Question
Solve the following differential equations:

$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}}(\sin^2\text{x}+\sin2\text{x})}{\text{y}(2\log\text{y}+1)}$

Answer

$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}}(\sin^2\text{x}+\sin2\text{x})}{\text{y}(2\log\text{y}+1)}$
$\Rightarrow\text{y}(2\log\text{y}+1)\text{dy}=\text{e}^{\text{x}}(\sin^2\text{x}+\sin2\text{x})\text{dx}$
$\Rightarrow(2\text{y}\log\text{y+y})\text{dy}=(\text{e}^{\text{x}}\sin^2\text{x + e}^{\text{x}}\sin2\text{x})\text{dx}$
$\Rightarrow2\text{y}\log\text{y}\text{ dy}+\text{y dy}=\text{e}^{\text{x}}\sin^2\text{x dx}+\text{e}^{\text{x}}\sin2\text{x}\text{ dx}$
Integrating both sides, we get
$2\int\text{y}\log\text{y dy}+\int\text{y dy}=\int\text{e}^{\text{x}}\sin^2\text{x dx}+\int\text{e}^{\text{x}}\sin2\text{x dx}$
$\Rightarrow2\Big[\log\text{y}\int\text{y dy}-\int\Big\{\frac{\text{d}}{\text{dy}}(\log\text{ y})\int\text{y dy}\Big\}\Big]\text{dy}+\int\text{y dy}\\=\sin^2\text{x}\int\text{e}^{\text{x}}\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\sin^2\text{x})\int\text{e}^{\text{x}}\text{dx}\Big]\text{dx}+\int\text{e}^{\text{x}}\sin2\text{x dx} $
$\Rightarrow2\Big[\log\text{y}\Big(\frac{\text{y}^2}{2}\Big)-\int\Big(\frac{1}{\text{y}}\Big)\frac{\text{y}^2}{2}\text{dy}\Big]+\int\text{y dy}\\=\sin^2\text{x }\text{e}^{\text{x}}-\int\big[2\sin\text{x}\cos\text{x}\text{ e}^{\text{x}}\big]\text{dx}+\int\text{e}^{\text{x}}\sin2\text{x dx + C}$
$\Rightarrow\text{y}^2\log\text{ y}-\int\text{y dy}+\int\text{y dy}\\=\text{e}^{\text{x}}\sin^2\text{x}-\int\text{e}^{\text{x}}\sin2\text{x dx}+\int\text{e}^{\text{x}}\sin2\text{x dx + C}$
$\Rightarrow\text{y}^2\log\text{y}=\text{e}^{\text{x}}\sin^2\text{x + C}$

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