Am I correct in thinking that the answer to this derivative problem is B?dy/dx = y/x^2dy/y = dx/x^2ln y * ln c = -1/xcy = e^(-1/x)y = ce^(-1/x)
Accepted Solution
A:
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