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y′(x)−y(x)2−1x2−1=0 ✓ Mathematica : cpu = 0.0574839 (sec), leaf count = 143
{{y(x)→−12e−c1e4c1(2x2+2x2−1x−1)+2e2c1+2x2−2x2−1x−1},{y(x)→12e−c1e4c1(2x2+2x2−1x−1)+2e2c1+2x2−2x2−1x−1}}
✓ Maple : cpu = 0.021 (sec), leaf count = 29
{ln(x+x2−1)−ln(y(x)+(y(x))2−1)+_C1=0}
(1)y′=±y2−1x2−1
Separable. For the positive case
dydx1y2−1=1x2−1dy(y2−1)12=dx(x2−1)12
Integrating
∫dy(y2−1)12=∫dx(x2−1)12+C
But ∫dy(y2−1)12=tanh−1y(y2−1)12=ln(y+y2−1), hence
ln(y+y2−1)=ln(x+x2−1)+C
For the negative case
dydx1y2−1=−1x2−1dy(y2−1)12=−dx(x2−1)12
∫dy(y2−1)12=−∫dx(x2−1)12+C
ln(y+y2−1)=−ln(x+x2−1)+C
Therefore
ln(y+y2−1)=±ln(x+x2−1)+C
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