 Hi and welcome to the session. I am Deepika here. Let's discuss the question which says show that the given differential equation is homogeneous and solid. x cos y over x plus y sin y over x into y dx is equal to y sin y over x minus x cos y over x into x dy. Let's start the solution. Now the given differential equation is x cos y over x plus y sin y over x into y dx is equal to y sin y over x minus x cos y over x into x dy. This can be written as x dy over y dx is equal to x cos y over x plus y sin y over x over y sin y over x minus x cos y over x or this can be written as dy by dx is equal to y into x cos y over x plus y sin y over x over into y sin y over x minus x cos y over x. Now we will try to make the right hand side of this equation as a function of y over x. So for this, divide the numerator as well as the denominator by x. We can divide where dx is equal to y over x into cos y over x plus y over x into sin y over x over y over x into sin y over x minus cos y over x. Now let us give this equation as number 2. Now the right hand side of the above equation that is equation 2 is of the form g of y over x and which is a homogeneous function of degree 0. Therefore, the given differential equation is a homogeneous differential equation. Now we will solve this homogeneous differential equation by putting v is equal to y over x or put y is equal to vx. Therefore dy by dx is equal to v plus x into dv over dx. Now on substituting the value of y and dy over dx in equation 2, so from equation 2 we have v plus x dv over dx is equal to v into cos v plus v sin v over v sin v. minus cos v or e plus x dv over dx is equal to v cos v plus v square sin v over v sin v minus cos v x dv over dx is equal to v cos v plus v square sin v over v sin v minus cos v x dv over dx is equal to v cos v plus v square sin v over v sin v minus cos v minus v or x dv over dx is equal to v cos v plus v square sin v minus v square sin v plus v cos v over v sin v minus cos v x dv over dx is equal to v cos v plus v cos v that is 2 v cos v over v sin v minus cos v. Now on separating the variables we have v sin v minus cos v over v cos v dv is equal to 2 over x into dx. Now integrating both sides we have integral of v sin v minus cos v over v cos v dv is equal to 2 integral of dx over x. Now the integral on the left hand side can be written as integral of v sin v over v cos v dv that is 10 v dv minus integral of cos v over v cos v dv that is 1 over v dv and this is equal to twice integral of dx over x. Now we know that integral of 10 v dv is equal to minus log mod of cos v and integral of 1 over v dv is minus log mod v and this is equal to 2 log of mod x minus log c where log c represents the constant of integration. Now on the left hand side if we take the negative sign common we have log of mod cos v plus log of mod v which is log of mod v cos v and this is equal to log of x square minus log c or this can be written as log c is equal to log x square into v cos v or c is equal to x square into v cos v. Now on replacing v by y over x we have c square into v cos v and this is equal to x square into v cos v. Now on replacing v by y over x we have c is equal to x square into y over x into cos y over x or c is equal to x y cos y over x hence the general solution of the given differential equation that is of differential equation 1 is x y cos y over x is equal to c. So this is the answer for the above question. I hope the solution is clear to you and you have enjoyed the session. Bye and take care.