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    This is the differential equation:
    \frac{d^2s}{d\lambda} + \frac{\lambda}{mL}s = g + \frac{\lambda}{m}

    Let the particular integral, s* be:
    s^* = C\lambda + D
    \frac{ds^*}{d\lambda}= C
    \frac{d^2s^*}{d\lambda}= 0

    Hence:
    \frac{\lambda}{mL}(C\lambda + D) = g + \frac{\lambda}{m}

    So:
    C\lambda + D = \frac{mgL}{\lambda} + L

    Comparing coeffficients:
    D = L

    But, is:
    C\lambda = \frac{mgL}{\lambda}?

    I know that they are the only variables remaining, but C\lambda's \lambda is raised to the power of 1, while \frac{mgL}{\lambda}'s \lambda is raised to the power of -1. So, I don't see how they are equal.

    Anyways, the pdf I am using for reference tells that the particular integral is:

     s^* = \frac{mgL}{\lambda} + L

    Which implies that the relationship of C with \frac{mgL}{\lambda} is true.
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    "Let the particular integral s* be: s*=C\lambda + D"

    Wait...are you sure the \lambda s are the same?

    EDIT: don't assume I know anything about diffeqs.
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    (Original post by aznkid66)
    "Let the particular integral s* be: s*=C\lambda + D"

    Wait...are you sure the \lambda s are the same?

    EDIT: don't assume I know anything about diffeqs.
    It's the trial function for the particular integral.
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    Right, just ignore that.

    Yeah, that's weird, it looks a lot like the book is assuming that relationship.

    And if s*=mgL/x + L, then:

    s*' = -mgL/x^2
    s*'' = 2mgL/x^3

    Which implies that, after plugging it into the original diffeq:

    (2mgL/x^3)+(x/mL)(mgL/x+L)=g+x/m

    2mgL/x^3 + g + x/m = g + x/m

    Which clearly isn't true when m,g,L,lambda≠0?
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    (Original post by Alpha-Omega)
    This is the differential equation:
    \frac{d^2s}{d\lambda} + \frac{\lambda}{mL}s = g + \frac{\lambda}{m}

    Let the particular integral, s* be:
    s^* = C\lambda + D
    \frac{ds^*}{d\lambda}= C
    \frac{d^2s^*}{d\lambda}= 0

    Hence:
    \frac{\lambda}{mL}(C\lambda + D) = g + \frac{\lambda}{m}

    So:
    C\lambda + D = \frac{mgL}{\lambda} + L

    Comparing coeffficients:
    D = L

    But, is:
    C\lambda = \frac{mgL}{\lambda}?

    I know that they are the only variables remaining, but C\lambda's \lambda is raised to the power of 1, while \frac{mgL}{\lambda}'s \lambda is raised to the power of -1. So, I don't see how they are equal.

    Anyways, the pdf I am using for reference tells that the particular integral is:

     s^* = \frac{mgL}{\lambda} + L

    Which implies that the relationship of C with \frac{mgL}{\lambda} is true.
    I'm very rusty on these things, and I didn't spot it at first because of the unfamiliar variables (s instead of y and \lambda instead of x), but are you sure your DE is the correct one?

    The reason I ask is that I've only ever seen a Particular Integral constructed for second order linear DEs and your DE is non-linear because of the presence of the \lambda s term on the LHS.
 
 
 
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