 Hello and welcome to the session. The given question says, solve the following differential equation and let's start with the solution. And the given equation is, y square minus x square into dy is equal to 3xy into dx. Or it can further be written as, on dividing both the sides by y square minus x square at dx. On the left hand side here, dy divided by dx and on the right hand side we have 3xy divided by y square minus x square. Now as we can see, this is the homogeneous function since the degree of the numerator is also 2 and the degree of the denominator is also 2. Therefore, quoting y is equal to vx here and first let us find its first derivative is equal to v plus x into dv divided by dx. So now substituting these two values in this equation, let this be equation number one. So one implies that on the left hand side we have v plus x into dv divided by dx is equal to 3 into x into vx whole divided by y square is v square x square minus x square. Or this is further equal to v plus x into dv divided by dx is equal to 3v divided by v square minus 1. Or this is further equal to x into dv divided by dx is equal to 3v divided by v square minus 1 minus v or before the half x into dv divided by dx is equal to in the denominator we have v square minus 1 and here we have 3v minus v cube plus v or before the half x into dv divided by dx is equal to 4v sorry 3v plus v gives 4v minus v cube divided by v square minus 1 or this is further equal to v square minus 1 divided by v cube minus 4v is equal to minus dx divided by x. Or before the half v square minus 1 divided by taking v common we are later with v square minus 4 which can be written as v square minus 2 square and this is equal to minus dx divided by x or we have v square minus 1 divided by v into v minus 2 into v plus 2 and this is equal to minus dx divided by x. Now let us solve this by partial fraction this can be written as v square minus 1 divided by v into v minus 2 into v plus 2 as a divided by v plus b divided by v minus 2 plus c divided by v plus 2 or this is further equal to a to v square minus 4 plus b into v square plus 2v plus c into v square minus 2v divided by this whole v into v minus 2 into v plus 2. So in comparing this we find here that v square minus 1 is equal to a plus b plus c v square we have plus 2b minus 2c into v then we have minus 4a and I am comparing the constants here we find that minus 1 is equal to minus 4a which implies that a is equal to 1 by 4. Now from here as we can see on the left hand side we do not have any coefficient of v therefore we have 2b minus 2c is equal to 0 so this implies that b is equal to c and here on comparing the coefficient of v square on both the sides we get a plus b plus c is equal to 1 or we have here a as 1 by 4 b and c are equal so we have 2 times of b is equal to 1 minus 1 by 4 which gives 3 by 4 this is further implies that b is equal to 3 by 8 and since b is equal to c so c is also equal to 3 by 8 therefore it can further be written as a divided by v and here a is 1 by 4 so we have 1 by 4 divided by v plus b divided by v minus 2 b is 3 by 8 so we have 3 by 8 divided by v minus 2 plus c divided by v plus 2 and c is again 3 by 8 divided by v plus 2 and this is equal to minus dx divided by x now integrating both the sides on the left hand side we have 1 by 4 integral 1 by v into dv plus 3 by 8 integral 1 upon v minus 2 into dv plus 3 by 8 integral dv divided by v plus 2 and this is equal to minus integral dx divided by x or this is further equal to 1 by 4 log b plus 3 by 8 log v minus 2 plus 3 by 8 log v plus 2 is equal to minus log x plus log c or this is further equal to 1 by 4 log v plus 3 by 8 taking common from these two terms we have log v minus 2 into log v plus 2 and this is equal to log v square minus 4 since log a plus log b is equal to log a into b and on the right hand side we have minus log x plus log c or this is further equal to 1 by 4 log v plus 3 by 8 log v square minus 4 is equal to log of c divided by x or this can further be written as 1 by 4 log now we have assumed that y is equal to vx and the beginning so v is equal to y divided by x so here we have y divided by x plus 3 divided by 8 log y square divided by x square minus 4 is equal to log c divided by x or further we have log y divided by x raise to the power 1 by 4 plus log y square divided by x square minus 4 raise to the power 3 by 8 is equal to log c divided by x or we have log y divided by x raise to the power 1 by 4 into y square divided by x square minus 4 raise to the power 3 by 8 is equal to log c divided by x or we have y divided by x raise to the power 1 by 4 into y square divided by x square minus 4 raise to the power 3 by 8 into x is equal equal to c. Hence, on solving the given differential equation, we get y divided by x raise to the power 1 by 4 into y square divided by x square minus 4 divided by 3 by 8 into x is equal to c. So, this completes the session by intake care.