MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade:
        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).
9 years ago

Answers : (2)

vikas askiitian expert
509 Points
							

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...

9 years ago
vikas askiitian expert
509 Points
							

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

9 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 51 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details