The solution of dydx - 1 - ex - y is-

The correct option is C e^ (x y) = x + c dydx 1 = e^x y ⇒ dy dx = e^x y . dx ⇒ d(x y)e^x y = dx ⇒ dx + e^ (x y). d(x ...

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Standard XII

Mathematics

Equations Reducible to Standard Forms

The solution ...

Question

The solution of $\frac{d y}{d x} - 1 = e^{x - y}$ is:

e^{x - y} + x = c

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

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

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e^{x - y} = x + c

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Solution

The correct option is C $e^{- (x - y)} = x + c$
$\frac{d y}{d x} - 1 = e^{x - y}$

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

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

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

$\Rightarrow x - e^{- (x - y)} = c$ , Integrate

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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. If

y = e^{x} + e^{- x} + log x^{2}

, find

\frac{d y}{d x}

Q. The general solution of

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

is .

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

prove that

\frac{d y}{d x} = \sqrt{y^{2} - 4}

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The solution of dydx−1=ex−y is:

The correct option is C e−(x−y)=x+cdydx−1=ex−y⇒dy−dx=ex−y.dx⇒−d(x−y)ex−y=dx⇒dx+e−(x−y).d(x−y)=0⇒x−e−(x−y)=c, Integrate

The solution of $\frac{d y}{d x} - 1 = e^{x - y}$ is:

The correct option is C $e^{- (x - y)} = x + c$
$\frac{d y}{d x} - 1 = e^{x - y}$

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

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

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

$\Rightarrow x - e^{- (x - y)} = c$ , Integrate