The solution of dy-dx 2- dy-dx ex-e-x-1-0 is-

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Particular Solution of a Differential Equation

The solution ...

Question

The solution of ${(\frac{d y}{d x})}^{2} - (\frac{d y}{d x}) (e^{x} + e^{- x}) + 1 = 0$ is:

y + e^{- x} = c o n s t

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x = log (c + y)

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y = e^{- x} + c o n s t

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y = e^{x} + c o n s t

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Solution

The correct options are
A $y + e^{- x} = c o n s t$
C $x = log (c + y)$
D $y = e^{x} + c o n s t$
The given equation can be factored as
$(\frac{d y}{d x} - e^{x}) (\frac{d y}{d x} - e^{- x}) = 0$

$\frac{d y}{d x} - e^{x} = 0 \Rightarrow c + y = e^{x} \Rightarrow x = log (c + y)$

$\frac{d y}{d x} - e^{- x} = 0 \Rightarrow c = - e^{- x} \Rightarrow y + e^{- x} + c = 0 \Rightarrow y + e^{- x} = c$

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Similar questions

Q. Solutions of the differential equation

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

are given by:

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

is/are

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) $e^{- x} + e^{- y} = C$

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Q. The solution of

e^{x - y} d x + e^{y - x} d y = 0

is:

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The solution of (dydx)2−(dydx)(ex+e−x)+1=0 is:

The correct options are A y+e−x=const C x=log(c+y) D y=ex+constThe given equation can be factored as (dydx−ex)(dydx−e−x)=0dydx−ex=0⇒c+y=ex⇒x=log(c+y)dydx−e−x=0⇒c=−e−x⇒y+e−x+c=0⇒y+e−x=c

The solution of ${(\frac{d y}{d x})}^{2} - (\frac{d y}{d x}) (e^{x} + e^{- x}) + 1 = 0$ is: