Am i correct in thinking that the answer to this derivative problem is b? dy/dx = y/x^2 dy/y = dx/x^2 ln y * ln c = -1/x cy = e^(-1/x) y = ce^(-1/x)

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



Checking the

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.

There are two bc a studen that equals to 2