y ' = x(1+y)
dy/dx = x(1+y)
dy/(1+y) = x dx .... separate variables
int[dy/(1+y)] = int[x dx] ... apply integral to both sides
ln(1+y) = (1/2)x^2+C ... see note below
1+y = e^{(1/2)x^2+C}
1+y = e^C*e^{(1/2)x^2}
1+y = Ce^{(1/2)x^2}
y = Ce^{(1/2)x^2}-1
Answer is actually choice C (not choice B)
note: we would use absolute value bars for the natural log on the left side, but because y > -1, this means 1+y > 0. So there's no need to worry about if 1+y is negative
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Checking the answer:
y = Ce^{(1/2)x^2}-1
dy/dx = d/dx[ Ce^{(1/2)x^2}-1 ]
dy/dx = d/dx[(1/2)x^2]*Ce^{(1/2)x^2}
dy/dx = (1/2)*2x*Ce^{(1/2)x^2}
dy/dx = x*Ce^{(1/2)x^2} ... we get some messy expression
x*(1+y) = x*(1+Ce^{(1/2)x^2}-1)
x*(1+y) = x*Ce^{(1/2)x^2} .... we get the same messy expression as before
So this shows that dy/dx = x*(1+y) is a true equation for y > -1 and y = Ce^{(1/2)x^2}-1, which confirms the right answer.
