What is the general solution of y′=1xy' = \frac{1}{x}y′=x1?
y=ln∣x∣+Cy = \ln|x| + Cy=ln∣x∣+C
y=1x2+Cy = \frac{1}{x^2} + Cy=x21+C
y=ex+Cy = e^x + Cy=ex+C
y=ln∣x2∣+Cy = \ln|x^2| + Cy=ln∣x2∣+C