The general solution of the differential equa | vedclass

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(D) The given differential equation is $e^{x} dy + (ye^{x} + 2x) dx = 0$. Dividing by $dx$,we get $e^{x} \frac{dy}{dx} + ye^{x} + 2x = 0$. Dividing by

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$ye^{x} + x^{2} = C$

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(D) The given differential equation is $e^{x} dy + (ye^{x} + 2x) dx = 0$. Dividing by $dx$,we get $e^{x} \frac{dy}{dx} + ye^{x} + 2x = 0$. Dividing by $e^{x}$,we get $\frac{dy}{dx} + y = -2x e^{-x}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1$ and $Q = -2x e^{-x}$. The integrating factor $(I.F.)$ is $e^{\int P dx} = e^{\int 1 dx} = e^{x}$. The general solution is given by $y(I.F.) = \int (Q 	imes I.F.) dx + C$. Substituting the values,$y e^{x} = \int (-2x e^{-x} \cdot e^{x}) dx + C$. $y e^{x} = \int -2x dx + C$. $y e^{x} = -x^{2} + C$. Therefore,$y e^{x} + x^{2} = C$.

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

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