 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says show that the general solution of the differential equation dy upon dx plus y square plus y plus 1 upon x square plus x plus 1 is equal to 0 is given by x plus y plus 1 is equal to a multiplied by 1 minus x minus y minus 2xy. Where a is parameter, let us now start with the solution. Now we have to show that x plus y plus 1 is equal to a multiplied by 1 minus x minus y minus 2xy is a solution of the given differential equation. We also know that this will be our solution of the given differential equation only when it satisfies the given differential equation. Now first of all we will consider the given solution. Now dividing both the sides of this equation by this bracket we get x plus y plus 1 upon 1 minus x minus y minus 2xy is equal to a. Now let us name this equation as equation 1. Now differentiating both sides of 1 with respect to x we get 1 minus x minus y minus 2xy multiplied by derivative of x plus y plus 1 minus x plus y plus 1 multiplied by derivative of 1 minus x minus y minus 2xy upon square of 1 minus x minus y minus 2xy is equal to 0. To find the derivative of this term we have applied quotient rule here and we know derivative of a constant term is equal to 0. Now this further implies 1 minus x minus y minus 2xy multiplied by 1 plus dy upon dx minus x plus y plus 1 multiplied by minus 1 minus dy upon dx minus 2x dy upon dx minus 2y is equal to 0. We know derivative of x is 1, derivative of y is dy upon dx and derivative of 1 is 0. Similarly here derivative of 1 is 0, derivative of minus x is minus 1, derivative of minus y is minus dy upon dx and derivative of minus 2xy is minus 2x dy upon dx minus 2y. To find the derivative of minus 2xy we have applied product rule. So clearly we can see multiplying both the sides of this equation by square of this bracket and finding out derivatives of these brackets we get this equation. Now this further implies 1 minus x minus y minus 2xy plus dy upon dx multiplied by 1 minus x minus y minus 2xy plus x plus y plus 1 plus dy upon dx multiplied by x plus y plus 1 plus 2x dy upon dx multiplied by x plus y plus 1 plus 2y multiplied by x plus y plus 1 is equal to 0. Clearly we can see multiplying these two brackets we get these two terms and multiplying these two brackets we get rest of the terms. Now in these three terms dy upon dx is common. So we can write this equation as dy upon dx multiplied by 1 minus x minus y minus 2xy plus x plus y plus 1 plus 2x square plus 2xy plus 2x plus 1 minus x minus y minus 2xy plus x plus y plus 1 plus 2xy plus 2y square plus 2y is equal to 0. Now we will simplify this bracket minus x plus x is 0. So they will cancel each other. Now minus 2xy plus 2xy is equal to 0. So we will cancel these two terms minus y plus y will also cancel each other. Now here minus x plus x will cancel each other similarly minus 2xy plus 2xy will cancel each other minus y plus y will also cancel each other. Now this equation becomes dy upon dx multiplied by 2 plus 2x plus 2x square plus 2 plus 2y plus 2y square is equal to 0. Now this implies dy upon dx multiplied by 2 plus 2x plus 2x square is equal to minus 2 plus 2y plus 2y square. Subtracting this bracket from both the sides of this equation we get this equation. Now dividing this equation by this bracket on both the sides we get dy upon dx is equal to minus 2 plus 2y plus 2y square upon 2 plus 2x plus 2x square. Now clearly we can see 2 is common in this bracket as well as in this bracket. So we can write dy upon dx is equal to minus 2 multiplied by 1 plus 5 plus 5 square upon 2 multiplied by 1 plus x plus x square. Now we will cancel common factor 2 from numerator and denominator both and we get dy upon dx is equal to minus 1 plus y plus y square upon 1 plus x plus x square. Now let us name this equation as equation 2. Now given differential equation is dy upon dx plus y square plus y plus 1 upon x square plus x plus 1 is equal to 0. Now we will substitute this value of dy upon dx in given differential equation. Now left hand side of the given differential equation becomes minus y square plus y plus 1 upon x square plus x plus 1 plus y square plus y plus 1 upon x square plus x plus 1. Now adding these 2 terms we get 0 and we know 0 is right hand side of the given differential equation. So we get LHS of the given differential equation is equal to RHS. So we get x plus y plus 1 is equal to a multiplied by 1 minus x minus y minus 2xy is a general solution of the given differential equation. Hence proved this completes the session. Hope you understood the solution. Take care. Have a nice day.