Consider the equation dydx=1x+2\frac{dy}{dx} = \frac{1}{x+2}dxdy=x+21. What is the general solution?
y=ln∣x+2∣+Cy = \ln|x+2| + Cy=ln∣x+2∣+C
y=1x+2+Cy = \frac{1}{x+2} + Cy=x+21+C
y=ln∣x∣+2+Cy = \ln|x| + 2 + Cy=ln∣x∣+2+C
y=(x+2)2+Cy = (x+2)^2 + Cy=(x+2)2+C