a.) solve:dy/dx+(2x+(tan^-1y)-(x^3))(1+y²)=0 b.) A particle of unit mass moves in a straight line in a resisting medium which produces resistance kV.If the particle starts with a velocity u from the position S=Sa,show that , as time goes on , the particle approaches the position S=Sa+(u/k).

ans a)  dy/dx  + [2x + (tan^-1y)- (x³) ][1+y²]

 now divide the equation by 1+y²

        dy/dx(1/1+y²)  +  [2x + tan^-1y - x³ ]       ................1

now put tan^-1y=t

 differentiate wrt x

          dy/dx(1/1+y²) =dt/dt .............2

now substitute 2 in 1

so

       dt/dx + t +2x-x³ = 0

this eq is linear differential equation and its integral factor is e^x ...

now u can solve this easily...

retardation of particle = kv

                  a=kv ..............1

       a=dv/dt   &   v=dx/dt

 putting these eq 1 becomes

                dv/dt = kds/dt

                dv =k ds

 now integrate both sides by taking proper limits

            u-0= k(s-s_a)

             s-s_a =u/k       or

             s= s_a+u/k