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(A) Given the differential equation: $(y + x \frac{dy}{dx}) \sin(xy) = \cos x$ Let $u = xy$. Then,differentiating with respect to $x$,we get $\frac{du

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$\sin x + \cos(xy) = 1$

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(A) Given the differential equation: $(y + x \frac{dy}{dx}) \sin(xy) = \cos x$ Let $u = xy$. Then,differentiating with respect to $x$,we get $\frac{du}{dx} = y + x \frac{dy}{dx}$. Substituting this into the equation,we have: $\sin(u) \frac{du}{dx} = \cos x$. Integrating both sides with respect to $x$: $\int \sin(u) du = \int \cos x dx$. This yields: $-\cos(u) = \sin x + C$. Substituting $u = xy$ back,we get: $-\cos(xy) = \sin x + C$. At $x = 0$,the value of $y$ is not explicitly given,but since $\cos(xy) = -\sin x - C$,at $x=0$,$\cos(0) = -\sin(0) - C$,which implies $1 = 0 - C$,so $C = -1$. Thus,$-\cos(xy) = \sin x - 1$,which simplifies to $\sin x + \cos(xy) = 1$.

The particular solution of the differential equation $(y + x \cdot \frac{dy}{dx}) \cdot \sin(xy) = \cos x$ at $x = 0$ is

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