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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).

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).

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2 Answers

vikas askiitian expert
509 Points
12 years ago

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 ex ...

now u can solve this easily...

vikas askiitian expert
509 Points
12 years ago

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-sa)

             s-sa =u/k       or

             s= sa+u/k

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