The solution of the differential equation sin | vedclass

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

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$\sin \left( \frac{y - 1}{x} \right) = a$

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(B) Given the differential equation $\sin \left( \frac{dy}{dx} \right) = a$. Taking the inverse sine on both sides,we get $\frac{dy}{dx} = \sin^{-1}(a)$. This is a variable separable differential equation. Integrating both sides with respect to $x$,we have $\int dy = \int \sin^{-1}(a) \, dx$. Since $\sin^{-1}(a)$ is a constant,we get $y = x \sin^{-1}(a) + c$. Using the initial condition $y(0) = 1$,we substitute $x = 0$ and $y = 1$: $1 = 0 \cdot \sin^{-1}(a) + c \implies c = 1$. Thus,the equation becomes $y = x \sin^{-1}(a) + 1$. Rearranging the terms,$y - 1 = x \sin^{-1}(a)$,which implies $\frac{y - 1}{x} = \sin^{-1}(a)$. Taking the sine of both sides,we obtain $\sin \left( \frac{y - 1}{x} \right) = a$.

The solution of the differential equation $\sin \left( \frac{dy}{dx} \right) = a$ with the initial condition $y(0) = 1$ is:

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