once the integral is found for dy/dx=y^2+1, how do you find the equation that satisfies the initial condition of y(4)=0?
i put y=tan(x+4), but it's wrong.
oh. you're inputing answers into some online application? is y = tan(x - 4) also wrong?
we have $\displaystyle y' = y^2 + 1$
$\displaystyle \Rightarrow \frac {y'}{y^2 + 1} = 1$
$\displaystyle \Rightarrow \arctan y = x + C$
$\displaystyle \Rightarrow y = \tan(x + C)$
Now $\displaystyle y(4) = 0$
$\displaystyle \Rightarrow 0 = \tan (4 + C)$
$\displaystyle \Rightarrow \arctan0 = 4 + C$
$\displaystyle \Rightarrow 0 = 4 + C$
$\displaystyle \Rightarrow C = -4$
Thus our solution is $\displaystyle y = \tan (x - 4)$