Find the general solution of given differential equation.
$y (2 x y + e^{x}) d x - e^{x} d y = 0$

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

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

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y^{- 1} e^{x} = c + x^{2}

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y^{- 1} e^{x} = c - x^{2}

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Solution

The correct option is D $y^{- 1} e^{x} = c - x^{2}$
$y^{- 1} e^{x} = c - x^{2}$
$\Rightarrow \frac{d y}{d x} - y = \frac{2 x y^{2}}{e^{x}} \Rightarrow \frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} = \frac{2 x}{e^{x}}$
Substitute $- \frac{1}{y} = t \Rightarrow \frac{1}{y^{2}} \frac{d y}{d x} = \frac{d t}{d x}$
$\Rightarrow \frac{d t}{d x} + t = \frac{2 x}{e^{x}}$

I.F. is $e^{\int d x} = e$

$\Rightarrow t e^{x} = \int \frac{2 x}{e^{x}} e^{x} d x \Rightarrow - \frac{1}{y} e^{x} = x^{2} + c$
$\Rightarrow y^{- 1} e^{x} = - x^{2} - c$

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Similar questions

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

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

y (2 x y + e^{x}) d x - e^{x} d y = 0

e^{x} d y + (y e^{x} + 2 x) d x = 0

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

c

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