Let y = y(x) be the solution of the differential equation sec xdy + {2(1 – x) tan x + x(2 – x)} dx = 0 such that y(0) = 2. Then y(2) is equal to [2024]

Question

Let y = y(x) be the solution of the differential equation sec xdy + {2(1 – x) tan x + x(2 – x)} dx = 0 such that y(0) = 2. Then y(2) is equal to [2024]

Smart Achievers · Accepted Answer

(4)

sec xdy + {2(1 – x) tan x + x(2 – x)} dx = 0

$\Rightarrow \frac{d y}{d x} = \frac{- 2 (1 - x) \tan x - x (2 - x)}{\sec x}$

$\Rightarrow \frac{d y}{d x} = - 2 (1 - x) \sin x - x \cos x (2 - x)$

$\Rightarrow \int d y = - 2 \int (1 - x) \sin x d x - \int \cos x (2 x - x^{2}) d x$

$\Rightarrow y = - 2 \int (1 - x) \sin x d x - (2 x - x^{2}) \sin x + \int (2 - 2 x) \sin x d x + c$

$\Rightarrow y = - (2 x - x^{2}) \sin x + c$

Now, $y (0) = 2 \Rightarrow 2 = c$

$\Rightarrow y = - (2 x - x^{2}) \sin x + 2 \Rightarrow y (2) = 2$

Solution

Q.
Let y = y(x) be the solution of the differential equation sec xdy + {2(1 – x) tan x + x(2 – x)} dx = 0 such that y(0) = 2. Then y(2) is equal to [2024]

1 2{1 – sin (2)}

2 1

3 2{sin (2) + 1}

4 2

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Solution

Q. Let y = y(x) be the solution of the differential equation sec xdy + {2(1 – x) tan x + x(2 – x)} dx = 0 such that y(0) = 2. Then y(2) is equal to [2024]

1 2{1 – sin (2)}

2 1

3 2{sin (2) + 1}

4 2

Want Us to Email you About Special Offers & Updates?

Q.
Let y = y(x) be the solution of the differential equation sec xdy + {2(1 – x) tan x + x(2 – x)} dx = 0 such that y(0) = 2. Then y(2) is equal to [2024]