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I'm currently finding a solution (i.e. y) for y′′ +t(y′)2 = 0.

The answer is:

y=(1/k)ln|(k−t)/(k+t)|+c2 if c1 =k^2 >0; y=(2/k)arctan(t/k)+c2 if c1 =−k^2 <0; y=−2t−1 +c2 if c1 =0; also y=c

(NOTE: c1 or c2 is just c_1 or c_2)

I've gotten the solutions as y = 0 or y=(2/k)arctan(t/k)+c2, depending on the initial conditions, but don't seem to understand how the first solution (i.e. y=(1/k)ln|(k−t)/(k+t)|+c2) provided in the answer was derived. Also, I don't quite understand why the restrictions are being stated (i.e. if c1 > k^2). If anyone could please explain why there are multiple restrictions and different solutions for each, and also how the first solution was derived, that would be extremely helpful.

Thank you!

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# 2nd order DE w/ various solutions

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