 Hello and welcome to the session. Let's discuss the following question. It says solve the following differential equation. So let's now start the solution. The given differential equation is 1 plus y square dx is equal to tan inverse y minus x into dy. Now we need to rewrite this differential equation. So we have 1 plus y square dx upon dy is equal to tan inverse y minus x. So this implies dx upon dy plus x upon 1 plus y square is equal to tan inverse y upon 1 plus y square. Now this is a linear differential equation where y is the independent variable and x is the dependent variable. So we need to find the integrating factor if is given by e to the power integral of the coefficient of x here which is 1 upon 1 plus y square dy. This is equal to integral of e to the power integral of 1 upon 1 plus y square is tan inverse y. Now since this is a linear differential equation where y is the independent variable and x is the dependent variable we first obtain the integrating factor and after obtaining the integrating factor we multiply x with the integrating factor and we put it equal to the integral of qy into the integrating factor. So we have x into the integrating factor which is e to the power tan inverse y is equal to the integral of integrating factor into qy which is tan inverse y upon 1 plus y square dy. Now again it is x into tan inverse y is equal to now we know that the derivative of tan inverse y is 1 upon 1 plus y square. So we will put tan inverse y is equal to t. So 1 upon 1 plus y square dy is equal to dt. So this integral becomes e to the power t upon since 1 upon 1 plus y square dy is dt. So here we will have e to the power t into t dt. Now again this is equal to x into e to the power tan inverse y. Now to integrate this we will use the by parts. So we have t into integral of e to the power t which is e to the power t minus integral of the derivative of t with respect to t which is 1 into integral of e to the power t dt which is again equal to t into e to the power t minus integral of e to the power t with respect to t is e to the power t plus c where c is the constant of integration. Now t is tan inverse y so we have x into e to the power tan inverse y is equal to tan inverse y e to the power tan inverse y minus e to the power tan inverse y plus c dividing both sides where e to the power tan inverse y we have x is equal to tan inverse y minus 1 plus c into e to the power minus tan inverse y hence x is equal to tan inverse y minus 1 plus c into e to the power minus tan inverse y is the required solution of the differential equation. So this completes the question and the session. Bye for now. Take care. Have a good day.