If 2 - sin x d y d x - y - 1 cos x -0 and y 0 - 1- then y - 2 is equal to -

We have, [ ( 2+sin #160; x ) dy dx +ycos ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Particular Solution of a Differential Equation

If 2 + sin...

Question

If $(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0$ and $y (0) = 1,$ then $y (\frac{π}{2})$ is equal to :

Open in App

Solution

Suggest Corrections

Similar questions

(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0

y (0) = 1

y = (\frac{π}{2})

(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0 and y (0) = 1,

y (\frac{π}{2})

y' + y = e^{x}, y (0) = \frac{1}{2}

x \frac{d y}{d x} - y = \log x, y (1) = 0

\frac{d y}{d x} + 2 y = e^{- 2 x} \sin x, y (0) = 0

x \frac{d y}{d x} - y = (x + 1) e^{- x}, y (1) = 0

(1 + y^{2}) d x + (x - e^{- \tan^{- 1} y}) d x = 0, y (0) = 0

\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x, y (0) = 1

x \frac{d y}{d x} + y = x \cos x + \sin x, y (\frac{π}{2}) = 1

\frac{d y}{d x} + y \cot x = 4 x cosec x, y (\frac{π}{2}) = 0

\frac{d y}{d x} + 2 y \tan x = \sin x; y = 0 when x = \frac{π}{3}

\frac{d y}{d x} - 3 y \cot x = \sin 2 x; y = 2 when x = \frac{π}{2}

\frac{d y}{d x} + y \cot x = 2 \cos x, y (\frac{π}{2}) = 0

d y = \cos x (2 - y \cos e c x) d x

\tan x (\frac{d y}{d x}) = 2 x \tan x + x^{2} - y; \tan x \neq 0

x = \frac{π}{2}

\frac{d y}{d x} = y + 3 > 0

y (0) = 2

y (ln 2)

Join BYJU'S Learning Program