Determine the general solution of y′=1x+2y' = \frac{1}{x+2}y′=x+21 for x>−2x > -2x>−2.
y=ln∣x+2∣+Cy = \ln|x+2| + Cy=ln∣x+2∣+C
y=(x+2)−1+Cy = (x+2)^{-1} + Cy=(x+2)−1+C
y=ex+2+Cy = e^{x+2} + Cy=ex+2+C
y=12(x+2)2+Cy = \frac{1}{2}(x+2)^2 + Cy=21(x+2)2+C