 Hi, and welcome to the session. I am Deepika here. Let's discuss the question It says for the fine differential equation find the particular solutions satisfying the given condition x squared dy plus x y plus y squared into ds is equal to 0 y is equal to 1 when x is equal to 1 Let's start the solution now the given differential equation is x squared dy plus x y plus y squared into ds is equal to 0 now this can be written as x squared dy is equal to minus of x y plus y squared into ds or dy over dx is equal to minus of x y plus y squared over x squared dy by dx is equal to minus of y over x plus y squared over x squared y over x in this equation as number one now we have expressed our given differential equation as a function of y over x so the above differential equation homogeneous differential equation now to solve this differential equation we will put y is equal to vx or v is equal to y over x therefore dy over dx is equal to v plus x into dv over dx now on substituting the value of y and dy by dx in equation one so from equation one we have x dv over dx is equal to minus v minus v squared dv over dx is equal to minus v minus v squared minus v or x dv over dx is equal to minus 2v minus v squared now on separating the variables we have dv over 2v plus v squared is equal to minus dx over x Integrating both sides we have integral of dv over 2v plus v squared is equal to negative integral of dx over x Let us get this equation as number two now We will solve the integral on the left hand side by using partial friction Here the integrand is a proper rational function Therefore by using the form of partial friction we can write the integrand 1 over 2v plus v squared is equal to 1 over v into 2 plus v Let 1 over v into 2 plus v is equal to a over v plus v over 2 plus v So we have 1 is equal to a into 2 plus v plus pv Now on equating the coefficient of v and constant term on both sides we have 2 is equal to 1 this implies a is equal to 1 over 2 and a plus v equal to 0 this implies b is equal to minus 1 over 2 therefore 1 over 2 v plus v squared is equal to 1 over 2 v minus 1 over 2 into 2 plus v So from equation 2 we have 1 over 2 v minus 1 over 2 into 2 plus v dv is equal to negative integral of dx over x or The integral on the left hand side can be written as integral of 1 over 2 v dv minus integral of 1 over 2 into 2 plus v dv is equal to negative integral of dx over x or This can be written as 1 over 2 log mod v minus 1 over 2 log of mod 2 plus v and this is equal to minus log mod x plus log 7 where log 7 represents a constant of integration or We can write this equation as 1 over 2 log of mod v over 2 plus v and this is equal to log 7 over mod x or u over 2 plus v is equal to 2 log 2 plus v is equal to log c o again we can write this equation as v over 2 plus v is Replacing v by y over x we have y over x over 2 plus y o this can be written as y at the value of this constant c when on substituting equation 3 we get equal to c equal to 1 over 3 the value of c in equation 3 we have y x square equal to 1 over 3 into 2x plus y y x square The particular solution of the given differential equation is y plus 2x is equal to 3x square y And this is the answer for the above question. So this completes our session I hope the solution is clear to you and you have enjoyed the session. Bye and take care