What is the general solution of y′=1xy' = \frac{1}{x}y′=x1 for x>0x > 0x>0?
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∣x∣+Cy = \ln|x| + Cy=ln∣x∣+C