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(D) Given the differential equation: $\frac{dy}{dx} = e^{2y} \cos x$. Separating the variables,we get: $e^{-2y} dy = \cos x dx$. Integrating both side

Vedclass · Accepted Answer

$2 \sin x + e^{-2y} - 2 = 0$

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(D) Given the differential equation: $\frac{dy}{dx} = e^{2y} \cos x$. Separating the variables,we get: $e^{-2y} dy = \cos x dx$. Integrating both sides: $\int e^{-2y} dy = \int \cos x dx$. This gives: $-\frac{1}{2} e^{-2y} = \sin x + C$. Given the condition $y(\frac{\pi}{6}) = 0$,substitute $x = \frac{\pi}{6}$ and $y = 0$: $-\frac{1}{2} e^{0} = \sin(\frac{\pi}{6}) + C$. $-\frac{1}{2} = \frac{1}{2} + C$,which implies $C = -1$. Substituting $C$ back into the equation: $-\frac{1}{2} e^{-2y} = \sin x - 1$. Multiplying by $-2$: $e^{-2y} = -2 \sin x + 2$. Rearranging the terms: $2 \sin x + e^{-2y} - 2 = 0$.

The particular solution of the differential equation $\frac{dy}{dx} = e^{2y} \cos x$,when $y(\frac{\pi}{6}) = 0$ is

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