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The solution of the differential equation $\frac{d y}{d x}-2 y \tan 2 x=e^x \sec 2 x$ is
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Verified Answer
The correct answer is:
$y \cos 2 x=e^x+C$
Given, differential equation is
$$
\frac{d y}{d x}-2 y \tan 2 x=e^x \sec 2 x
$$
Here, $P=2 \tan 2 x, Q=e^x \sec 2 x$
$$
\begin{aligned}
\therefore \quad \mathrm{IF} & =e^{\int-2 \tan 2 x d x} \\
& =e^{\frac{-2 \log \sec 2 x}{2}} \\
& =e^{-\log \sec 2 x} \\
& =\frac{1}{\sec 2 x}
\end{aligned}
$$
$\therefore$ Required solution is
$$
\begin{aligned}
& \frac{y}{\sec 2 x}=\int e^x \cdot \frac{1}{\sec 2 x} \cdot \sec 2 x d x+C \\
\Rightarrow \quad y \cos 2 x & =\int e^x \cdot 1 d x+C \\
\Rightarrow \quad & y \cos 2 x=e^x+C
\end{aligned}
$$
where, $C$ is the constant of integration.
$$
\frac{d y}{d x}-2 y \tan 2 x=e^x \sec 2 x
$$
Here, $P=2 \tan 2 x, Q=e^x \sec 2 x$
$$
\begin{aligned}
\therefore \quad \mathrm{IF} & =e^{\int-2 \tan 2 x d x} \\
& =e^{\frac{-2 \log \sec 2 x}{2}} \\
& =e^{-\log \sec 2 x} \\
& =\frac{1}{\sec 2 x}
\end{aligned}
$$
$\therefore$ Required solution is
$$
\begin{aligned}
& \frac{y}{\sec 2 x}=\int e^x \cdot \frac{1}{\sec 2 x} \cdot \sec 2 x d x+C \\
\Rightarrow \quad y \cos 2 x & =\int e^x \cdot 1 d x+C \\
\Rightarrow \quad & y \cos 2 x=e^x+C
\end{aligned}
$$
where, $C$ is the constant of integration.
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