The solution of frac{dy}{dx} = frac{e^x(sin^2 | vedclass

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{e^x(\sin^2 x + \sin 2x)}{y(2\log y + 1)}$By separating the variables,we get: $(2y \log y +

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$y^2(\log y) - e^x \sin^2 x + c = 0$

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{e^x(\sin^2 x + \sin 2x)}{y(2\log y + 1)}$By separating the variables,we get: $(2y \log y + y) dy = e^x(\sin^2 x + \sin 2x) dx$Integrating both sides: $\int (2y \log y + y) dy = \int e^x(\sin^2 x + \sin 2x) dx$For the $LHS$: Let $u = \log y$,then $du = \frac{1}{y} dy$. This integral is $\int y(2 \log y + 1) dy$. Using integration by parts on $\int 2y \log y dy$,we get $y^2 \log y - \int y^2 \cdot \frac{1}{y} dy = y^2 \log y - \frac{y^2}{2}$. Thus,$\int (2y \log y + y) dy = y^2 \log y - \frac{y^2}{2} + \frac{y^2}{2} = y^2 \log y$.For the $RHS$: $\int e^x(\sin^2 x + \sin 2x) dx$. Note that $\frac{d}{dx}(e^x \sin^2 x) = e^x \sin^2 x + e^x(2 \sin x \cos x) = e^x(\sin^2 x + \sin 2x)$.Therefore,the integral is $e^x \sin^2 x + c$.Equating both sides: $y^2 \log y = e^x \sin^2 x + c$,which can be written as $y^2 \log y - e^x \sin^2 x + c = 0$.

The solution of $\frac{dy}{dx} = \frac{e^x(\sin^2 x + \sin 2x)}{y(2\log y + 1)}$ is

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