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(B) The given differential equation is $\frac{dy}{dx} - 2y 	an 2x = e^x \sec 2x$. This is a linear differential equation of the form $\frac{dy}{dx} +

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$y \cos 2x = e^x + C$

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(B) The given differential equation is $\frac{dy}{dx} - 2y 	an 2x = e^x \sec 2x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = -2 	an 2x$ and $Q = e^x \sec 2x$. The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int -2 	an 2x dx}$. $IF = e^{-\ln(\sec 2x)} = e^{\ln(\cos 2x)} = \cos 2x$. The general solution is $y \cdot (IF) = \int Q \cdot (IF) dx + C$. Substituting the values, we get $y \cos 2x = \int (e^x \sec 2x) \cdot \cos 2x dx + C$. Since $\sec 2x \cdot \cos 2x = 1$, we have $y \cos 2x = \int e^x dx + C$. Therefore, the solution is $y \cos 2x = e^x + C$, where $C$ is the constant of integration.

The solution of the differential equation $\frac{dy}{dx} - 2y \tan 2x = e^x \sec 2x$ is

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