 Hi and welcome to the session. I am Deepika here. Let's discuss a question which says show that the given differential equation is homogeneous and solid x dy minus y dx is equal to under root of x square plus y square into tx. Now we know that when the differential equation is expressed as a homogeneous function of degree 0 then it is called as homogeneous differential equation. So using this key idea, let's start the solution. The given differential equation is x dy minus y dx is equal to under root of x square plus y square dx. Let us give this equation as number one. Now this can be written as x dy is equal to y plus under root of x square plus y square into dx or dy by dx is equal to y over x plus under root of x square plus y square over x or dy by dx is equal to y over x plus under root of 1 plus y square over x square. Let us give this equation as number two. Now the right hand side of the above equation that is of equation two is of the form g of y over x that is a function of y over x and so it is a homogeneous function of degree zero. Therefore the given differential equation that is equation one is a homogeneous differential equation by putting y is equal to vx v is equal to y over x therefore dy by dx is equal to v plus x into dv over dx. Now on substituting the value of y and dy by dx in equation two we have plus x into dv over dx is equal to v plus under root of 1 plus v square or this can be written as x dv over dx is equal to v plus under root of 1 plus v square minus v or x dv over dx is equal to under root of 1 plus v square. Now on separating the variables we have dv over under root of 1 plus v square is equal to dx over x. Now integrating both sides we have integral of dv over under root of 1 plus v square is equal to integral of vx over x. Now the integral on the left hand side is of the form integral of dx over under root of x square plus a square and this is equal to log of mod x plus under root of x square plus a square plus c. So by using this result we have integral of dv over under root of 1 plus v square is log of mod v plus under root of v square plus 1 is equal to log of mod x plus log c where log c represents the constant of integration. Now replacing v by y over x we have log of mod y over x plus under root of y square over x square plus 1 is equal to log of mod x plus log c or this can be written as log of mod y plus under root of x square plus y square over x is equal to log c mod x y plus under root of x square plus y square over x is equal to cx or this can be written as y plus under root of x square plus y square is equal to cx square. Hence the general solution of the given differential equation is y plus under root of x square plus y square is equal to cx square. So this is the answer for the above question and this completes our session. I hope the solution is clear to you and you have enjoyed this session. Bye and have a nice day.