 Hello and welcome to the session. In this session we discussed the following question that says, find the particular solution of the differential equation satisfying the given conditions x square d y plus x y plus y square d whole dx equal to 0, y is equal to 1 when x is equal to 1. Let's proceed with the solution now. The given differential equation is x square d y plus x y plus y square d whole dx is equal to 0, dividing both sides by x square dx. We get d y by dx plus x y plus y square this whole upon x square is equal to 0. This is possible when we get x not equal to 0. This gives us d y by dx plus x y upon x square plus y square upon x square is equal to 0 and x is not equal to 0. And so here we get d y by dx plus y upon x plus y square upon x square is equal to 0 where x is not equal to 0. So we can further write this as d y by dx is equal to minus of y upon x plus y square upon x square where this x is not equal to 0. This left hand side is a function of y upon x. So it is an homogeneous function of 0 degree. We let this as equation 1. Now putting i equal to vx in equation 1 and on putting y equal to vx we get d y by dx is equal to v plus x into dv by dx. So we have the equation as v plus x into dv by dx is equal to minus of vx upon x plus v square x square upon x square. That is we put y equal to vx in this equation 1 and on putting y equal to vx we get d y by dx is equal to v plus x into dv by dx. So on putting these two values we get this. Now this x cancels with x, x square cancels with x square. So we get v plus x into dv by dx is equal to minus of v plus v square or you can say x into dv by dx is equal to minus v minus v square minus v. So now we have x into dv by dx is equal to minus of v square plus 2v. On separating the variables we get dv upon v square plus 2v is equal to minus dx upon a integrating we get integral of dv upon v square plus 2v is equal to minus integral of dx upon x. Further we can write this as dv upon v square plus 2v plus 1 minus 1 is equal to minus integral dx upon x. Now this can also be written as integral dv upon this v square plus 2v plus 1 can be written as v plus 1v whole square minus 1 can be written as 1 square. This is equal to minus integral dx upon x. The formula integral of dx upon x square minus a square is equal to 1 upon 2a into log of modulus x minus a upon x plus a plus the constant of integration c. This formula on the left hand side we get 1 upon 2 into log modulus v plus 1 minus 1 upon v plus 1 plus 1 is equal to and integral of dx upon x is log of modulus x. So here we have minus log modulus x plus log c minus the constant of integration where this 1 cancels with minus 1. So here we have 1 upon 2 into log of modulus v upon v plus 2 is equal to log of v upon modulus x. That is we now have log of modulus v upon v plus 2 is equal to 2 into log t upon modulus x which gives us log of modulus v upon v plus 2 is equal to log of this gives us is equal to c square upon x. Now we had assumed y to be equal to vx so this means v is equal to y upon a equal to y upon x we get is equal to c square upon x. Here we have y upon is equal to c square upon x square this gives us c square equal to x square y upon y plus 2x. In this equation we have a condition given that when x is equal to 1 we get y equal to 1. So now here putting equal to 1 and y equal to 1 we get c square equal to 1 upon 1 plus 2 that is equal to 1 upon 3 that is c square is equal to 1 upon 3. Now we have assumed this equation to be equation number 2 and now we got value of c square as 1 upon 3. So putting 1 equal to 1 upon 3 in equation 2 we get 1 upon 3 is equal to x square y upon y plus 2x which gives us is equal to 3x square y. This is the required particular solution that is y plus 2x equal to 3x square y is the particular solution of the given differential equation. So this is our final answer this completes the session but we have understood the solution of this question.