# Solve the differential equation.1) y'=(y^2)+1

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### 2 Answers

If you have more than one question you need to make separate posts.

In order to solve this equation, we can recognize it as a separable differential equation. This means that each part of the equation is written in only one variable, so the equation can be separated into two pars, and each part is integrated.

In this case, we have:

`{dy}/{dx}=y^2+1` now move the y-parts to the left and the x-parts to the right

`{dy}/{y^2+1}=dx` integrate both sides

`int{dy}/{y^2+1}=int dx` left left integral is tan inverse, and the right integral is a polynomial

`tan^{-1}y=x+C` where C is a constant of integration, now take tan of both sides

`y=tan(x+C)`

**The differential equation has solution `y=tan(x+C)` .**

You need to solve the equation `y' = y^2 + 1` such that:

`(y')/(y^2+1) = 1`

You need to integrate both sides with respect to x such that:

`int (y')/(y^2+1)dx = int dx`

`arctan(y) = x + c `

You need to remember that `tan(arctan a) = a` , hence, reasoning by analogy, yields:

`tan(arctan y) = tan(x + c) => y = tan(x+c)`

**Hence, evaluating the general solution to the given first order nonlinear differential equation yields `y = tan(x+c).` **