 Hello and welcome to this session. In this session we discuss the following question which says solve the differential equation x plus pi y minus 3 the whole dx is equal to 3x plus 4y minus 2 whole dy. If we are given the equation of a kind dy by dx equal to ax plus by plus c upon capital ax plus capital by plus capital c then such equations are reduced to homogeneous form by taking the small x as capital x plus h and small y as capital y plus k where these h and k are constants which are so chosen to make the given equation homogeneous. And we know that the differential equation of the form dy by dx equal to f of xy upon phi of xy where this f of xy and phi of xy are the homogeneous functions of x and y and are of same degree. So this is the key idea that we use for this question. Let's proceed with the solution now. The given differential equation is x plus 5y minus 3 the whole dx is equal to 3x plus 4y minus 2 the whole dy. Now this equation can be written as dy by dx is equal to x plus 5y minus 3 the whole upon 3x plus 4y minus 2. So this equation is of the form dy by dx is equal to ax plus dy plus c upon capital ax plus capital by plus capital c and now we will reduce this given equation to homogeneous form by taking small x equal to capital x plus h and small y equal to capital y plus k. So now putting small x equal to capital x plus h and small y equal to capital y plus k. So we now have b capital y upon b capital x is equal to capital x plus h that is in place of x we put capital x plus h plus 5 into capital y plus k the whole minus 3 this whole upon 3 into capital x plus h plus 4 into capital y plus k the whole minus 2. This gives us b capital y upon b capital x is equal to capital x plus h plus 5 capital y plus 5k minus 3 this whole upon 3 capital x plus 3h plus 4 capital y plus 4k minus 2. Or we can write this as b capital y upon b capital x is equal to capital x plus 5 capital y plus h plus 5k minus 3 this whole upon 3 capital x plus 4 capital y plus 3h plus 4k minus 2 now we need to find h and k and for this we take h plus 5k minus 3 is equal to 0 that is we take this equal to 0 and 3h plus 4k minus 2 is equal to 0 and this expression is equal to 0. Now we will solve these two equations to get the values of h and k. So from here we have h upon 5 into minus 2 is minus 10 minus of 4 into minus 3 that is minus of minus 12 which becomes plus 12 is equal to minus k upon minus 2 into 1 that is minus 2 into minus of 3 into minus 3 that is minus of minus 9 and that becomes plus 9 is equal to 1 upon 4 into 1 is 4 minus 3 into 5 is 15. So h upon 2 is equal to minus k upon 7 is equal to 1 upon minus 11 and from here we can say that h is equal to minus 3 into 5 is 15. So h upon minus 2 upon 11 and k is equal to 7 upon 11. Consider this equation to be equation number 1. Now from equation 1 we have d capital y upon d capital x is equal to capital x plus 5 capital y plus 0. So in the numerator we have capital x plus 5 capital y now we have taken this term to be 0 but in the same way in the denominator we have 3 capital x plus 4 capital y plus 0 at this term is also taken as 0. So here we have 3 capital x plus 4 capital y. Now this equation is a homogeneous equation. So we have reduced the given equation to homogeneous form. Now to solve this we take capital y equal to b into capital x. Now differentiating both sides with respect to capital x we have d capital y by d capital x is equal to b plus capital x into dv by d capital x. Now let us just be equation number 2. Now substituting these values of capital y and d capital x is equal to dv by d capital y by d capital x in equation 2 we get v plus capital x into dv by d capital x is equal to capital x plus 5 into point place of y we put v capital x upon 3 capital x plus 4 into capital y that is v capital x. So we now have v plus capital x into dv by d capital x is equal to capital x plus 5 v capital x upon 3 capital x plus 4 v capital x. Now further we have v plus capital x into dv by d capital x is equal to taking capital x common in the numerator capital x into 1 plus 5 v the whole and this whole upon taking capital x common from the denominator we have capital x into 3 plus 4 v the whole. Now this x x cancels and a further we have capital x into dv by d capital x is equal to 1 plus 5 v upon 3 plus 4 v minus v which gives us capital x into dv by d capital x is equal to 1 plus 5 v minus 3 v minus 4 v square this whole upon 3 plus 4 v then capital x into dv by d capital x is equal to minus of 4 v square minus 4 v square by d capital x is equal to minus 2 v minus 1 v whole and this whole upon 3 plus 4 v. Now separating the variables we have 3 plus 4 v upon 4 v square minus 2 v minus 1 the whole dv is equal to minus d capital x upon capital x. Next integrating both sides we have integral of 3 plus 4 v upon 4 v square minus 2 v