4 -arctan(x)]
    y = [tan(pi/4) -tan(arctan(x))] / [1 +tan(pi/4)*tan(arctan(x))]
    y = (1-x)/(1 +1*x)
    y = (1-x)/(1+x) ------------------answer.

    --------------------
    Find the name of the family of curves if the slope of the tangent line at any point (x,y) is (1-2x)/(1+y).

    so, dy/dx = (1-2x)/(1+y).

    Cross multiply,
    (1+y)dy = (1-2x)dx
    Integrate both sides,
    y +(y^2)/2 +C1 = x -2(x^2)/2 +C2
    Multiply both sides by 2, and combine 2C1 and 2C2 into C only,
    2y +y^2 = 2x -2x^2 +C
    2x^2 -2x +y^2 +2y -C = 0
    (2x^2 -2x) +(y^2 +2y) -C = 0
    Umm, they are ellipses --------------answer.
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  4. #4
    Forum Admin topsquark's Avatar
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    \frac{dy}{dx}-\frac{1}{x}y=-xy^2.

    In general a Bernoulli equation is of the form:

    \frac{dy}{dx}+P(x)y=Q(x)y^n

    In all cases the substitution v=y^{1-n} is useful.

    So, let v=y^{1-2}=\frac{1}{y}

    Thus y=\frac{1}{v} and \frac{dy}{dx}=-\frac{1}{v^2} \frac{dv}{dx} so the differential equation becomes:

    -\frac{1}{v^2} \frac{dv}{dx}-\frac{1}{x}\frac{1}{v}=-x \left ( \frac{1}{v} \right ) ^2

    Or
    x \frac{dv}{dx}+v=x^2

    There are various and sundry methods to solve this equation. As it's a Cauchy-Euler equation, my favorite method is the substitution x=e^t, even though it's a bit of overkill for a first order equation.

    x=e^t
    dx=e^t dt
    \frac{d}{dx}=e^{-t}\frac{d}{dt}

    So the equation becomes:
    e^te^{-t}\frac{dv}{dt}+v=e^{2t}

    or
    \frac{dv}{dt}+v=e^{2t}

    The homogeneous solution of this equation is:
    m+1=0 implies m=-1 implies
    v_h=Ae^{-t}

    And the particular solution:
    v_p=Be^{Ct}
    \frac{dv_p}{dt}=BCe^{Ct}
    BCe^{Ct}+Be^{Ct}=e^{2t}

    Gives C=2 and BC+B=1 or B=1/3.

    So v(t)=Ae^{-t}+\frac{1}{3}e^{2t}

    Converting t back to x:
    v(x)=\frac{A}{x}+\frac{1}{3}x^2

    Converting v back to y:
    y(x)=\frac{1}{\frac{A}{x}+\frac{1}{3}x^2}

    or
    y(x)=\frac{3x}{x^3+3A}

    -Dan
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  5. #5
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    Quote Originally Posted by ticbol
    Why not post them separately. One per posting if possible.
    I thought about doing that first. But then I felt that it would take up a little too much server space. If you like, I can delete the questions that havent been answered and post them seperately.

    A Big thanks to topsquark and ticbol
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  6. #6
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    Quote Originally Posted by askmemath
    I thought about doing that first. But then I felt that it would take up a little too much server space. If you like, I can delete the questions that havent been answered and post them seperately.

    A Big thanks to topsquark and ticbol
    What too much server space? Not that much, I suppose.

    Nobody wants to answer too many questions/problems in one sitting. Almost everybody here does give answers for fun only---when spare time permits.

    One, or two, problem/problems per posting is good enough.
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