$\displaystyle \dfrac{dx}{dt}$ = $\displaystyle x^2 + \dfrac{1}{81}$
x(0) = 5
the answer i got was
$\displaystyle
\dfrac{1}{9} tan( \dfrac{t}{9} + \dfrac{13.94}{9}) $
but seems to be wrong and i cant seem to figure out what im doing wrong
$\displaystyle \dfrac{dx}{dt}$ = $\displaystyle x^2 + \dfrac{1}{81}$
x(0) = 5
the answer i got was
$\displaystyle
\dfrac{1}{9} tan( \dfrac{t}{9} + \dfrac{13.94}{9}) $
but seems to be wrong and i cant seem to figure out what im doing wrong
$\displaystyle \dfrac{dx}{dt} = x^2 + \frac{1}{81} \Rightarrow \dfrac{dx}{x^2 + \frac{1}{81}}= dt \Rightarrow \int\!\dfrac{dx}{x^2 + \frac{1}{81}}= \int\!\,dt \Rightarrow 9 \tan^{-1} (9x) = t + c \Rightarrow 9\tan^{-1} 45 = c \dots$
So it appears that your solution is correct.