1) The solution of the differential equation ydx−xdy+3x2y2ex3dx=0 satisfying y=1 when x=1, is A) y(ex3−(1+2e))−x=0 B) y(ex3+(1−e))+x=0 C) y(ex3+(1+e))−x=0 D) y(ex3−(1+e))+x=0 Answer: Option DExplanation: We have, ydx−xdy+3x2y2ex3dx=0 ydx−xdyy2+3x2ex3dx=0 ⇒ d(xy)+d(ex3)=0 On integrating , we get xy+ex3=c putting x=y=1, we get c=1+e ∴ xy+ex3 =1+e x+y(ex3−(1+e))+x=0