# 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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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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