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Question

(A) Given the differential equation: $\cos \left(\frac{dy}{dx}\right) = a$. Taking $\cos^{-1}$ on both sides: $\frac{dy}{dx} = \cos^{-1}(a)$. Integrat

Vedclass · Accepted Answer

$y = x \cos^{-1}(a) + 1$

Vedclass · Answer

(A) Given the differential equation: $\cos \left(\frac{dy}{dx}\right) = a$. Taking $\cos^{-1}$ on both sides: $\frac{dy}{dx} = \cos^{-1}(a)$. Integrating both sides with respect to $x$: $\int dy = \int \cos^{-1}(a) dx$. This gives: $y = x \cos^{-1}(a) + C$ $(1)$. Given the condition $y = 1$ when $x = 0$: $1 = 0 \cdot \cos^{-1}(a) + C$,which implies $C = 1$. Substituting $C = 1$ into equation $(1)$,we get: $y = x \cos^{-1}(a) + 1$.

Find a particular solution satisfying the given condition: $\cos \left(\frac{dy}{dx}\right) = a$ $(a \in R)$; $y = 1$ when $x = 0$.

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