 Hello and welcome to the session. In this session we discussed the following question which says, solve the following differential equation x square minus y squared dx plus 2xydy equal to 0, given that y is equal to 1 when x equal to 1. Now let's move on to the solution. The given differential equation is x square minus y square dx plus 2xydy is equal to 0. So from here we get dy by dx is equal to y square minus x square upon 2xy. Now this given differential equation can be written in the form dy by dx equal to y upon 2x minus x upon 2y. So we say that this differential equation is a homogeneous differential equation. Let this equation be equation 1. Now we know that to solve homogeneous differential equation we put y equal to vx. Now differentiate both the sides with respect to x. So we get dy by dx is equal to v plus x into dv by dx. Now we put y equal to vx and dy by dx equal to this in equation 1. So we get v plus x into dv by dx is equal to v square x square minus x square over 2v x square. This implies that we have v plus x into dv by dx is equal to v square minus 1 over 2v. Now we transpose this v to the right hand side that is we get x into dv by dx is equal to v square minus 1 over 2v minus v. This gives us x into dv by dx is equal to 2v in the denominator then v square minus 1 minus 2v square. This further implies that x into dv by dx is equal to minus v square minus 1 over 2v. Now on separating the variables we get minus 2v over v square plus 1 dv is equal to dx over x. Now this can be further written as 2v over v square plus 1 dv is equal to minus dx over x. Then in the next step we integrate both the sides. So we get integral 2v over v square plus 1 dv is equal to minus integral dx over x. Now to solve this integral we put v square plus 1 equal to t. On differentiating both the sides with respect to t we get 2v dv is equal to dt. Then we put 2v dv equal to dt in this integral and v square plus 1 equal to t. So we get integral dt by t is equal to minus integral dx over x. That is this gives us log modulus t is equal to minus log modulus x plus log modulus c. Now we put the value for t equal to v square plus 1 in this. So this gives us log modulus v square plus 1 plus log modulus x is equal to log modulus c. That is we get log modulus v square plus 1 into x is equal to log modulus c. So this implies we get v square plus 1 into x is equal to plus minus c. Or we can also write this as v square plus 1 into x is equal to c1. Now replacing v by y over x we get y square over x square plus 1 into x is equal to c1. That is we get y square plus x square is equal to c1 x. Now it's given in the question that when x is equal to 1 then we have y is equal to 1. Therefore putting x equal to 1 and y equal to 1 in this equation we get 1 square plus 1 square is equal to c1 into 1. That is this gives us c1 is equal to 2. Let's name this equation as equation 2. Now we substitute c1 equal to 2 in equation 2 and we get y square plus x square is equal to 2x. That is we can also write this as x square plus y square is equal to 2x. So this is the required solution. Hence our final answer is x square plus y square equal to 2x. This completes the session. Hope you have understood the solution for this question.