The solution of $e^{x - y} d x + e^{y - x} d y = 0$ is:

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

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e^{2 x} + e^{2 y} = c

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

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

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Solution

The correct option is B $e^{2 x} + e^{2 y} = c$
$\frac{e^{x}}{e^{y}} d x + \frac{e^{y}}{e^{x}} d y = 0$

$\Rightarrow \int e^{2 x} d x + \int e^{2 y} = c_{1}$

$\Rightarrow \frac{e^{2 x}}{2} + \frac{e^{2 y}}{2} = \frac{c_{1}}{2}$

$\Rightarrow e^{2 x} + e^{2 y} = 2 c_{1} = c$

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Similar questions

The solution of the equation $\frac{d y}{d x} - e^{x - y} + x^{2} e^{- y}$ is
[MP PET 2004]

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. If

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

, prove that

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

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

Solve the differential equations.

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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The solution of ex−ydx+ey−xdy=0 is:

The correct option is B e2x+e2y=cexeydx+eyexdy=0⇒∫e2xdx+∫e2y=c1⇒e2x2+e2y2=c12⇒e2x+e2y=2c1=c

The solution of $e^{x - y} d x + e^{y - x} d y = 0$ is:

The correct option is B $e^{2 x} + e^{2 y} = c$
$\frac{e^{x}}{e^{y}} d x + \frac{e^{y}}{e^{x}} d y = 0$

$\Rightarrow \int e^{2 x} d x + \int e^{2 y} = c_{1}$

$\Rightarrow \frac{e^{2 x}}{2} + \frac{e^{2 y}}{2} = \frac{c_{1}}{2}$

$\Rightarrow e^{2 x} + e^{2 y} = 2 c_{1} = c$