We have, (3y−7x+7)dx+(7y−3x+3)dy=0 ⇒(7x−3y−7)dx+(3x−7y−3)dy=0 Given equation is non-homogeneous and a1b2−a2b1=−40≠0 On solving 7x−3y−7=0 and 3x−7y−3=0 we get x=1,y=0 Now, substitude x=1+u and y=0+v=v ∴dx=du and dy=dv... in Eq. (i), we get (7u−3v)du+(3u−7v)dv=0‌‌... (i) This homogeneous equation in u and v. Put, u=tv⇒du=tdv+vdt in Eq. (ii), we get (7tv−3v)(tdv+vdt)+(3tv−7v)dv=0 ⇒‌‌(7t−3)(tdv+vdt)+(3t−7)dv=0 ⇒(7t2−7)dv+v(7t−3)dt=0⇒‌‌∫‌