The solution of ex-e-x dy-dx-ex-e-x is-

The correct option is B y =log | e^x+e^ x |+c dydx=e^x e^ xe^x+e^ x let v=e^x+e^ x ⇒ dv=(e^x e^ x)dx ⇒∫ dy =∫1vdv+c ⇒ y=log v+c ...

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Byju's Answer

Standard XII

Mathematics

Existence of Limit

The solution ...

Question

The solution of $(e^{x} + e^{- x}) \frac{d y}{d x} = e^{x} - e^{- x}$ is:

y = log | e^{x} - e^{- x} | + c

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y = sinh x + c

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y = log | e^{x} + e^{- x} | + c

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y = sec x + c

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Solution

The correct option is B $y = log | e^{x} + e^{- x} | + c$
$\frac{d y}{d x} = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$

let $v = e^{x} + e^{- x}$

$\Rightarrow d v = (e^{x} - e^{- x}) d x$
$\Rightarrow \int d y = \int \frac{1}{v} d v + c$

$\Rightarrow y = log v + c$

$y = log (e^{x} + e^{- x}) + c$

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Similar questions

Q. The general solution of the differential equation

\frac{d y}{d x} = e^{x + y}

, is
(a) e^x+ e^−y = C
(b) e^x + e^y = C
(c) e⁻^x + e^y = C
(d) e^−x + e^−y = C

The general solution of the differential equation $\frac{d y}{d x} = e^{x + y}$ is
(a) $e^{x} + e^{- y} = C$
(b) $e^{x} + e^{y} = C$
(c) $e^{- x} + e^{y} = C$
(d) $e^{- x} + e^{- y} = C$

Q. Find the area enclosed the curves :

y = e x log x

and

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where

log e = 1

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

Q. Compute the area of the region bounded by the curves

y = e x log x

and

y = \frac{log x}{e x}

, where

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The solution of (ex+e−x)dydx=ex−e−x is:

The correct option is B y=log|ex+e−x|+cdydx=ex−e−xex+e−xlet v=ex+e−x⇒dv=(ex−e−x)dx⇒∫dy=∫1vdv+c⇒y=log v+cy=log(ex+e−x)+c

The solution of $(e^{x} + e^{- x}) \frac{d y}{d x} = e^{x} - e^{- x}$ is:

The correct option is B $y = log | e^{x} + e^{- x} | + c$
$\frac{d y}{d x} = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$

let $v = e^{x} + e^{- x}$

$\Rightarrow d v = (e^{x} - e^{- x}) d x$
$\Rightarrow \int d y = \int \frac{1}{v} d v + c$

$\Rightarrow y = log v + c$

$y = log (e^{x} + e^{- x}) + c$