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a
y2+1+cey2=1
b
x2+1+cex2y2=1
c
y2+1+cey2=x
d
x+1+cey2x2=1
answer is B.
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Detailed Solution
G.E dydx+xy=x3y3⇒1y3dydx+xy2=x3 Put 1y2=z⇒dzdx−2xz=−2x3;I.F=e−x2 General solution ze−x2∫−2x3e−x2dx Put x2=t⇒ze−x2=−∫te−tdt ⇒1y2e−x2=x2e−x2+e−x2+c⇒x2+1+cex2y2=1