 Hello and welcome to the session. In this session we discussed the following question which says, solve the following differential equation dy by dx equal to x into 2y minus x upon x into 2y plus x if y is equal to 1 when x is equal to 1. Let's move on to the solution. The given differential equation is dy by dx equal to x into 2y minus x upon x into 2y plus x. Now this given differential equation can be written in the form dy by dx equal to 2 into y upon x minus 1 upon 2 into y upon x plus 1. We got this by dividing the numerator and denominator of this by x square. So this shows that the given differential equation is a homogeneous differential equation. So we say this is a homogeneous differential equation. Now to solve the homogeneous differential equation we put y equal to vx. Let's consider this as equation 1 and this as equation 2. Now in the next step we differentiate 2 with respect to x. So we get dy by dx is equal to v plus x into dv by dx. Now we substitute this value of dy by dx that is this and y equal to vx in equation 1 we get v plus x into dv by dx is equal to x into 2vx minus x upon x into 2vx plus x. Now this x and this x gets cancelled. So we get v plus x into dv by dx is equal to 2vx minus x upon 2vx plus x. Then again in the next step we have v plus x into dv by dx is equal to x into 2v minus 1 upon x into 2v plus 1. Now this x and this x gets cancelled and so we have v plus x into dv by dx equal to 2v minus 1 over 2v plus 1. Then in the next step we transpose this v to the right hand side. So we get x into dv by dx is equal to 2v minus 1 over 2v plus 1 minus v. Then next we have x into dv by dx is equal to 2v plus 1 in the denominator 2v minus 1 in the numerator minus v into 2v plus 1. So we get x into dv by dx is equal to 2v minus 1 minus 2v square minus v upon 2v plus 1. That is we get x into dv by dx is equal to minus 2v square plus v minus 1 over 2v plus 1. Now we separate the variables in this equation. So we get 2v plus 1 over minus 2v square plus v minus 1 dv equal to dx upon x. Or we could also write this as 2v plus 1 over 2v square minus v plus 1 dv is equal to minus dx over x. Then in the next step we need to integrate both the sides. So we get integral 2v plus 1 over 2v square minus v plus 1 dv is equal to integral minus dx over x. Now let i be equal to integral 2v plus 1 over 2v square minus v plus 1 dv. First let's find out the value for this i and let's name this equation as 3. Now we have this i be equal to 1 upon 2 integral 4v plus 2 over 2v square minus v plus 1 dv. Here we have multiplied the numerator and denominator by 2. So we get this. Then in the next step we have this would be equal to 1 upon 2 integral 4v minus 1 plus 3 that is we have taken 2 as minus 1 plus 3 upon 2v square minus v plus 1 dv. Now this is equal to 1 upon 2 integral 4v minus 1 upon 2v square minus v plus 1 dv plus 1 upon 2 integral 3 dv upon 2v square minus v plus 1. We take this term as i1. So this is equal to i1 plus this term as i2. So we have i is equal to i1 plus i2. First let's find out the value for i1 that is i1 equal to 1 upon 2 integral 4v minus 1 upon 2v square minus v plus 1 dv. We take let 2v square minus v plus 1 v equal to t on differentiating both the sides with respect to t we get 4v minus 1 dv is equal to dt. So this gives us i1 is equal to 1 upon 2 integral dt by t that is we have 1 upon 2 log modulus t plus c1. This is the value for i1 on substituting the value for t that is 2v square minus v plus 1. We get i1 is equal to 1 upon 2 log modulus 2v square minus v plus 1 plus c1 where the c1 is the constant of integration. Next we consider i2 which is equal to 1 upon 2 integral 3 dv by 2v square minus v plus 1 that is this is equal to 3 upon 2 integral dv by 2v square minus v plus 1. This is equal to 3 upon 4 integral dv by v square minus v upon 2 plus 1 upon 2. This is equal to 3 upon 4 integral dv by v minus 1 upon 4 the whole square plus root 7 by 4 the whole square. We know that integral dx over x square plus a square is equal to 1 upon a tan inverse x upon a plus c. So using this here we get this is equal to 3 upon 4 into 1 upon a and a in this case is root 7 upon 4. So it would be 3 upon 4 into 4 upon root 7 tan inverse v minus 1 upon 4 over root 7 upon 4 plus c2. That is we get this is equal to 3 upon root 7 tan inverse 4v minus 1 over root 7 plus c2. This is the value for i2 where we have the c2 is the constant of integration. Now we have i is equal to i1 plus i2. So we substitute the values for i, i1 and i2. So we get integral 2v plus 1 over 2v square minus v plus 1 dv is equal to 1 upon 2 log modulus 2v square minus v plus 1 plus c1 plus 3 upon root 7 tan inverse 4v minus 1 upon root 7 plus c2. We take c1 plus c2 v equal to c dash is the constant of integration. That is we get integral 2v plus 1 over 2v square minus v plus 1 dv is equal to 1 upon 2 log modulus 2v square minus v plus 1 plus 3 upon root 7 tan inverse 4v minus 1 over root 7 plus c dash. Now from 3 we get this that is 1 upon 2 log modulus 2v square minus v plus 1 plus 3 upon root 7 tan inverse 4v minus 1 upon root 7 plus c dash is equal to minus integral dx by x. This implies 1 upon 2 log modulus 2v square minus v plus 1 plus 3 upon root 7 tan inverse 4v minus 1 over root 7 plus c dash is equal to minus log modulus x plus log c double dash. This gives us 1 upon 2 log modulus 2v square minus v plus 1 plus 3 upon root 7 tan inverse 4v minus 1 over root 7 is equal to minus log modulus x plus log c where we have this log c is equal to log c double dash minus c dash. Now we substitute y is equal to vx that is v is equal to y upon x. So we get 1 upon 2 log modulus 2v square over x square minus y over x plus 1 plus 3 upon root 7 tan inverse 4v minus x upon x into root 7 is equal to minus log modulus x plus log c. It's given in the question that when x is equal to 1 we get y equal to 1. So using this we get the value for log c that is we have when x is equal to 1 we have y is equal to 1. On putting x equal to 1 and y equal to 1 in this we get 1 upon 2 log modulus 2 minus 1 plus 1 plus 3 upon root 7 tan inverse 4 minus 1 upon root 7 equal to minus log modulus 1 plus log c. So this implies we get 1 upon 2 log modulus 2 plus 3 upon root 7 tan inverse 3 upon root 7 is equal to log c. Now let us mark this equation as equation 4. Now putting this value of log c in equation 4 that is substituting this value of log c in equation 4. We get 1 upon 2 log modulus 2y square upon x square minus y upon x plus 1 plus 3 upon root 7 tan inverse 4v minus x upon x root 7 is equal to minus log modulus x plus 1 upon 2 log modulus 2 plus 3 upon root 7 tan inverse 3 upon root 7. That is we have 1 upon 2 log modulus 2y square minus xy plus x square upon x square plus 3 upon root 7 tan inverse 4y minus x upon x root 7 is equal to minus log modulus x plus 1 upon 2 log modulus 2 plus 3 upon root 7 tan inverse 3 upon root 7. So this is the required solution of the given differential equation. Hence our final answer is 1 upon 2 log modulus 2y square minus xy plus x square upon x square plus 3 upon root 7 tan inverse 4y minus x upon x root 7 is equal to minus log modulus x plus 1 upon 2 log modulus 2 plus 3 upon root 7 tan inverse 3 upon root 7. So this completes the session. Hope you have understood the solution for this question.