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The general solution of the differential equation $\mathrm{e}^x d y+$ $\left(y e^x+2 x\right) d x=0$ is
(a) $x^y+x^2=C$
(b) $x e^y+y^2=C$
(c) $\mathrm{ye}^{\mathrm{x}}+\mathrm{x}^2=\mathrm{C}$
(d) $y^{e^y+x^2}=$ C
(a) $x^y+x^2=C$
(b) $x e^y+y^2=C$
(c) $\mathrm{ye}^{\mathrm{x}}+\mathrm{x}^2=\mathrm{C}$
(d) $y^{e^y+x^2}=$ C
Solution:
1385 Upvotes
Verified Answer
The differential equation is
$$
\begin{aligned}
&e^x d y+\left(y e^x+2 x\right) d x=0 \\
&\text { or } e^x \frac{d y}{d x}+e^x y=-2 x \text { or } \frac{d y}{d x}+1 . y=-2 x e^{-x} \\
&\text { I.F. }=e^{\text {lpdx }}=e^x \\
&\therefore \quad \text { Solution is } y e^x=\int(-2 x) e^{-x} \times e^x d x+c \\
&=-\int 2 x d x+c=-x^2+c \\
&\Rightarrow y e^x+x^2=c
\end{aligned}
$$
Option (c) is correct
$$
\begin{aligned}
&e^x d y+\left(y e^x+2 x\right) d x=0 \\
&\text { or } e^x \frac{d y}{d x}+e^x y=-2 x \text { or } \frac{d y}{d x}+1 . y=-2 x e^{-x} \\
&\text { I.F. }=e^{\text {lpdx }}=e^x \\
&\therefore \quad \text { Solution is } y e^x=\int(-2 x) e^{-x} \times e^x d x+c \\
&=-\int 2 x d x+c=-x^2+c \\
&\Rightarrow y e^x+x^2=c
\end{aligned}
$$
Option (c) is correct
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