We have given the differential equation,
(2x−3y−1)dx+(x−3y−5)dy=0
dxdy+(x−3y−5)(2x−3y−1)=0
Putting x=X+h and y=Y+k
→2h−3k−1=0and h−3k−5=0
Hence, dXdY+(X+3Y)(2X−3Y)=0
Putting Y=vX
v+XdXdv+1−3v2−3v=0
XdXdv=(1−3v)(3v−2)−v
=(1−3v)(2v−2−3v2)
∫(2v−2−3v21−3v)dv+∫XdX=logC
21log(2v−2−3v2)+logX=logc
Hence, this is the final solution of the given differential equation.
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