Solve `(dy)/(dx)=(x+2y+3)/(2x+3y+4)`

Explanation

The given differential equation is not homonogenous. So, we use the following solution technique.
Put x=X+h, y=Y+k
We have `(dY)/(dx) = (X+2Y+(h+2k+3))/(2X+3Y+(2h+3k+4))`
To determine h and k, we wrtie.
`h+2k+3=0, 2h+3k+4=0`
`rArr h=1, k=-2`
so that `(dY)/(dx) = (X+2y)/(2X+3Y)`
Putting `Y=VX`, we get
`V+X(dV)/(dX) = (1+2V)/(2+3V)`
or `(2+sqrt(3))/(2sqrt(3))log(sqrt(3V)-1)-(2-sqrt(3))/(2sqrt(3))log(sqrt(3)Y+X)=A`.
where A is another constant and `X=x-1, Y=y+2`