minus 1 the whole dv is equal to minus integral of d capital x upon capital x. Now multiplying both sides we get integral of 6 plus 8 v upon 4 v square minus 2 v minus 1 the whole dv is equal to minus 2 integral d capital x upon capital x. Now further we have integral 6 plus 8 v can be written as 8 v minus 2 plus 8 the whole dv upon 4 v square minus 2 v minus 1 is equal to minus 2 integral d capital x upon capital x. And further we can write this as integral 8 v minus 2 upon 4 v square minus 2 v minus 1 the whole dv plus 8 integral dv upon 4 v square minus 2 v minus 1 is equal to minus 2 integral d capital x upon capital x. Now for solving this integral we take let 4 v square minus 2 v minus 1 be equal to t. Now differentiating both sides with respect to t we have 8 v minus 2 the whole dv is equal to dt. So we further have integral dt by t that is in place of 8 v minus 2 dv we write dt. So dt by t plus 8 upon 4 integral dv upon now we have taken 4 common from the denominator. So in the denominator we now have v square minus 1 upon 2 v minus 1 upon 4 this is equal to minus 2 integral v capital x upon capital x. So this gives us log of modulus t plus 2 into integral dv upon now this denominator that is v square minus 1 upon 2 v minus 1 upon 4 can be written as v square minus 1 upon 2 v plus 1 upon 16 minus 1 upon 16 that is you have added and subtracting 1 upon 16 minus 1 upon 4. So this is further equal to v square minus 1 upon 2 v plus 1 upon 16 minus 5 upon 16 and so we can write this as v minus 1 upon 4 the whole square minus 5 upon 16. So here we write in the denominator v minus 1 upon 4 whole square minus 5 upon 16 and so this is equal to minus 2 log of modulus capital x plus the constant of integration c. Now in place of t where we put this value we have log modulus of 4 v square minus 2 v minus 1 plus 2 now integral of dv upon v minus 1 upon 4 whole square minus 5 upon 16 would be equal to integral of dv upon v minus 1 upon 4 whole square minus root 5 upon 4 whole square. Now the value of this integral is equal to 1 upon 2 into a which is root 5 upon 4 into log modulus of v minus 1 upon 4 minus root 5 upon 4 this one upon v minus 1 upon 4 plus root 5 upon 4. Here we have used the formula of integration integral dx upon x square minus a square is equal to 1 upon 2 a log of modulus x minus a upon x plus so that is here we have 2 into 1 upon 2 root 5 upon 4 so we can write 4 here into log modulus v minus 1 upon 4 minus root 5 upon 4 this one upon v minus 1 upon 4 plus root 5 upon 4 and this is equal to minus 2 log modulus capital x plus the constant of integration c. So this gives us log modulus 4 v square minus 2 v minus 1 now this 2 and 2 cancels and so we have plus 4 upon root 5 into log modulus 4 v minus root 5 minus 1 upon 4 v minus root 5 minus 1 upon 4 v plus root 5 minus 1 and this is equal to minus 2 log of modulus x plus the constant of integration c. We have taken capital y is equal to v into capital x this means that v is equal to capital y by capital x. So now putting v equal to capital y by capital x we get log modulus 4 into capital y square upon capital x square minus 2 into capital y upon capital x minus 1 plus 4 upon root 5 into log modulus 4 into capital y upon capital x minus root 5 minus 1 this whole upon 4 into capital x upon capital x plus root 5 minus 1 is equal to minus 2 log modulus capital x plus the constant of integration c. So this gives us log modulus 4 capital y square minus 2 into capital x into capital y minus capital x square this whole upon capital x square plus 4 upon root 5 into log modulus 4 into capital y minus root 5 into capital x minus capital x minus capital x this whole upon 4 into capital y plus root 5 into capital x minus 1 and this is equal to minus 2 log modulus capital x plus the constant of integration c. This further gives us log modulus 4 capital y square minus 2 capital x capital y minus capital x square this whole upon capital x square plus 4 upon root 5 log modulus 4 into capital y minus 1 plus root 5 into capital x upon 4 into capital y minus 1 minus 2 log modulus 4 into capital y minus root 5 the whole into capital x is equal to minus 2 log modulus capital x plus the constant of integration c. Now earlier we had got the value of h as minus 2 by 11 and value of k as 7 upon 11. So this means the small x or you can say capital x is written as small x minus the value of h. So minus minus becomes plus 2 upon 11 in the same way from here we have capital y is equal to y minus the value of capital k and so as k is 7 upon 11. So here we have y minus 7 upon 11. This is the minus 2 log modulus required solution when we have capital x equal to small x plus 2 upon 11 and capital y is equal to small y minus 7 upon 11. So this is our final answer. This completes the session but we have understood the solution of this question.