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If ∫ 2 e x + 3 e − x 4 e x + 7 e − x d x = 1 1 4 ( u x + v l o g e ( 4 e x + 7 e − x ) ) + C where C is a constant of integration, then u + v is equal to_____.

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    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago

    Let l = ∫ 2 e x + 3 e − x 4 e x + 7 e − x d x = ∫ 2 e 2 x + 3 4 e 2 x + 7 d x  

    = ∫ 2 e 2 x 4 e 2 x + 7 d x + ∫ 3 e − 2 x 4 + 7 e − 2 x d x               

    Put 4 e 2 x + 7 = t 4 + 7 e − 2 x = λ  

    8 e 2 x d x = d t − 1 4 e − 2 x d x = d λ              

    = 1 4 ∫ d t t − 3 1 4 ∫ d λ λ = 1 4 l n     t − 3 1 4 l n   λ + c

    u = 1 3 2 , v = 1 2              

    ->so, u + v = 7

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If y = y(x) is the solution of the differential equation, dy/dx + 2ytanx = sinx, y(π/3) = 0, then the maximum value of the function y(x) over R is equal to :
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Raj Pandey

dy/dx + 2y tan (x) = sin (x)
I.F. = e^ (∫2tan (x)dx) = e^ (2ln (sec (x) = sec² (x)
Solution is y sec² (x) = ∫sin (x)sec² (x)dx = ∫sec (x)tan (x)dx
⇒ y sec² (x) = sec (x) + c
y (π/3) = 0 ⇒ 0 * sec² (π/3) = sec (π/3) + c ⇒ 0 = 2 + c ⇒ c = -2.
∴ y = (sec (x) - 2) / sec² (x)
Now let g (t) = (t - 2)/t² = 1/t - 2/t² for |t| ≥ 1.
g' (t) = -1/t² + 4/t³
g' (t) = 0 ⇒ t = 4.
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If the solution curve of the differential equation  ( 2 x − 1 0 y 3 ) d y + y d x = 0 , passes through the points (0, 1) and (2, b), then b is a root of the equation:
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alok kumar singh

( 2 x − 1 0 y 3 ) d y + y d x = 0              

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

d x d y + 2 y x = 1 0 y 2 → Linear differential equation

P = 2 y , Q = 1 0 y 2              

N o w     p u t     x = 2 , y = β     t h e n     2 β 2 = 2 β 5 − 2              

or β 5 − β 2 − 1 = 0  

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Let y(x) be the solution of the differential equation 2x2dy + (ey -2x)dx = 0, x> 0.

If y(e) = 1, then y(1) is equal to:
A
alok kumar singh

d y d x + e y − 2 x 2 x 2 = 0  

e − y d y d x − e − y x = − − 1 2 x 2              

Put e − y = t ⇒ − e − y d y d x = d t d x  

d t d x + t x = 1 2 x 2              

d t d x + t x = 1 2 x 2

 I. F. = e ∫ d x x = e l n x = x  

Soln. tx = ∫ x 2 x 2 d x = 1 2 l n x + c  

x = 1 ⇒ e − y = 1 2 ⇒ y = l n 2              

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