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Math Help - differential equations

  1. #1
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    differential equations

    a sphere with radius of 1meter has a temp = 15degrees. It lies inside a concentric sphere with radius of 2meters and temp = 25degrees. The temp T(r) at a distance r from the common center of the spheres satisfies the differential equation

    (d^2T)/dr^2 +(2/r)(dT/dr) = 0

    if we let S=dT/dr then S satisfies a first order differential equation. Solve it to find an expression for the temperature T(r) between the spheres.
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  2. #2
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    Quote Originally Posted by noscbs View Post
    a sphere with radius of 1meter has a temp = 15degrees. It lies inside a concentric sphere with radius of 2meters and temp = 25degrees. The temp T(r) at a distance r from the common center of the spheres satisfies the differential equation

    (d^2T)/dr^2 +(2/r)(dT/dr) = 0

    if we let S=dT/dr then S satisfies a first order differential equation. Solve it to find an expression for the temperature T(r) between the spheres.
    S = \frac{dT}{dr}

    so

    \frac{dS}{dr} + \frac{2}{r}S = 0.


    Integrating Factor = e^{\int{\frac{2}{r}\,dr}}

     = e^{2\ln{r}}

     = e^{\ln{r^2}}

     = r^2.


    So multiply through by the Integrating Factor

    r^2\left(\frac{dS}{dr} + \frac{2}{r}S\right) = 0r^2

    r^2\frac{dS}{dr} + 2rS = 0

    \frac{d}{dr}(r^2 S) = 0

    r^2S = \int{0\,dr}

    r^2S = C_1

    S = C_1r^{-2}.



    Now remembering that S = \frac{dT}{dr}

    \frac{dT}{dr} = C_1r^{-2}

    T = \int{C_1r^{-2}\,dr}

     = C_2 - C_1r^{-1}.


    Now use your initial conditions to find C_1 and C_2.
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  3. #3
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    I do not understand where the integrating factor came from, and I also do not understand how you jumped from to
    if you would please explain these further that would be great. Also i do not understand why my post was moved out of calculus section, because this is only a calc2 problem, so perhaps this math is higher than my current level
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  4. #4
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    nvm sorry if you had to waste your time reading these replies, i understand everything perfectly.
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  5. #5
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    If you are having trouble with the interating factor technique, in this case, the first order DE is also seperable.

    \frac{dS}{dr} + \frac{2}{r}S = 0

    \frac{dS}{dr} = -\frac{2}{r}S

    \frac{1}{S}\,\frac{dS}{dr} = -\frac{2}{r}

    \int{\frac{1}{S}\,\frac{dS}{dr}\,dr} = \int{-\frac{2}{r}\,dr}

    \int{\frac{1}{S}\,dS} = -2\ln{|r|} + C_1

    \ln{|S|} + C_2 = -\ln{|r^2|} + C_1

    \ln{|S|} + \ln{|r^2|} = C, where C = C_1 - C_2

    \ln{|Sr^2|} = C

    |Sr^2| = e^C

    Sr^2 = \pm e^C

    Sr^2 = A, where A = \pm e^C

    S = Ar^{-2}


    which is the same answer as given before.
    Last edited by Prove It; November 2nd 2009 at 05:02 PM.
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