The particular solution of the differential e | vedclass

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Question

(A) Given the differential equation: $\cos \left(\frac{dy}{dx}\right) = 0.5$. Taking the inverse cosine on both sides: $\frac{dy}{dx} = \cos^{-1}(0.5)

Vedclass · Accepted Answer

$y = \frac{\pi}{3}x + 1$

Vedclass · Answer

(A) Given the differential equation: $\cos \left(\frac{dy}{dx}\right) = 0.5$. Taking the inverse cosine on both sides: $\frac{dy}{dx} = \cos^{-1}(0.5)$. Since $\cos(60^{\circ}) = 0.5$,we have $\frac{dy}{dx} = \frac{\pi}{3}$. Integrating both sides with respect to $x$: $\int dy = \int \frac{\pi}{3} dx$. This gives $y = \frac{\pi}{3}x + C$. Using the initial condition $y = 1$ at $x = 0$: $1 = \frac{\pi}{3}(0) + C$,which implies $C = 1$. Therefore,the particular solution is $y = \frac{\pi}{3}x + 1$.

The particular solution of the differential equation $\cos \left(\frac{dy}{dx}\right) = 0.5$ with the condition $y = 1$ at $x = 0$ is:

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