 Hi, and welcome to the session I am Deepika here. Let's discuss a question which says for the following differential equation find the particular solution satisfying the given condition x plus y into dy plus x minus y into dx 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 plus y into dy plus x minus y into dx is equal to 0 and we have to find the particular solution of this differential equation satisfying the condition y is equal to 1 when x is equal to 1 or we can rewrite this differential equation as x plus y into dy is equal to minus of x minus y into dx or dy by dx is equal to y minus x over x plus y or this can be written as dy by dx is equal to y over x minus 1 over 1 plus y over x. Let us get this equation as number 1. Now the right hand side of equation 1 is of the form g of y over x that is a function of y over x hence it is a homogeneous function of degree 0 and 4 the given differential equation is a homogeneous differential equation. Now to solve it we will put y is equal to v x therefore dy by dx is equal to v plus x into dv over dx. Now we will substitute the value of y and dy by dx in equation 1. So from equation 1 we have v plus x dv over dx is equal to v minus 1 over 1 plus v x dv over dx is equal to v minus 1 over 1 plus v minus v dv over dx is equal to v minus 1 minus v minus v square over 1 plus v x dv over dx is equal to minus of 1 plus v square over 1 plus v. Now on separating the variables we can write this equation as 1 plus v over 1 plus v square into dv is equal to minus dx over x. Now on integrating both sides we have integral of 1 plus v over 1 plus v square dv is equal to negative integral of dx over x or this integral that is the integral on the left hand side can be written as integral of 1 over 1 plus v square dv plus integral of v over 1 plus v square dv is equal to negative integral of dx over x. Now integral of 1 over 1 plus v square dv is tan inverse v here. Let us substitute 1 plus v square is equal to t then 2 v dv is equal to dt that is v dv is equal to 1 over 2 dt. So the integral of v over 1 plus v square dv is 1 by 2 into log t that is log 1 plus v square and this is equal to minus log mod x plus c. Now replace v by y over x we have tan inverse y over x plus 1 over 2 log 1 plus y square over x square and this is equal to minus log mod x plus c or tan inverse y over x plus now 1 by 2 log 1 plus y square over x square can be written as 1 over 2 log x square plus y square minus 1 by 2 log x square and this is equal to minus log mod x plus c or tan inverse y over x plus 1 over 2 log x square plus y square minus 2 into 1 by 2 that is minus log mod x is equal to minus log mod x plus c or tan inverse y over x plus 1 over 2 log of x square plus y square is equal to c. Let us give this equation as number 2. Now this is a general solution of the given differential equation. We have to find the particular solution when x is equal to 1, y is equal to 1 so on substituting x is equal to 1 and y is equal to 1 in equation 2 we get tan inverse 1 over 1 that is 1 plus 1 over 2 log of 1 square plus 1 square is equal to c pi by 4 because tan inverse 1 is pi by 4 plus 1 by 2 log 2 is equal to c. Now on substituting the value of c in equation 2 we get tan inverse y over x plus 1 over 2 log of x square plus y square is equal to pi by 4 plus 1 over 2 log 2. Now on multiplying this equation by 2 we have 2 tan inverse y over x plus log of x square plus y square is equal to pi by 2 plus log 2. Hence the particular solution of the given differential equation log of x square plus y square plus 2 tan inverse y over x is equal to pi by 2 plus log 2 so this is our answer for the above question. This completes our session. I hope the solution is great to you and you have enjoyed the session. Bye and take care.