 Hi and welcome to the session. I am Deepika here. Let's discuss the question we says. Prove that x square minus y square is equal to c into x square plus y square whole square is the general solution of differential equation x cube minus 3 x y square into dx is equal to y cube minus 3 x square y into dy where c is a parameter. Let's start the solution. Now the given differential equation is x cube minus 3 x y square into dx is equal to y cube minus 3 x square y into dy or this can be written as dy by dx is equal to x cube minus 3 x y square over y cube minus 3 x square y or this can be written as dy by dx is equal to x cube into 1 minus 3 y square over x square upon x cube into y cube over x cube minus 3 y over x. Here we have taken x cube common from the numerator as well as from the denominator. Let us give this equation as number one. Now right hand side of equation one is a function of y over x. So this is a homogeneous differential equation. To solve this equation we will put y is equal to vx of 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 one we have plus x into dv over dx is equal to 1 minus 3 v square over v cube minus 3 v this can be written as x into dv over dx is equal to 1 minus 3 v square over v cube minus 3 v minus v or over dx is equal to 1 minus 3 v square minus v raise to power 4 plus 3 v square over v cube minus 3 v over dx is equal to 1 minus v raise to power 4 over v cube minus 3 v. Now on separating the variables we have v cube over 1 minus v raise to power 4 dv is equal to dx over x. Now integrating both sides we have integral of plus 3 v over 1 minus v raise to power 4 into dv is equal to integral of dx over x. Let us give this equation as number 2. Now we will solve the integral on the left hand side by using partial fractions. Let the integrant on the left hand side which is v cube minus 3 v over 1 minus v raise to power 4 is equal to a over 1 minus v plus b over 1 plus v plus over 1 plus v square. Therefore v cube minus is equal to a into 1 plus v into 1 plus v square b into 1 minus v into 1 plus v square plus into 1 minus v into 1 plus v that is 1 minus v square v is equal to 1 and minus 1. We have let us put v is equal to 1 then 1 minus 3 that is minus 2 is equal to is equal to minus 1 over 2 is equal to minus 1 then we have 2 is equal to 4b or b is equal to 1 over 2 to minus 1 over 2 b is equal to 1 over 2. Now we will find the value of sufficient of on both the sides we get 1 is equal to putting the value of a is equal to minus 4 over 2 which is a plus b plus v into 1 plus v square is equal to v 1 over 2 into 1 plus v minus v raise to power 4 dv is equal to integral of 1 over 2 dv is equal to integral of v h plus log n is equal to log log of 1 plus v square x square minus x square is equal to power 4 is e plus y square square x square let us say to x square plus y square the given differential equation we have proved that x square minus y square is equal to c into x square plus y square whole square is the general solution of the given differential equation. I hope the solution is clear to you and you have enjoyed the session. Bye and have a nice day.