Solve the differential equation: $y (2 x y + e^{x}) d x - e^{x} d y = 0.$

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

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

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

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

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

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

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Solution

The correct option is D $y (x^{2} + c) + e^{x} = 0.$
$y (2 x y + e^{x}) d x - e^{x} d y = 0 \Rightarrow e^{x} \frac{d y}{d x} - y e^{x} = 2 x y^{2}$
$\Rightarrow \frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} = 2 x e^{- x}$
Put $- \frac{1}{y} = v \Rightarrow \frac{1}{y^{2}} d y = d v$
$∴ \frac{d v}{d x} + v = 2 x e^{- x}$ ...(1)
Here $P = 1 \Rightarrow \int P d x = \int d x = x$
$∴ I . F . = e^{x}$
Multiplying (1) by $I . F .$ we get
$e^{x} \frac{d v}{d x} + e^{x} v = 2 x$
Integrating both sides we get
$e^{x} v = \int 2 x d x = x^{2} + c$
$\Rightarrow y (x^{2} + c) + e^{x} = 0$

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