The general solution of the differential equation ${log}_{e} (\frac{d y}{d x}) = x + y$ is:

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

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

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

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

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Solution

The correct option is B $e^{x} + e^{- y} = C$
Given ${log}_{e} (\frac{d y}{d x}) = x + y$

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

On integrating, we get

$- e^{- y} = e^{x} - C$
$\Rightarrow C = e^{x} + e^{- y}$

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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

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

The general solution of differential equation $\frac{d y}{d x} = e^{x - y} + x^{2} e^{- y}$ is

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 loge(dydx)=x+y is:

The correct option is B ex+e−y=CGiven loge(dydx)=x+y⇒dydx=ex+y=exey⇒e−ydy=exdxOn integrating, we get−e−y=ex−C⇒C=ex+e−y

The general solution of the differential equation ${log}_{e} (\frac{d y}{d x}) = x + y$ is:

The correct option is B $e^{x} + e^{- y} = C$
Given ${log}_{e} (\frac{d y}{d x}) = x + y$

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

On integrating, we get

$- e^{- y} = e^{x} - C$
$\Rightarrow C = e^{x} + e^{- y}$