(i) Consider the differential equation of the form `(dy)/(dx) =(ax+by+c)/(a'x +b'y +c')` where `a/(a') ne b/(b')`
This equation is not homogeneous. In order to reduced this equation to the homogeneous form substitute `x= X + h, y = Y + k`, where h and `k`, are constants which are to be determined.
`=> (dy)/(dx) =(dY)/(dX)`
and the above equation becomes
`(dY)/(dX) =(a (X+h) +b(Y+k)+c)/(a' (X+h) +b'(Y+k)+c')`................(i)
`(dY)/(dX)= ((aX+bY)+ah+bk+c)/((a'X+b'Y)+ a'h+b'k+c')`...................(ii)
Now, `h` and `k` will be chosen such that
`ah+bk+c=0`
`a'h +b'k +c'=0`
`h/(bc' b'c)= k/(ca' - c'a)`
`= 1/(ab' -a'b)`
For these values of `h` and `k` in Eq. (ii) reduces to
`(dY)/(dX) =(aX+bY)/(a'X +b'Y)`
which is a homogeneous equation in `X, Y` and can be solved by the substitution `Y = vX`.
`(dY)/(dX) = v+ X * (dV)/(dX)`
Replacing, `X` and `Y` in the solution so obtained by `x - h` and `y - k`, respectively, we can obtain the required solution in term of `x` and `y`.
(ii) If `(dy)/(dx) =(ax+by+c)/(a'x+b'y + c')` and `a/(a') = b/(b') =m`
Then, `(dy)/(dx) =(m(a'x +b'y)+c)/(a'x +b'y +c')`
where, `m` is any number.
In such case substitute `a' x + b' y = v`
So, that `a' +b' (dy)/(dx) =(dv)/(dx)`
Transform the differential equation of the form
`1/(b') ((dv)/(dx) -a') =(mv+c)/(v+c')` we get
`(dv)/(dx) = a' +b' ((mv+c)/(v+c'))`
which is a differential equation in variable separable form and it can easily be solved.
(i) Consider the differential equation of the form `(dy)/(dx) =(ax+by+c)/(a'x +b'y +c')` where `a/(a') ne b/(b')`
This equation is not homogeneous. In order to reduced this equation to the homogeneous form substitute `x= X + h, y = Y + k`, where h and `k`, are constants which are to be determined.
`=> (dy)/(dx) =(dY)/(dX)`
and the above equation becomes
`(dY)/(dX) =(a (X+h) +b(Y+k)+c)/(a' (X+h) +b'(Y+k)+c')`................(i)
`(dY)/(dX)= ((aX+bY)+ah+bk+c)/((a'X+b'Y)+ a'h+b'k+c')`...................(ii)
Now, `h` and `k` will be chosen such that
`ah+bk+c=0`
`a'h +b'k +c'=0`
`h/(bc' b'c)= k/(ca' - c'a)`
`= 1/(ab' -a'b)`
For these values of `h` and `k` in Eq. (ii) reduces to
`(dY)/(dX) =(aX+bY)/(a'X +b'Y)`
which is a homogeneous equation in `X, Y` and can be solved by the substitution `Y = vX`.
`(dY)/(dX) = v+ X * (dV)/(dX)`
Replacing, `X` and `Y` in the solution so obtained by `x - h` and `y - k`, respectively, we can obtain the required solution in term of `x` and `y`.
(ii) If `(dy)/(dx) =(ax+by+c)/(a'x+b'y + c')` and `a/(a') = b/(b') =m`
Then, `(dy)/(dx) =(m(a'x +b'y)+c)/(a'x +b'y +c')`
where, `m` is any number.
In such case substitute `a' x + b' y = v`
So, that `a' +b' (dy)/(dx) =(dv)/(dx)`
Transform the differential equation of the form
`1/(b') ((dv)/(dx) -a') =(mv+c)/(v+c')` we get
`(dv)/(dx) = a' +b' ((mv+c)/(v+c'))`
which is a differential equation in variable separable form and it can easily be solved.