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2 December, 06:08

Show that A (t) = 300-250e0.2-0.02t satisfies the differential equation ⅆAⅆt=6-0.02A with initial condition A (10) = 50.

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  1. 2 December, 06:15
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    Step-by-step explanation:

    dA/dt = 6 - 0.02A

    dA/dt = - 0.02 (A - 300)

    Separate the variables.

    dA / (A - 300) = - 0.02 dt

    Integrate.

    ln (A - 300) = - 0.02t + C

    Solve for A.

    A - 300 = Ce^ (-0.02t)

    A = 300 + Ce^ (-0.02t)

    Use initial condition to find C.

    50 = 300 + Ce^ (-0.02 * 10)

    50 = 300 + Ce^ (-0.2)

    -250 = Ce^ (-0.2)

    C = - 250e^ (0.2)

    A = 300 - 250e^ (0.2) e^ (-0.02t)

    A = 300 - 250e^ (0.2 - 0.02t)
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