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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 $Q = e^x \sec 2x$. First,we find the integrating factor $(IF)$: $IF = e^{\int P dx} = e^{\int -2 	an 2x dx} = e^{-\ln(\sec 2x)} = e^{\ln(\cos 2x)} = \cos 2x$. The general solution is given by $y \cdot IF = \int (Q \cdot IF) dx + C$. Substituting the values: $y \cos 2x = \int (e^x \sec 2x \cdot \cos 2x) dx + C$. Since $\sec 2x \cdot \cos 2x = 1$,we have: $y \cos 2x = \int e^x dx + C$. $y \cos 2x = e^x + C$,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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Observe the following statements:
$A$. Integrating factor of $\frac{dy}{dx} + y = x^2$ is $e^x$.
$R$. Integrating factor of $\frac{dy}{dx} + P(x)y = Q(x)$ is $e^{\int P(x) dx}$.
Then,the true statement among the following is:

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