 Hi, and welcome to the session. When I discuss the following questions, the question says, solve the following differential equation, 1 plus x squared into d y by dx minus 2 x y equals 2 x squared plus 2 into x squared plus 1. Let's now begin with the solution. Given the differential equation is 1 plus x squared into d y by dx minus 2 x y equals 2 x squared plus 2 into x squared plus 1. Now this equation can be written as d y by dx 2 x by 1 plus x squared into y equals to x squared plus 2 into x squared plus 1 by 1 plus x squared. We have divided the both sides of this equation by 1 plus x squared. And this implies d y by dx minus 2 x by 1 plus x squared into y is equal to x squared plus 2. And if the differential equation is of the form d y by dx plus p y equals to q, then its solution is given by y into integrating factor equals to integral q into integrating factor. This integrating factor is equal to e to the power integral of p with respect to x. Now let's say this equation is equation number 1. Clearly equation 1 is of the form d y by dx plus p y equals to q is equal to minus 2 x divided by 1 plus x squared. And q is equal to squared plus 2. Fine, integrating factor is equal to e to the power integral of p with respect to x. Now here p is equal to minus 2 x by 1 plus x squared. So we have integral of minus 2 x by 1 plus x squared with respect to x. And this is equal to e to the power minus law 1 plus x squared. This is equal to 1 plus x squared to the power minus 1. And this is equal to 1 by 1 plus x squared. The solution of this type of differential equation is given by y into integrating factor equals to integral of q into integral factor with respect to x. So we have y into 1 by 1 plus x squared equals to integral of equal to integral plus 1 plus 1 by 1 plus x squared dx 1 by 1 plus x squared to 1 by 1 plus x squared is less than 3. This implies y is equal to 1 plus x squared. It's a required solution is y equals to 1 plus x squared into x plus tan inverse x plus c. So this completes the session. I take care.