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(B) The given linear differential equation is of the form $\frac{dy}{dx} + Py = Q$, where $P = -	an x$ and $Q = e^x \sec x$. First, we find the integ

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$y \cos x = e^x + c$

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(B) The given linear differential equation is of the form $\frac{dy}{dx} + Py = Q$, where $P = -	an x$ and $Q = e^x \sec x$. First, we find the integrating factor $(IF)$: $IF = e^{\int P dx} = e^{\int -	an x dx} = e^{\ln(\cos x)} = \cos x$. The general solution is given by $y \cdot (IF) = \int Q \cdot (IF) dx + c$. Substituting the values: $y \cos x = \int (e^x \sec x) \cdot \cos x dx + c$. Since $\sec x \cdot \cos x = 1$, the equation simplifies to: $y \cos x = \int e^x dx + c$. Integrating $e^x$, we get: $y \cos x = e^x + c$.

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

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