What is the solution to the differential equation y' = y^2 - 1?

What is the solution to the differential equation y' = y^2 - 1?

Step 1: Start with the differential equation y' = y^2 - 1.
Step 2: Rewrite y' as dy/dx, so the equation becomes dy/dx = y^2 - 1.
Step 3: Separate the variables by moving all y terms to one side and x terms to the other side: dy / (y^2 - 1) = dx.
Step 4: Factor the left side: dy / ((y - 1)(y + 1)) = dx.
Step 5: Integrate both sides. The left side requires partial fraction decomposition: 1/((y - 1)(y + 1)) = A/(y - 1) + B/(y + 1).
Step 6: Solve for A and B to find the integral: ∫(1/(y - 1) - 1/(y + 1)) dy = ∫dx.
Step 7: The integrals give ln|y - 1| - ln|y + 1| = x + C, where C is the constant of integration.
Step 8: Combine the logarithms: ln|(y - 1)/(y + 1)| = x + C.
Step 9: Exponentiate both sides to eliminate the logarithm: (y - 1)/(y + 1) = e^(x + C).
Step 10: Let K = e^C, then (y - 1)/(y + 1) = Ke^x.
Step 11: Solve for y: y - 1 = Ke^x(y + 1).
Step 12: Rearrange to isolate y: y(1 + Ke^x) = 1 + Ke^x.
Step 13: Finally, solve for y: y = (1 + Ke^x) / (1 + Ke^x).
Step 14: This simplifies to y = 1/(C - x), where C is a constant.

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