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(C) Given the differential equation $\cos \left(\frac{dy}{dx}\right) = a$. Taking $\cos^{-1}$ on both sides,we get $\frac{dy}{dx} = \cos^{-1} a$. Inte

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$\cos \left(\frac{y-2}{x}\right) = a$

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(C) Given the differential equation $\cos \left(\frac{dy}{dx}\right) = a$. Taking $\cos^{-1}$ on both sides,we get $\frac{dy}{dx} = \cos^{-1} a$. Integrating both sides with respect to $x$,we have $\int dy = \int \cos^{-1} a \, dx$. This gives $y = x \cos^{-1} a + c$ .... $(1)$. Given the initial condition $y(0) = 2$,substitute $x = 0$ and $y = 2$ into equation $(1)$: $2 = 0 \cdot \cos^{-1} a + c \Rightarrow c = 2$. Substituting $c = 2$ back into equation $(1)$,we get $y = x \cos^{-1} a + 2$. Rearranging the terms,$y - 2 = x \cos^{-1} a$,which implies $\frac{y - 2}{x} = \cos^{-1} a$. Taking $\cos$ on both sides,we obtain $\cos \left(\frac{y - 2}{x}\right) = a$.

The particular solution of the differential equation $\cos \left(\frac{dy}{dx}\right) = a$,under the conditions $a \in \mathbb{R}$ and $y(0) = 2$ is

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