The general solution of the differential equation dydx-ex-y is -

The correct option is D e^x+e^y=c dydx=e^x+y dydx=e^x.e^y dye^y=e^xdx On integrating both side we get ∫dye^y=∫ e^xdx ⇒ e^ y=e^x+c ...

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Standard XII

Mathematics

Variable Separable Method

The general s...

Question

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

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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

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

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Solution

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

$\frac{d y}{d x} = e^{x} . e^{y}$

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

On integrating both side we get

$\int \frac{d y}{e^{y}} = \int e^{x} d x$

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

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

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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

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

Q. Which of the following relation in x and y is general solution of the differential equation

\frac{d y}{d x} = y

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The general solution of the differential equation dydx=ex+y is :

The correct option is D ex+ey=cdydx=ex+ydydx=ex.eydyey=exdxOn integrating both side we get∫dyey=∫exdx⇒−e−y=ex+ce−y+ex=c.

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

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

$\frac{d y}{d x} = e^{x} . e^{y}$

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

On integrating both side we get

$\int \frac{d y}{e^{y}} = \int e^{x} d x$

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

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