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Let y = y(x) be the solution curve of the differential equation d y d x + ( 2 x 2 + 1 1 x + 1 3 x 3 + 6 x 2 + 1 1 x + 6 ) y = ( x + 3 ) x + 1 , x > − 1 ,  which passes through the point (0, 1). Then y(1) is equal to:

Option 1 -

1 2

Option 2 -

3 2

Option 3 -

5 2

Option 4 -

7 2

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

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago
    Correct Option - 2


    Detailed Solution:

    d y d x + ( 2 x 2 + 1 1 x + 1 3 x 3 + 6 x 2 + 1 1 x + 6 ) y = x + 3 x + 1 , x > − 1

    IF = e ∫ p d x = ( x + 1 ) 2 ( x + 2 ) x + 3

    ∫ P d x = ∫ 2 x 2 + 1 1 x + 1 3 x 3 + 6 x 2 + 1 1 x + 6 d n = ∫ ( 2 x + 1 + 1 x + 2 − 1 x + 3 ) d x

    = l n ( ( x + 1 ) 2 ⋅ ( x + 2 ) / ( x + 3 ) )

    2 x 2 + 1 1 x + 1 3 ( x + 1 ) ( x + 2 ) ( x + 3 ) = A x + 1 + B x + 2 + C x + 3

    2 x 2 + 1 1 x + 1 3 = A ( x + 2 ) ( x + 3 ) + B ( x + 1 ) ( x + 3 ) + C ( x + 1 ) ( x + 2 )

    x = -1

    ->4 = 2A Þ A = 2

    x = -2

    -> -1 = -B Þ B = 1

    x2 – 3 Þ -2 = 2c

    c = -1

    y ⋅ ( x + 1 ) 2 ( x + 2 ) x + 3 = ∫ x + 3 x + 1 ⋅ ( x + 1 ) 2 ( x + 2 ) x + 3 d x

    = ∫ ( x + 1 ) ( x + 2 ) d x

    = x 3 3 + 3 x 2 2 + 2 x + c

    ( 0 , 1 ) ⇒ 1 ⋅ 2 3 = c

    x = 1  y ⋅ ( 3 ) = 1 3 + 3 2 + 2 + 2 3 = 3 2 + 3 = 9 2

    y =  3 2

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V
Vishal Baghel

I = ∫ (e? (x²+1)/ (x+1)² dx = f (x)e? + c
I = ∫ (e? (x²-1+1+1)/ (x+1)² dx
I = ∫e? [ (x-1)/ (x+1) + 2/ (x+1)² ] dx
for x = 1
f' (1) = 12/24 - 12/16 = 3/4

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:
A
alok kumar singh

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              

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

C = 3 2 − 3 = − 3 2

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

= c o s ( 2 n − 1 ) x ( 2 n − 1 ) ] 0 π / 2 = 1 2 n − 1 [ 0 − 1 ] = − 1 2 n − 1

b n − 1 − b n = 1 2 n − 1

b 1 − b 2 = 1 3 ( A ) − 1 5 , . . . . . .

b 2 − b 3 = 1 3 ( B ) − 5 , − 7 , − 9 , . . . . .

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                

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

y 2 ( x ) = − ( x + 1 ) + 2 e x                

y2 > y1, no solution.

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