 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 square dy by dx is equal to x square minus 2 y square plus xy. Let's start the solution. Now the given differential equation is dy by dx is equal to x square minus 2 y square plus xy. Let us give this equation as number one. Or this can be written as dy by dx is equal to x square minus 2 y square plus xy over x square. Now if we express the right hand side of this equation as a function of y over x which is a homogeneous function of degree 0 then the given differential equation is a homogeneous differential equation. So we will try to make the right hand side of this equation as a function of y over x. Now this equation can be written as dy by dx is equal to 1 minus 2 into y square over x square plus y over x. Let us give this equation as number two since right hand side of above equation that is equation two is of the form 0 y over x and so it is a homogeneous function of degree 0. Therefore the given differential equation that is the equation one is a homogeneous differential equation. We will solve equation two 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 we will substitute the value of y and dy by dx in equation two. So from equation two we have v plus x into dv over dx is equal to 1 minus 2 v square plus v or x into dv over dx is equal to 1 minus 2 v square plus v minus 3 or x into dv over dx is equal to 1 minus 2 v square or now on separating the variables we have dv over 1 minus 2 v square is equal to dx over x now integrating both sides we have integral of dv over 1 minus 2 v square is equal to integral of dx over x or integral of dv over now let us take two common. So we have this is 1 by 2 minus v square is equal to integral of dx over x now we can rewrite this integral as 1 by 2 integral of dv over 1 over root 2 square minus v square is equal to integral of dx over x let us give this equation as number 3 now the left hand side of equation 3 is of the form integral of dx over a square minus x square and this is equal to 1 over 2 a into log of mod a plus x over a minus x plus c now by using this integral the left hand side of equation 3 is 1 over 2 into 2 a now a is 1 over root 2 so 2 a is 2 into 1 over root 2 which is root 2 into log of mod a plus x that is 1 over root 2 plus v over 1 over root 2 minus v the right hand side is log of mod x plus c now on replacing v by y over x we get 1 over 2 root 2 into log of mod 1 over root 2 plus y over x over 1 over root 2 minus y over x and this is equal to log of mod x plus c or this can be written as 1 over 2 root 2 into log of mod x plus root 2 y over x minus root 2 y and this is equal to log of mod x plus c hence the general solution of the given differential equation is 1 over 2 root 2 into log of mod x plus root 2 y over x minus root 2 y and this is equal to log of mod x plus c so this is our answer for the above question hope the solution is clear to you and you have enjoyed the session bye and take care