≤ft( dy/dx ) x = y ² x + x, find general solution

Correct Answer is : y=tan⁡x(log⁡∣csc⁡x−cot⁡x∣+cos⁡x+c)y = x ≤ft( | x - x| + x + c )y=tanx(log∣cscx−cotx∣+cosx+c)

y = x ≤ft( | x - x| + x + c )

Correct Answer is : y=tan⁡x(log⁡∣csc⁡x−cot⁡x∣+cos⁡x+c)y = x ≤ft( | x - x| + x + c )y=tanx(log∣cscx−cotx∣+cosx+c)

Test Index

BITSAT 2022 Paper

Section: Mathematics

Question : 102 of 130

Marks: +1, -0

\left( \frac{dy}{dx} \right) \tan x = y \sec^2 x + \sin x

Solution: 👈: Video Solution

Given that

\left( \frac{dy}{dx} \right) \tan x = y \sec^2 x + \sin x

\tan x \left( \frac{dy}{dx} \right) - y \sec^2 x = \sin x

\frac{dy}{dx} + \frac{-\sec^2 x}{\tan x} y = \frac{\sin x}{\tan x}

Here,

P = \frac{\sec^2 x}{\tan x} \text{ and } Q = \cos x

IF = e^{\int P \, dx} = e^{\int \frac{-\sec^2 x}{\tan x} \, dx}

IF = e^{-\log(\tan x)} = e^{\log \cot x} = \cot x

Now,

y \times IF = \int (Q \times IF) \, dx + C

y \cot x = \int \cos x \cot x \, dx + C

y \cot x = \int \frac{1 - \sin^2 x}{\sin x} \, dx + C

y \cot x = \log |\csc x - \cot x| + \cos x + C

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