2xydx+(x2−y2)dy=0⇒2xydx=(y2−x2)dy⇒dxdy=y2−x22xy ...(i)
Put y=vx
dxdy=v+xdxdv
Using these in (i).
v+xdxdv=v2x2−x22vx2⇒xdxdv=v2−12v−v
⇒xdxdv=1−v22v−v3+v=1−v23v−v3⇒xdxdv=1−v2v(3−v2)⇒v(3−v2)(1−v2)dv=xdx
⇒v(3−v2)1dv−v(3−v2)v2dv=xdx
Using partial fractions,
[3v1−6(v+3)1−6(v−3)1]dv−3−v2vdv=xdx⇒[3v1−6(v+3)1−6(v−3)1]dv+21.3−v2−2vdv=xdx
On integrating,
31ln∣v∣−61ln∣∣v+3∣∣−61ln∣∣v−3∣∣−(−21ln∣∣3−v2∣∣)=ln∣x∣+lnC
⇒31ln∣∣xy∣∣−61ln∣∣xy+3∣∣−61ln∣∣xy−3∣∣+21ln∣∣3−x2y2∣∣=ln∣Cx∣⇒ln∣xy∣1/3−ln∣xy+3∣1/6−ln∣xy−3∣1/6+ln∣3−x2y2∣1/2=ln∣Cx∣⇒ln∣(xy+3)1/6(xy−3)1/6(xy)1/3(3−x2y2)1/2∣=ln∣Cx∣
⇒ln∣(3−x2y2)1/6(xy)1/3(3−x2y2)1/2∣=ln∣Cx∣⇒ln∣(xy)1/3(3−x2y2)1/3∣=ln∣Cx∣⇒(xy)1/3(3−x2y2)1/3=Cx⇒(xy)(3−x2y2)=Kx3 [∵K=C3]⇒3yx2−y3=Kx6
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