 Welcome to the session. I am Deepika here. Let's discuss a question it says. Show that the given differential equation is homogeneous and solid. x square minus y square into dx plus 2xy dy is equal to 0. So, let's start the solution. Now, the given differential equation is x square minus y square into dx plus 2xy dy is equal to 0. Let us give this as number 1 or we can rewrite this as xy dy is equal to minus of x square minus y square into dx or dy by dx is equal to minus of x square minus y square over 2xy or we can write dy by dx is equal to y square minus x square over 2xy. Now, we know that if a differential equation of the form dy by dx is equal to f of xy can be written in the form dy by dx is equal to g of y over x. Where g of y over x is a homogeneous function of degree 0, then the given differential equation is homogeneous. So, we will try to make this differential equation as a function of y over x. So, we can rewrite this as dy by dx is equal to y square over x square minus 1 over 2 into y over x. Let us give this as number 2 since right hand side of the above equation is of the form g of y over x and so it is a homogeneous function of degree 0. Therefore, the given differential equation is a homogeneous differential equation. We will solve this differential equation by putting y is equal to vx. So, 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 g1 by dx in equation 2. So, from equation 2 we have v plus x into dv over dx is equal to v square minus 1 over 2v. Because y is equal to vx implies v is equal to y over x, we have x into dv over dx is equal to v square minus 1 over 2v minus v. x into dv over dx is equal to v square minus 1 minus 2v square over 2v. x into dv over dx is equal to minus 1 minus v square over 2v. Now, on separating the variables, we have 2v over 1 plus v square into dv is equal to minus dx over x. Now, on substituting the value of y and g1 by dx is equal to minus 1 minus 2v minus 1 over 2v. We have 2v over 1 plus v square into dv is equal to minus dx over x. Now, on integrating both sides we have integral of 2v over 1 plus v square dv is equal to minus integral of dx over x. Now, here put 1 plus v square is equal to t. So, we have 2v dv is equal to dt. So, integral of dt over t is log t which is log of 1 plus v square is equal to minus log of mod x plus log c. We have 1 plus v square is equal to c over x. Now, on replacing the value of v by y over x, we get 1 plus y square over x square is equal to c over x. x square plus y square over x square is equal to c over x. r r plus y square is equal to c x square over x plus y square is equal to c x. So, the general solution the given differential equation is x square plus y square is equal to c x. So, this is our answer for the above question. I hope the solution is clear to you and you have enjoyed the session. Bye and take care.