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2nd ord ODE, non-const cffts watch

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    Now let

    p(x) = \exp (k(x))

    so

    p'(x) = k'(x) \exp (k(x))

    ERRORS from here on down, please ignore!

    p''(x) = \left( k''(x) + k'(x) \right) \exp (k(x))

    \frac{p''(x)}{p(x)} = k''(x) + k'(x)

    So we need to find K(x) s.t.
    K'(x) + K(x) = -r(x)

    where K(x) = k'(x).

    But this ones 'easy' :rolleyes: ...

    \frac{\mathrm{d}}{\mathrm{d}x}\l  eft(e^x K(x) \right) = -e^x r(x)

    K(x) = -e^{-x} \int{ e^x r(x) \mathrm{d}x }

    k(x) = -\int\left( e^{-x} \int e^x r(x) \mathrm{d}x \right) \mathrm{d}x

    So...

    p(x) = \exp \left( -\int\left( e^{-x} \int e^x r(x) \mathrm{d}x \right) \mathrm{d}x \right)

    Please check my working!
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    (Original post by Mathmoid)
    p'(x) = k'(x) \exp (k(x))

    p''(x) = \left( k''(x) + k'(x) \right) \exp (k(x))
    Oopsie.

    p''(x) = \left( k''(x) + (k'(x))^2) \exp (k(x))
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    (Original post by generalebriety)
    Oopsie.

    p''(x) = \left( k''(x) + (k'(x))^2) \exp (k(x))
    Oh carp!
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    (Original post by Mathmoid)
    Oh carp!
    In conclusion (as my brain is rapidly switching off) to find p(x) we need to solve:

    K'(x) + (K(x))^2 = r(x)

    As this is non-linear I think it will very much depend on r(x) as to whether we can find solutions for K and what these will look like.

    If it is possible to find solutions, I think my method works in principle, but there is no guarantee that the integrals will be easy / possible even from this point onwards.
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    Here's what I did.

    I had

    \frac{d^2y}{dx^2}+r(x)y=0

    I thought I could maybe simplify this by finding a substitution that would turn it into an ode with constant coeffts.

    let x=f(u)

    Then,

    \frac{dx}{du}=f'(u)\mbox{  and  }\frac{du}{dx}=(f'(u))^{-1}


    \frac{dy}{dx}=\frac{dy}{du}\cdot  \frac{du}{dx}= \frac{dy}{du} \cdot(f'(u))^{-1}


    \frac{d^2y}{dx^2}=\frac{d}{dx}\l  eft(\frac{dy}{dx} \right)=\frac{d}{du}\left(\left(  f'(u)\right)^{-1}\cdot\frac{dy}{du}\right)\cdot  \frac{du}{dx}

    \frac{d^2y}{dx^2}=(f'(u))^{-1}\{(f'(u))^{-1}\cdot \frac{d^2y}{dx^2}-(f'(u))^{-2}\cdot f''(u)\cdot \frac{dy}{du} \}

    \frac{d^2y}{dx^2}=(f'(u))^{-2}\cdot \frac{d^2y}{dx^2} - f''(u) \cdot (f'(u))^{-3} \cdot \frac{dy}{du}

    Substituting for \frac{d^2y}{dx^2} into the original ode, and rearranging, gives,

    \frac{d^2y}{du^2}-\frac{f''(u)}{f'(u)} \cdot \frac{dy}{du}+r(u) \cdot f'(u)^2 \cdot y = 0

    Now, for a const-coefft ode, I need \frac{f''}{f'}=const \mbox{  and  } r \cdot f'^2 = const

    For \frac{f''}{f'}=const, I ended up with f(u)=De^{ku}+C

    Even if I ignored the const of integration, this solution would only give a const-coefft ode for r(x) = an exponential function - not too much of an improvement beyond standard methods.

    For q \cdot f'^2 = const ... well, this is where I stopped and came here looking for help.

    It was starting to get a bit hairy now. Even for a simple poynomial such as r(x) = ax² + bx + c, solving for f wasn't going to be easy. And then there was no guarantee that \frac{f''}{f'} would end up a constant!!

    I am sooo glad there was a simpler solution to my prob
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    That's some brave work in the face of sheer hopelessness. :p:
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    I don't think you're going to get nice solutions most of the time.

    After all when r(x) = -x this is Airy's equation, which has special function solutions.

    I think your best approach is with power series given that r is a polynomial.
 
 
 
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