x2y3+xydydx=1,y(1)=0,y=2 when x=

dxdy=xy+x2y3⇒−1x2dxdy+yx=−y3⇒ddy1x+1x⋅y=−y3I.F=exp⁡∫y⋅dy=ey2/2 Soln is ey2/2x=∫−y3ey2/2⋅dy⇒1x=2−y2+c.e−y2/2x=1,y=0⇒c=−1 ⇒x=−e