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

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

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

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y e^{x} - x^{2} = C

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

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Solution

The correct option is D $y e^{x} + x^{2} = C$
$e^{x} d y + (y e^{x} + 2 x) d x = 0$
$\Rightarrow e^{x} d y = - (y e^{x} + 2 x) d x$
$\Rightarrow \frac{d y}{d x} + y = - 2 x . e^{- x}$ which is an exact DE
$∴ I F = e^{\int 1 d x} = e^{x}$
General solution is given by

$y I F = \int I F . Q (x) d x + C$

$y e^{x} = \int e^{x} (- 2 x e^{- x} d x) + C$
$y e^{x} = - x^{2} + C$

$\Rightarrow y e^{x} + x^{2} = C$

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Similar questions

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

The general solution of the differential equation is

A. xe^y + x² = C

B. xe^y + y² = C

C. ye^x + x² = C

D. ye^y+ x² = C

(1 + e^{\frac{x}{y}}) d x + (1 - \frac{x}{y}) e^{\frac{x}{y}} d y = 0

c

y e^{\frac{x}{y}} d x = (x e^{\frac{x}{y}} + y^{2}) d y, (y \neq 0)

^{'} C^{'}

y e^{x / y} d x = (x e^{x / y} + y^{2} sin y) d y

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