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Fill in the blanks for the following (I to xi)

(i) The degree of the differential equation d2ydx2+edydx+0  is _________.

(ii) The degree of the differential equation  is _________.

(iii) The number of arbitrary constants in the general solution of a differential equation of order three is _________.

(iv) dydx+yx logx=1x is an equation is of the type _________.

(v) The general solution of a differential equation of the type dydx+P1x=Q1 is given by _________.

(vi) The solution of xdydx+2y=x2  is _________.

(vii) The solution of (1+x2)dydx+2xy−4x2=0  is _________.

(viii) The solution of ydx+(x+xy)dy=0  is _________.

(ix) The general solution of dydx+y=sinx  is _________.

(x) The solution of cotydx=xdy  is _________.

(xi) The integrating factor of dydx+y=1+yx=0  is _________.

 

0 4 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

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    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

    This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

    ( v i i )     T h e     g i v e n     d i f f e r e n t i a l     e q u a t i o n     i s     ( 1 + x 2 ) d y d x + 2 x y − 4 x 2 = 0 ⇒                                         d y d x + 2 x y 1 + x 2 = 4 x 2 1 + x 2 Since,  it  is  a  linear  differential  equation  where  P=2x1+x2  and  Q=4x21+x2                                                 I . F . = e ∫ P d x = e ∫ 2 x 1 + x 2 d x = e l o g ( 1 + x 2 ) = 1 + x 2 ∴       S o l u t i o n     i s       y × I . F . = ∫ Q × I . F . d x + c ⇒                                             y . ( 1 + x 2 ) = ∫ 4 x 2 1 + x 2 × ( 1 + x 2 ) d x + c ⇒                                               y . ( 1 + x 2 ) = ∫ 4 x 2 d x + c                 ⇒ y . ( 1 + x 2 ) = 4 3 x 3 + c ⇒                                                                                   y = 4 3 x 3 ( 1 + x 2 ) + c ( 1 + x 2 ) − 1 H e n c e ,     t h e     s o l u t i o n     i s     y = 4 3 x 3 ( 1 + x 2 ) + c ( 1 + x 2 ) − 1 ( v i i i )     T h e     g i v e n     d i f f e r e n t i a l     e q u a t i o n     i s     y d x + ( x + x y ) d y = 0 ⇒         ( x + x y ) d y = − y d x       ⇒ x ( 1 + y ) d y = − y d x ⇒                     1 + y y d y = − 1 x d x I n t e g r a t i n g     b o t h     s i d e s ,     w e     g e t                       ∫ 1 + y y d y = − ∫ 1 x d x ⇒ ∫ ( 1 y + 1 ) d y = − ∫ 1 x d x ⇒                   l o g y + y = − l o g x + l o g c ⇒                   l o g y + l o g x + l o g e y = l o g c ⇒                                                           l o g ( x y . e y ) = l o g c ⇒                   x y = c . e − y H e n c e ,     t h e     s o l u t i o n     i s     x y = c . e − y

     

     

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