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The differential equation of the family of curves y2=4a(x+a) is:

(A) y24dydx=(x+dydx)

(B) 2ydydx=4a

(C) yd2ydx2+dydx2=0

(D) 2xdydx+y(dydx)2−y

0 2 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

    This is an Objective Type Questions as classified in NCERT Exemplar

    Sol:

    The  given  equation  of  family  of  curves  is                                  y2=4a(x+a)⇒                             y2=4ax+4a2                                          …(i)Differentiating  both  sides,  w.r.t.x,  we  get                           2y.dydx=4a⇒                         y.dydx=2a          ⇒y2.dydx=aNow,  putting  the  value  of  a  in  eqn.(i)  we  get                    y2=4x(y2.dydx)+4(y2.dydx)2⇒               y2=2xydydx+y2(dydx)2⇒                 y=2xdydx+y(dydx)2⇒2x.dydx+y.(dydx)2−y=0Hence,  the  correct  option  is  (d).

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