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

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

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

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

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

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Solution

The correct options are
C $y^{- 1} e^{x} = c - x^{2}$
D $e^{x} = y (c - x^{2})$
$2 x y^{2} + e^{x} y = e^{x} \frac{d y}{d x}$
$\Rightarrow 2 x e^{- x} y^{2} + y = \frac{d y}{d x}$
$\Rightarrow \frac{d y}{d x} - y = 2 x e^{- x} y^{2}$
$\Rightarrow y^{- 2} . \frac{d y}{d x} - y^{- 1} = 2 x e^{- x}$
Let $\frac{1}{y} = t$
$\frac{- 1}{y^{2}} . d y = d t$
Hence
$\frac{d y}{d x} = - y^{2} . \frac{d t}{d x}$
Substituting in the above action, we get
$- \frac{d t}{d x} - t = 2 x e^{- x}$
$\Rightarrow \frac{d t}{d x} + t = - 2 x e^{- x}$
$I F = e^{\int 1. d x}$
$= e^{x}$
Hence
$t . e^{x} = \int - 2 x . d x$
$\Rightarrow t e^{x} = - x^{2} + c$
$\Rightarrow \frac{e^{x}}{y} = - x^{2} + c$
$\Rightarrow e^{x} = y (- x^{2} + c)$ .

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Similar questions

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

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

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

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

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