Solve dydx=1x2\frac{dy}{dx} = \frac{1}{x^2}dxdy=x21 for x≠0x \neq 0x=0.
y=ln∣x∣+Cy = \ln|x| + Cy=ln∣x∣+C
y=−1x+Cy = -\frac{1}{x} + Cy=−x1+C
y=1x+Cy = \frac{1}{x} + Cy=x1+C
y=−1x2+Cy = -\frac{1}{x^2} + Cy=−x21+C