 Hi and welcome to the session. Let us discuss the following question. Question says, find the general solution for the following differential equation. Given differential equation is, dy upon dx is equal to 1 plus x square multiplied by 1 plus y square. Let us now start with the solution. Given differential equation is, dy upon dx is equal to 1 plus x square multiplied by 1 plus y square. Now let us name this differential equation as 1. Now separating the variables of equation 1, we get dy upon 1 plus y square is equal to 1 plus x square dx. Now integrating both the sides of this equation, we get integral of dy upon 1 plus y square is equal to integral of 1 plus x square dx. Now we will find out this integral with respect to y and we will find this integral with respect to x. Now using this formula of integration, we get integral dy upon 1 plus y square is equal to tan inverse y plus c1 where c1 is the constant of integration. Now we will write this equal to sin as it is. Now we will find this integral. Now using these two formulas of integration, we can find this integral. Now we know integral of dx is equal to x. So here we can write x and integral of x square is equal to x raised to the power 2 plus 1 upon 2 plus 1. And here we can write c2 where c2 is the constant of integration. Now this further implies tan inverse y is equal to x plus x cube upon 3 plus c. Substituting c for c2 minus c1, we get this equation. Now the required solution of the given differential equation is tan inverse y is equal to x plus x cube upon 3 plus c. This completes the session. Hope you understood the solution. Take care and have a nice day.