help plese!! differential equation 4dt = t (t^2 -4)^(1/2) ds?

can somebody help me to solve this differential equation with step by step solution.

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  • 1 decade ago
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    First separate the variables (t to the LHS, s to the RHS)

    4dt = t·√(t² -4) ds

    4/[t·√(t² -4)] dt = ds

    Next step is to integrate LHS with respect to t and right hand side with respect to s:

    ∫ 4/[t·√(t² -4)] dt = ∫ ds

    An irrational integral of the type √(t² -a²) like on the LHS can be transformed to a rational integral solved by the substitution

    t = a·sec(x)

    here

    t = 2·sec(x)

    dt/dx = 2·sec(x)·tan(x) dx (because 1+tan²(x) = sec²(x))

    dt = 2·sec(x)·√(sec²(x) - 1)

    Substitute in the integral

    ∫ 4/[2·sec(x)·√(4·sec²(x) -4)] ·2·sec(x)·√(sec²(x) - 1)dx = ∫ ds

    ∫ 4/[2·√(sec²(x) -1)] ·√(sec²(x) - 1)dx = ∫ ds

    ∫ 2dx = ∫ ds

    2x = s + C

    Where C is the constant of integration.

    With the reverse substitution t = 2·sec(x) <=> x = arcsec(t/2) = arccos(2/t) you get:

    2·arccos(2/t) = s + C

    Hence the solution of the DE is

    s = 2·arccos(2/t) - C

    or

    t = 2/cos((s+C)/2)

    For the evaluation of the constant C you need an additional condition.

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