Calculus

Consider the differential equation dy/dx = x^2(y - 1). Find the particular solution to this differential equation with initial condition f(0) = 3.

I got y = e^(x^3/3) + 2.

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  1. assuming that your answer is correct
    dy/dx = x^2 e^(x^3/3)
    but
    e^(x^3/3) = y-2
    so
    dy/dx = x^2 (y-2) hummm

    I'll try
    dy/(y-1) = x^2 dx

    ln(y-1)= x^3/3 + c

    (y-1) = e^[x^3/3 + c]
    y-1 = e^(x^3/3) e^c
    y-1 = Ce^(x^3/3)
    y = 1 + Ce^(x^3/3)

    3 = 1 + C
    C = 2

    so
    y = 1 + 2e^(x^3/3)

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