MCQ
The solution of the equation $\frac{d y}{d x}=e^{x+y}+e^y x^2$ is :
  • A
    $e^x+e^y=\frac{x^3}{3}+c$
  • B
    $e^{-x}+e^y+\frac{x^3}{3}=c$
  • C
    $e^{-x}+e^{-y}-\frac{x^3}{3}=c$
  • $e^x+e^{-y}+\frac{x^3}{3}=c$

Answer

Correct option: D.
$e^x+e^{-y}+\frac{x^3}{3}=c$
(D)
$
\frac{d y}{d x}=e^{x+y}+e^y x^2
$
$
\begin{array}{rlrl}
\frac{d y}{d x} & =e^x \times e^y+e^y x^2 \\
\frac{d y}{d x} & =e^y\left(e^x+x^2\right) \\
\Rightarrow & \frac{d y}{e^y} & =\left(e^x+x^2\right) d x \\
\Rightarrow & e^{-y} d y & =\left(e^x+x^2\right) d x \\
\text { Hence } & \int e^{-y} d y & =\int\left(e^x+x^2\right) d x \\
& -e^{-y} & =e^x+\frac{x^3}{3}+c \\
\text { or } & -e^x+e^{-y}+\frac{x^3}{3} & =c
\end{array}
$
Therefore the correct choice is (D).

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