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Higher Order Derivatives

dydx =y tan 2...

Question

$\frac{d y}{d x} = y \tan 2 x, y (0) = 2$

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Solution

$We have, \frac{d y}{d x} = y \tan 2 x, y (0) = 2 \Rightarrow \frac{1}{y} d y = \tan 2 x d x Integrating both sides, we get \int \frac{1}{y} d y = \int \tan 2 x d x \Rightarrow \log |y| = \frac{1}{2} \log |\sec 2 x| + \frac{1}{2} \log C \Rightarrow y^{2} = C \sec 2 x . . . . . (1) It is given that at x = 0, y = 2 . ∴ C = 4 Substituting the value of C in (1), we get ∴ y^{2} = \frac{4}{\cos 2 x} \Rightarrow y = \frac{2}{\sqrt{\cos 2 x}} Hence, y = \frac{2}{\sqrt{\cos 2 x}} is the required solution .$

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Similar questions

Q. Solve each of the following initial value problems:
(i)

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}

Q. Solve each of the following initial value problems:
(i)

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 \tan x, y (0) = 1

2 x \frac{d y}{d x} = 5 y, y (1) = 1

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

\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}

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

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

x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1

Q. Solve each of the following initial value problems:
(i)

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

given that y = 0 when

x = \frac{π}{2}

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

2 x \frac{d y}{d x} = 5 y, y (1) = 1

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

\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}

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

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

x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1

\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}

when y = 0, x = 0 [NCERT EXEMPLAR]
(ix)

2 (y + 3) - x y \frac{d y}{d x} = 0

, y(1) = −2 [NCERT EXEMPLAR]

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