# What is a solution to the differential equation dy/dx=y^2+1 with y(1)=0?

Sep 9, 2016

$y = \tan \left(x - 1\right)$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} + 1$

grouping variables

$\frac{\mathrm{dy}}{{y}^{2} + 1} = \mathrm{dx}$

Integrating both sides

$\arctan \left(y\right) = x + C$

then

$y = \tan \left(x + C\right)$

Now, observing the initial condition

$y \left(1\right) = \tan \left(1 + C\right) = 0$ so

$1 + C = 0 + k \pi , k = \pm 1 , \pm 2 , \cdots$

and concluding

$y = \tan \left(x - 1\right)$