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Solve this ODE? watch

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    Does anyone have any idea how to solve this equation

    d2y/dx2 = K / y2

    where K is a constant?

    Or indeed, does anyone know why it is hard to solve/whether I should do it numerically....
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    maybe:

    => dx^2/d^2y = y^2/k

    int[d^2x] = (1/k) int[dy^2]

    x^2/2 = (1/k)y^4/12 +c
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    actually might be

    x^2/2 = (1/k)y^4/12 + ky + c

    or even:

    x^2/2 + nx = (1/k)y^4/12 + ky + c
    • Thread Starter
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    Good try. That was one particular erroneous route I hadn't considered!

    However, 1/(d^2y/dx^2) does not equal (d^2x/dy^2). That only works for dy/dx = 1/(dx/dy).

    I've just put it into Maple and got a horrid answer out. I'm actually trying to find the equation of motion for a space vehicle powered by a solar sail, which x is the distance from the sun:

    d2x/dt2 = K/x2

    So I did:

    v*dv/dx = K/x2

    v2/2 = K(1/R - 1/x2)

    where R is the radius of the sun (that's the closest u can start your solar sail vehicle!)

    So that's the furthest I got!
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    Good try. That was one particular erroneous route I hadn't considered!

    However, 1/(d^2y/dx^2) does not equal (d^2x/dy^2). That only works for dy/dx = 1/(dx/dy).

    I've just put it into Maple and got a horrid answer out. I'm actually trying to find the equation of motion for a space vehicle powered by a solar sail, which x is the distance from the sun:

    d2x/dt2 = K/x2

    So I did:

    v*dv/dx = K/x2

    v2/2 = K(1/R - 1/x2)

    where R is the radius of the sun (that's the closest u can start your solar sail vehicle!)

    So that's the furthest I got
    Ah well. What kind of level stuff is this?
    • Thread Starter
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    Well, I'm a 2nd year physics undergrad. But the maths is probably A-level. I think it might be a non-linear equation, and I have no idea how to solve them, or approximately solve them.

    I just wondered if any bright sparks had some ideas.
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    We never did these type of problems at A-level, but this is the sort of thing I've been doing in my first year of undergrad maths.

    Write p = dy/dx

    d2y/dx2 = d/dx(dy/dx) = dp/dx

    dp/dy = (dp/dx)*(dx/dy) , by the chain rule
    dp/dy = (d2y/dx2)(1/p)

    => d2y/dx2 = p(dp/dy)

    => p(dp/dy) = K/y2

    INT (p)dp = INT (K/y2)dy

    p2/2 = -K/y + c

    p = [2(c - K/y)]1/2
    => dy/dx = [2c - (2K)/y]1/2

    INT [2c - (2K)/y]-1/2 dy = INT dx

    (-1/K)[2c - (2K)/y]1/2 = x + c'


    And rearrange as necessary.
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    (Original post by ssmoose)
    maybe:

    => dx^2/d^2y = y^2/k

    int[d^2x] = (1/k) int[dy^2]

    x^2/2 = (1/k)y^4/12 +c
    There are several fundemental errors in your thinking:

    1) dy/dx = 1/(dx/dy) !=> d2y/dx2 = 1/(d2x/dy2)

    2) What do you assume "∫d2x" and "∫dy2" mean?

    Whatever you think it means, it's pretty far from the truth! For example, try doing it on the equation d2x/dy2 = 1 (which has a fairly obvious solution). This is just an exercise in using the usual A-Level trick:

    d2x/dt2 = v dv/dx

    Where v = dx/dt.

    • Thread Starter
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    (Original post by Wrangler)
    This is just an exercise in using the usual A-Level trick:

    d2x/dt2 = v dv/dx

    Where v = dx/dt.

    So how do you get x in terms of t for what i did?
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    v dv/dx = K/x2

    vdv = Kdx/x2

    v2/2 = -K/x + C

    dx/dt = sqrt[ 2C - 2K/x ] = sqrt[2Cx-2K]/sqrt[x]

    sqrt[x]dx/sqrt[Cx-K] = sqrt[2]dt

    Let Cx = Kcosh2y
    dx = (2K/C)coshy.sinhy.dy
    sqrt[x] = sqrt[K/C]coshy
    sqrt[Cx-K] = sqrt[K]sqrt[cosh2y-1] = sqrt[K]sinhy

    Gives an integral of cosh2y dy = dt (with the appropriate multiplicative constants).
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    its further maths i think its differential equations
    • Thread Starter
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    Awesome. Good work, AlphaNumeric.

    I think I might have just rearranged something wrong to being with which is why Maple had a nightmare, because you get to a pretty standard integral, so there's no reason why it shouldn't have solved it. Ah well.
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    (Original post by Worzo)
    So how do you get x in terms of t for what i did?
    Magic, of course.
 
 
 
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