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Partial Derivatives

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine if the following differential equation is exact. If it is exact solve it.

    2. Relevant equations

    [tex]\left(\frac{1}{t} + \frac{1}{t^{2}} - \frac{y}{t^{2} + y^{2}}\right)dt + \left(ye^{y} + \frac{t}{t^{2} + y^{2}}\right)dy = 0[/tex]

    3. The attempt at a solution

    I am a little rusty on my partial derivatives I am not sure if this is right.

    [tex]M(t, y) = \frac{1}{t} + \frac{1}{t^{2}} - \frac{y}{t^{2} + y^{2}}[/tex]

    [tex]N(t, y) = ye^{y} + \frac{t}{t^{2} + y^{2}}[/tex]

    [tex]\frac{\partial M}{\partial y} = -y[-(t^{2} + y^{2})^{-2}(2y)] - (t^{2} + y^{2})^{-1}[/tex]

    [tex]\frac{\partial N}{\partial t} = t[-(t^{2} + y^{2})^{-2}(2t)] + (t^{2} + y^{2})^{-1}[/tex]
     
  2. jcsd
  3. Oct 2, 2009 #2

    Dick

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    Looks good so far. You'll have to do some simplifying to see if they are the same. Put them over a common denominator.
     
  4. Oct 2, 2009 #3
    [tex]-y[-(t^{2} + y^{2})^{-2}(2y)] - (t^{2} + y^{2})^{-1} = t[-(t^{2} + y^{2})^{-2}(2t)] + (t^{2} + y^{2})^{-1}[/tex]

    [tex]-y\left[-\frac{2y}{(t^{2} + y^{2})^{2}}\right] - \frac{1}{t^{2} + y^{2}} = t\left[-\frac{2t}{(t^{2} + y^{2})^{2}}\right] + \frac{1}{t^{2} + y^{2}}[/tex]

    [tex]\frac{2y^{2}}{(t^{2} + y^{2})^{2}} + \frac{2t^{2}}{(t^{2} + y^{2})^{2}} = \frac{1}{t^{2} + y^{2}} + \frac{1}{t^{2} + y^{2}}[/tex]

    [tex]\frac{2y^{2} + 2t^{2}}{(t^{2} + y^{2})^{2}} = \frac{2}{t^{2} + y^{2}}[/tex]

    [tex]\frac{2y^{2} + 2t^{2}}{t^{2} + y^{2}} = 2[/tex]

    [tex]2y^{2} + 2t^{2} = 2y^{2} + 2t^{2}[/tex] They are exact.
     
    Last edited: Oct 2, 2009
  5. Oct 2, 2009 #4
    To finish it:

    [tex]\frac{\partial f}{\partial t} = \frac{1}{t} + \frac{1}{t^{2}} - \frac{y}{t^{2} + y^{2}}[/tex]

    [tex]\frac{\partial f}{\partial y} = ye^{y} + \frac{t}{t^{2} + y^{2}}[/tex]

    [tex]f(t,y) = \int\left(\frac{1}{t} + \frac{1}{t^{2}} - \frac{y}{t^{2} + y^{2}}\right)dt + \Phi(y)[/tex]

    [tex]f(t,y) = ln|t| - \frac{1}{t} - tan^{-1}\left(\frac{t}{y}\right) + \Phi(y)[/tex]

    [tex]\frac{\partial f}{\partial y} = 0 - 0 + \frac{t}{y^{2} + t^{2}} + \frac{d\Phi}{dy} = ye^{y} + \frac{t}{t^{2} + y^{2}}[/tex]

    [tex]\frac{d\Phi}{dy} = ye^{y}[/tex]

    [tex]\Phi = \int\left(ye^{y}\right)dy + c[/tex]

    [tex]\Phi = ye^{y} - e^{y} + c[/tex]

    [tex]f(t,y) = ln|t| - \frac{1}{t} - tan^{-1}\left(\frac{t}{y}\right) + ye^{y} - e^{y} + c[/tex]
     
  6. Oct 2, 2009 #5

    Dick

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    Sure. Well done!
     
  7. Oct 2, 2009 #6
    Thanks for the help.
     
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