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Question 167 of 177

Q.Find the particular solution of the differential equation dydx=e2ycos⁡x\dfrac{dy}{dx} = e^{2y}\cos x, when x=π6x = \dfrac{\pi}{6}, y=0y = 0.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2023Subjective· 3mImportance★★★★★
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Separate variables and use the initial condition x=π/6,y=0x=\pi/6,y=0.

dydx=e2ycos⁡x⇒e−2y dy=cos⁡x dx\dfrac{dy}{dx}=e^{2y}\cos x \Rightarrow e^{-2y}\,dy=\cos x\,dx

Integrating: −e−2y2=sin⁡x+C-\dfrac{e^{-2y}}{2}=\sin x+C

At x=π6, y=0x=\dfrac{\pi}{6},\ y=0: −12=12+C⇒C=−1-\dfrac12=\dfrac12+C \Rightarrow C=-1

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