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13 November, 16:38

Consider the initial value problem 2ty′=4y, y (2) = - 8. 2ty′=4y, y (2) = - 8. find the value of the constant cc and the exponent rr so that y=ctry=ctr is the solution of this initial value problem.

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  1. 13 November, 17:00
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    Let's rewrite the function:

    2ty ' = 4y

    2t (dy / dt) = 4y

    t (dy / dt) = 2y

    We use the separable variables method:

    (dy / y) = 2 (dt / t)

    We integrate both sides:

    ln (y) = 2ln (t) + C

    we apply exponential to both sides:

    exp (ln (y)) = exp (2ln (t) + C)

    Exponential properties

    exp (ln (y)) = exp (2ln (t)) * exp (C)

    Log properties:

    exp (ln (y)) = Cexp (ln (t ^ 2))

    Exponential properties:

    y = C * t ^ 2

    Initial conditions y (2) = - 8:

    -8 = C * (2) ^ 2

    We clear C:

    C = - 8 / (2 ^ 2) = - 8/4 = - 2

    The function is:

    y = - 2 * t ^ 2

    Therefore we have to compare:

    y = - 2 * t ^ 2

    y = ct ^ r

    The values of c and r are:

    c = - 2

    r = 2

    Answer:

    the value of the constant c and the exponent r are:

    c = - 2

    r = 2

    so that y = - 2 * t ^ 2 is the solution of this initial value problem
  2. 13 November, 18:45
    0
    2ty'=4y

    Replacing y'=dy/dt in the equation:

    2t (dy/dt) = 4y

    Grouping terms:

    dy/y=4dt / (2t)

    dy/y=2dt/t

    Integrating both sides:

    ln (y) = 2ln (t) + ln (c), where c is a constant

    Using property logarithm: b ln (a) = ln (a^b), with b=2 and a=t

    ln (y) = ln (t^2) + ln (c)

    Using property of logarithm: ln (a) + ln (b) = ln (ab), with a=t^2 and b=c

    ln (y) = ln (ct^2)

    Then:

    y=ct^2

    Using the initial condition: y (2) = - 8

    t=2→y=-8→c (2) ^2=-8→c (4) = - 8

    Solving for c:

    c=-8/4

    c=-2

    Then the solution is y=-2t^2

    Comparing with the solution: y=ct^r

    c=-2, r=2

    Answer: T he value of the constant c is - 2 (c=-2) and the exponent r is 2 (r=2)
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