If y=sin x+A cos x is the general solution of | vedclass

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(A) Given the general solution: $y = \sin x + A \cos x$  $(i)$ Differentiating with respect to $x$: $\frac{dy}{dx} = \cos x - A \sin x$  (ii) From $(i

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$\sec x$

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(A) Given the general solution: $y = \sin x + A \cos x$  $(i)$ Differentiating with respect to $x$: $\frac{dy}{dx} = \cos x - A \sin x$  (ii) From $(i)$, we have $A \cos x = y - \sin x$, so $A = \frac{y - \sin x}{\cos x}$. Substituting $A$ into (ii): $\frac{dy}{dx} = \cos x - (\frac{y - \sin x}{\cos x}) \sin x$ $\frac{dy}{dx} = \cos x - y 	an x + \sin x 	an x$ $\frac{dy}{dx} + y 	an x = \cos x + \sin x 	an x = \frac{\cos^2 x + \sin^2 x}{\cos x} = \frac{1}{\cos x} = \sec x$ Comparing this with the standard linear form $\frac{dy}{dx} + f(x)y = Q(x)$, we get $f(x) = 	an x$. The integrating factor ($I$.$F$.) is given by $e^{\int f(x) dx} = e^{\int 	an x dx} = e^{\ln |\sec x|} = \sec x$.

If $y=\sin x+A \cos x$ is the general solution of $\frac{dy}{dx}+f(x)y=\sec x$, then an integrating factor of the differential equation is

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