 Hello and welcome to the session. The given question says, solve the following differential equation and we are given 2xy into dx plus x square plus 2y square into dy is equal to 0. Let's start with the solution and we are given that 2x into y into dx plus x square plus 2y square into dy is equal to 0 or we have 2xy into dx is equal to minus x square plus 2y square into dy or dy divided by dx is equal to minus 2xy whole divided by x square plus 2y square. Now as we can see this is a homogeneous differential equation of order 2 therefore let y is equal to vx so this implies that dy divided by dx is equal to v into derivative of x is 1 plus x into derivative of v is dv divided by dx. Now substituting y is equal to vx and value of dy divided by dx in this equation let this be equation number one. So one implies on the left hand side we have v plus x into dv divided by dx is equal to minus 2 into x into vx whole divided by x square plus 2 times of y square is v square into x square. This further implies that v plus x into dv divided by dx is equal to minus 2v whole divided by 1 plus 2v square here we have cancelled x square from the numerator denominator. This further implies that x into dv divided by dx is equal to minus 2v divided by 1 plus 2v square minus v or we have x into dv divided by dx is equal to here we have minus 2v minus v into 1 plus 2v square whole divided by 1 plus 2v square here we have 1 plus 2v square as lcm we further have x into dv divided by dx is equal to minus 2v minus v gives minus 3v and then we have minus 2 times of v cube whole divided by 1 plus 2v square or we further have dx divided by x is equal to 1 plus 2v square whole divided by taking minus sign common we have 3v plus 2v cube into dv. Now integrating both the sides on the left hand side we have integral dx divided by x and on the right hand side we have minus integral 1 plus 2v square whole divided by 3v plus 2v cube into dv or we have log x is equal to minus integral. Now let us take t is equal to 3v plus 2v cube so this implies dt is equal to 3 plus 6v square into dv or we have dt divided by 3 taking 3 common from the right hand side we have here 1 plus 2v square into dv. So this can be written as 1 plus 2v square into dv is dt divided by 3 and in the denominator we have 3v plus 2v cube which is t so this is what plus a constant c. This is further equal to log x is equal to minus taking 1 by 3 outside the integral sign we have integral dt divided by t plus c or log x is equal to minus 1 by 3 log mod t plus a constant c1 or we have log x plus 1 by 3 log mod t is equal to c1 or this is further equal to log x plus log t raise to the power 1 by 3 is equal to c1 or we have log into x into t raise to the power 1 by 3 is equal to c1 or x into t raise to the power 1 by 3 is equal to e raise to the power c1 and let this constant denoted by d. So we have x into t raise to the power 1 by 3 is equal to d now let us write t in terms of x now we have assumed that t is 3v plus 2v cube so we have x into 3v plus 2v cube raise to the power 1 by 3 is equal to d now we have y is equal to vx so v is equal to y divided by x so here it can further be written as x into 3 into y by x plus 2 times of y cube divided by x cube raise to the power 1 by 3 is equal to d or cubing both the sides we have x cube into 3 times of y divided by x plus 2 times of y cube divided by x cube is equal to d cube or x cube here let us take the lcm x cube so we have 3 times of y into x square plus 2 y cube is equal to d cube and for the simplifying we have 3 y x square plus 2 y cube is equal to let us denote this d cube by some constant right hence our answer is 3 x square y plus 2 y cube is equal to some constant c so this completes the session why and take care