I have the following ODE Where given is $x(0)=1$:
$(t+3)dx=4x^2dt$
After separation of variables I got this:
$\frac{-1}{x} = 4\ln(t+3)+C$
I think this simplifies more as:
$x=\frac{-1}{\ln((t+3)^4)+C}$
Please tell me if this is correct, I also have problem finding C in this case, MapleTA does not accept my answer which happens to be:
$x=\frac{-1}{\ln((t+3)^4)+(-0.52)}$
Update:
I found that the C should e expresed also in $\ln$ (logarithmic) term, not a number. Any ideas?