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Differential Equations Help!

  1. Jun 16, 2005 #1
    Find an equation of the tangent line to the curve:
    [tex]xe^y + ye^x = 1[/tex]
    at the point:
    [tex]P(0,1)[/tex]

    Find the values for [tex]\lambda[/tex] for which:
    [tex]y = e^{\lambda x}[/tex]

    satisfies the equation:
    [tex]y + y' = y''[/tex]


    I have been assigned these two problems, which are not covered for another 3 chapters.

    I am uncertain how to solve problem 1.

    This is my first attempt at problem 2, uncertain if this is correct.
    [tex]e^{\lambda x} + e^{\lambda x} \lambda = e^{\lambda x} \lambda^2[/tex]
     
  2. jcsd
  3. Jun 16, 2005 #2

    saltydog

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    Differentiate implicitly throughout the expression. You can do that right? You know, when you do so for:

    [tex]xe^y[/itex]

    you get
    [tex] xe^y(dy)+e^y(dx)[/tex]

    Right? Now do the rest, then solve for the quotient dy/dx.

    Looks good. You can figure out what lambda is right?
     
  4. Jun 16, 2005 #3

    dextercioby

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    The theorem of implicit functions solves the first problem.

    Daniel.
     
  5. Jun 16, 2005 #4
    Implicit Function Theorem...



    The correct application of the Implicit Function Theorem:

    [tex]\frac{dy}{dx} = - \frac{F_x}{F_y} = - \frac{e^y + ye^x}{xe^y + e^x}[/tex]

    Solution 1:
    [tex]\boxed{\frac{dy}{dx} = - \frac{e^y + ye^x}{xe^y + e^x}}[/tex]

    Solution 2:
    [tex]y = e^{\lambda x} \; \; y + y' = y''[/tex]
    [tex]e^{\lambda x} + e^{\lambda x} \lambda = e^{\lambda x} \lambda^2[/tex]
    [tex]e^{\lambda x} (1 + \lambda) = e^{\lambda x} \lambda^2[/tex]
    [tex](1 + \lambda) = \lambda^2[/tex]
    [tex]\lambda^2 - \lambda = 1[/tex]
    [tex]\lambda^2 - \lambda - 1 = 0[/tex]
    [tex]\lambda = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{5}}{2}[/tex]
    [tex]\boxed{\lambda = \frac{1 \pm \sqrt{5}}{2}}[/tex]
     
    Last edited: Jun 16, 2005
  6. Jun 16, 2005 #5

    dextercioby

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    Nope

    [tex] \frac{dy}{dx}=-\frac{e^{y}+ye^{x}}{xe^{y}+e^{x}} [/tex]

    computed at the point (0,1) gives the slope [itex] -e-1 [/itex].

    Daniel.
     
  7. Jun 16, 2005 #6

    Slope formula:
    [tex](y - 1) = (-e - 1)(x - 0)[/tex]

    Solution 1:
    [tex]\boxed{y(x) = -x(e + 1) + 1}[/tex]
     
    Last edited: Jun 16, 2005
  8. Jun 16, 2005 #7

    dextercioby

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    Yes,it looks good now.

    Daniel.
     
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