Question

For each of the differential equation, find the general solution: $\frac{dy}{dx} =(1 +x^2) (1+y^2 )$

Answer 1

Answer:

$tan^{-1}y = x + \frac{x^{3}}{3} + c$

Step-by-step explanation:

Hi,

Given differential equation is

$\frac{dy}{dx} = (1 + x^{2})(1 + y^{2})$

Separating out y terms to one side and x terms

to other side , we get

$\dfrac{dy}{1 + y^{2}} = (1 + x^{2})dx$

Integrating on both sides, we get

$\int \dfrac{dy}{1 + y^{2}} = \int (1 + x^{2})dx$

$tan^{-1}y = x + \frac{x^{3}}{3} + c$

where c is an arbitrary constant.

Hence, the general solution of the differential equation is

$tan^{-1}y = x + \frac{x^{3}}{3} + c$

Hope, it helps !