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1) yy'-e^x= 0, and y=4 when x=0 means that:

2) The rate of change of P with respect to t is directly proportional to (10-t) and inversely proportional to the cube root of P. The differential equation that models this situation is:

  • Calculus -

    y dy = e^x dx
    1/2 y^2 = e^x + c
    if (0,4) is on the graph, then
    8 = 1+c
    c = 7
    1/2 y^2 = e^x + 7
    Looks like D to me

    dP/dt = k(10-t)/∛t
    Looks like E to me
    If you must use 3 to indicate the root, why say sqrt? The sq clearly implies square root. Try 3root of cbrt.

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