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A differential equation representing the family of parabolas with axis parallel to y – axis and whose length of latus rectum is the distance of the point (2, -3) form the line 3x + 4y = 5, is given by:

Option 1 -

1 0 d 2 x d y 2 = 1 1

Option 2 -

1 1 d 2 y d x 2 = 1 0

Option 3 -

1 1 d 2 x d y 2 = 1 0

Option 4 -

1 0 d 2 y d x 2 = 1 1

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago
    Correct Option - 2


    Detailed Solution:

    L . R . = | 3 × 2 + 4 x − 3 − 5 | 3 2 + 4 2 = 1 1 5              

    Equation of family of parabolas

    ( x − h ) 2 = 1 1 5 ( y − k )              

    Differentiate 2 (x – h) = 1 1 5 d y d x  

    Again differentiate 2 = 1 1 5 d 2 y d x 2  

    1 1 d 2 y d x 2 = 1 0              

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If y = y(x) is the solution of the differential 2 x 2 d y d x − 2 x y + 3 y 2 = 0 such that y(e) = e 3 ,  then y(1) equal to
A
alok kumar singh

2 x 2 d y d x − 2 x y + 3 y 2 = 0

d y d x = y x − 3 2 ( y x ) 2

Put y = vx

  d y d x = v + x d v d x

P o i n t ( e , e 3 )

⇒ − e ( e 3 ) + 3 2 l n e = C

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3 2 l n | x | − x y + 3 2 = 0

x = 1 ⇒ y = 2 3              

If b n = ∫ 0 π 2 c o s 2 n x s i n x dx, n ∈ N ,  then
A
alok kumar singh

b n = ∫ 0 π / 2 c o s 2 n x s i n x d x

= ∫ 0 π / 2 1 + c o s 2 n x 2 s i n x d x

b n − b n − 1 = ∫ 0 π / 2 c o s ( 2 n x ) − c o s ( 2 ( n − 1 ) x ) 2 s i n x d x

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b 3 − b 4 = 1 7 ( C )

Let y = y1 (x) and y = y2 (x) be two distinct solutions of the differential equation  = x + y, with y1 (0) = 0 and y2 (0) = 1 respectively. Then, the number of points of intersection of y = y1 (x) and y = y2 (x) is
A
alok kumar singh

d y d x − y = x ,  linear differential equation.

IF = e-x

y . e − x = ∫ x e − x d x = − e − x ( x + 1 ) + C                

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If   d y d x + 2 x − y ( 2 y − 1 ) 2 x − 1 = 0 , x , y > 0 ,     y ( 1 ) = 1 , then y(2) is equal to:
A
alok kumar singh

d y d x + 2 x − y ( 2 y − 1 ) 2 x − 1 = 0

x, y > 0, y(1) = 1

d y d x = − 2 x ( 2 y − 1 ) 2 y ( 2 x − 1 )         

∫ 2 y 2 y − 1 d y = − ∫ 2 x 2 x − 1 d x           

= l o g e ( 2 y − 1 ) l o g e 2 = − l o g e ( 2 x − 1 ) l o g e 2 + l o g e c l o g e 2  

Taking log of base 2.

∴  y = 2 – log2 

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