 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 square dy by dx is equal to X square minus 2 y square plus X y. Let's start the solution. Now the given differential equation is dy by dx is equal to X square minus 2 y square plus X y. Let us give this equation as number 1 or this can be written as dy by dx is equal to X square minus 2 y square plus X y 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 zero 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 2 since right hand side of above equation that is equation 2 is of the form zero y over x and so it is a homogeneous function of degree zero therefore the given differential equation that is the equation 1 is a homogeneous differential equation. We will solve equation 2 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 2. So from equation 2 we have v plus 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 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 2 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 2a 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 2a now a is 1 over root 2 so 2a 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 the solution is clear to you and you have enjoyed the session bye and take care