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    Hi,

    I think I may have forgotten some basic things so I think it's just a quick one.

    If you have a first order ODE I know you would do a integrating factor for something similar, but for this second order equation I have forgotten the trick or how to find the integrating factor:

     \frac{d^{2}c}{dr^2}+\frac{1}{r} \frac{dc}{dr}=0


    Many thanks for your help !
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    (Original post by xfootiecrazeesarax)
    Hi,

    I think I may have forgotten some basic things so I think it's just a quick one.

    If you have a first order ODE I know you would do a integrating factor for something similar, but for this second order equation I have forgotten the trick or how to find the integrating factor:

     \frac{d^{2}c}{dr^2}+\frac{1}{r} \frac{dc}{dr}=0


    Many thanks for your help !
    I believe the substitution  r = e^u will help to simplify this down for you to a fairly simple ODE

    Edit: Also I think the substitution  \frac{dc}{dr} = f(r) is another one you can use.
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    (Original post by xfootiecrazeesarax)
    Hi,

    I think I may have forgotten some basic things so I think it's just a quick one.

    If you have a first order ODE I know you would do a integrating factor for something similar, but for this second order equation I have forgotten the trick or how to find the integrating factor:

     \frac{d^{2}c}{dr^2}+\frac{1}{r} \frac{dc}{dr}=0
    There are various options here: If you have a DE of the form

    \displaystyle \sum a_k x^k \dfrac{d^k y}{dx^k} = 0, you can solve it by guessing at solutions of the form y = x^\alpha in a very similar way to how we solve \displaystyle \sum a_k \dfrac{d^k y}{dx^k} = 0. (Note that you can't "just" plug \alpha into the polynomial with the same coefficients, because the kth derivative of x^\alpha is \alpha(\alpha-1)...(\alpha+1-k) not \alpha^k, but otherwise it is very similar).

    [This is very similar to the r = e^u suggestion from the previous poster].

    If you have a 2nd order DE with no constant term, you can reduce it to a first order DE by the substitution y = dc/dr. Solve the first order DE tind y, then add an arbit constant to find c.

    Finally, you could recognize that r \dfrac{d^2c}{dr^2} + \dfrac{dc}{dr} = \dfrac{d}{dr} \left(r \dfrac{dc}{dr} \right). (This last is quite a "specific" trick, but derivatives of this form come up enough that you should be able to recognize them and variants without too much trouble).
 
 
 
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