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The general solution of the differential equation $\frac{d y}{d x}=1+x+y+x y$ is
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Verified Answer
The correct answer is:
$y=k e^{x+\frac{x^2}{2}}-1$
We have,
$$
\begin{aligned}
& \frac{d y}{d x}=1+x+y+x y \\
\Rightarrow & \frac{d y}{d x}=(1+x)(1+y) \Rightarrow \frac{d y}{1+y}=(1+x) d x
\end{aligned}
$$
On integrating both the sides, we get
$$
\begin{gathered}
\quad \int \frac{d y}{1+y}=\int(1+x) d x \\
\Rightarrow \log (1+y)=x+\frac{x^2}{2}+c \Rightarrow 1+y=e^{x+\frac{x^2}{2}+c} \\
\Rightarrow \quad y=e^{x+\frac{x^2}{2}} \cdot e^c-1
\end{gathered}
$$
$$
\Rightarrow \quad y=k e^{x+\frac{x^2}{2}}-1 \quad\left[\text { where, } k=e^c\right. \text { ] }
$$
$$
\begin{aligned}
& \frac{d y}{d x}=1+x+y+x y \\
\Rightarrow & \frac{d y}{d x}=(1+x)(1+y) \Rightarrow \frac{d y}{1+y}=(1+x) d x
\end{aligned}
$$
On integrating both the sides, we get
$$
\begin{gathered}
\quad \int \frac{d y}{1+y}=\int(1+x) d x \\
\Rightarrow \log (1+y)=x+\frac{x^2}{2}+c \Rightarrow 1+y=e^{x+\frac{x^2}{2}+c} \\
\Rightarrow \quad y=e^{x+\frac{x^2}{2}} \cdot e^c-1
\end{gathered}
$$
$$
\Rightarrow \quad y=k e^{x+\frac{x^2}{2}}-1 \quad\left[\text { where, } k=e^c\right. \text { ] }
$$
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