The general solution of the differential equation dy dx - e y-x - e y-x is

The correct option is A e ^ y = e ^ x e ^ x +c Given, dy dx = e ^ y ( e ^ x + e ^ x ) ⇒ e ^ y dy=( e ^ x + e ^ x ) dx On integrating ...

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Standard XII

Mathematics

Linear Differential Equations of First Order

The general s...

Question

The general solution of the differential equation $\frac{d y}{d x} = e^{y + x} + e^{y - x}$ is

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

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

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

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

where

c

is an arbitrary constant

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Solution

The correct option is A $e^{- y} = e^{- x} - e^{x} + c$
Given, $\frac{d y}{d x} = e^{y} (e^{x} + e^{- x})$
$\Rightarrow e^{- y} d y = (e^{x} + e^{- x}) d x$
On integrating both sides, we get
$- e^{- y} = e^{x} - e^{- x} - c$
$\Rightarrow e^{- y} = e^{- x} - e^{x} + c$

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Similar questions

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. 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. The general solution of the differential equation y

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

is ___ {c is arbitrary constant}

\int \frac{2}{{(e^{x} + e^{- x})}^{2}} d x

(a)

\frac{- e^{- x}}{e^{x} + e^{- x}} + C

(b)

- \frac{1}{e^{x} + e^{- x}} + C

(c)

\frac{- 1}{{(e^{x} + 1)}^{2}} + C

(d)

\frac{1}{e^{x} - e^{- x}} + C

The general solution of the differential equation $e^{x} d y + (y e^{x} + 2 x) d x = 0$ is

a) $x e^{y} + x^{2} = C$

b) $x e^{y} + y^{2} = C$

c) $y e^{x} + x^{2} = C$

d) $y e^{y} + x^{2} = C$

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The general solution of the differential equation dydx=ey+x+ey−x is

The correct option is A e−y=e−x−ex+cGiven, dydx=ey(ex+e−x)⇒e−ydy=(ex+e−x)dxOn integrating both sides, we get−e−y=ex−e−x−c⇒e−y=e−x−ex+c

The general solution of the differential equation $\frac{d y}{d x} = e^{y + x} + e^{y - x}$ is

The correct option is A $e^{- y} = e^{- x} - e^{x} + c$
Given, $\frac{d y}{d x} = e^{y} (e^{x} + e^{- x})$
$\Rightarrow e^{- y} d y = (e^{x} + e^{- x}) d x$
On integrating both sides, we get
$- e^{- y} = e^{x} - e^{- x} - c$
$\Rightarrow e^{- y} = e^{- x} - e^{x} + c$