I was working on the 2019-FE-P3, for the Problem3 of the Deferred section, I got C1=10e^(-t)+10arctan(e^t)+C1, C2=-4e^(-2t)+4ln(e^(-2t)+1)+C2
The answer shows C1=10e^(-t)+5arctan(e^t)+C1, C2=-4e^(-2t)-4ln(e^(2t)+1)+C2.
I was wondering if anyone got the same answer with me, thanks very much.