find the soluionof the differential equation:
sin2x dy/dx + y=tanx

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \sin \left(2\mathrm{x}\right).\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=\mathrm{tanx} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\sin \left(2\mathrm{x}\right)}=\frac{\mathrm{tanx}}{\sin \left(2\mathrm{x}\right)} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}.\mathrm{cosec}\left(2\mathrm{x}\right)=\frac{\mathrm{sinx}}{2.\mathrm{sinx}.\mathrm{cosx}.\mathrm{cosx}} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}.\mathrm{cosec}\left(2\mathrm{x}\right)=\frac{1}{2}{\sec }^2\mathrm{x} ;\left(\mathrm{i}\right) \\ \mathrm{On} \mathrm{comparing} \mathrm{with} \mathrm{first} \mathrm{order} \mathrm{linear} \mathrm{differential} \mathrm{equation}: \\ \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{P}\left(\mathrm{x}\right).\mathrm{dx}=\mathrm{Q}\left(\mathrm{x}\right) \\ \mathrm{Integrating} \mathrm{factor}={\mathrm{e}}^{\int \mathrm{P}\left(\mathrm{x}\right).\mathrm{dx}} \\ ={\mathrm{e}}^{\int \mathrm{cosec}\left(2\mathrm{x}\right).\mathrm{dx}} \\ ={\mathrm{e}}^{\left[-\frac{\ln \left(\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}\right)}{2}\right]} \\ ={\mathrm{e}}^{\left[-\frac{1}{2}\ln \left(\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}\right)\right]} \\ ={\mathrm{e}}^{\left[\ln {\left(\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}\right)}^{-\frac{1}{2}}\right]} \left[\mathrm{n}.\log \left(\mathrm{m}\right)=\log \left({\mathrm{m}}^{\mathrm{n}}\right)\right] \\ ={\left(\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}\right)}^{-\frac{1}{2}} \left[{\mathrm{e}}^{\ln \left(\mathrm{a}\right)}=\mathrm{a}\right] \\ =\frac{1}{\sqrt{\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}}} \\ \mathrm{Hence} \left(\mathrm{i}\right) \mathrm{becomes}: \\ \mathrm{y}.\frac{1}{\sqrt{\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}}}=\frac{1}{2}\int \frac{{\sec }^2\mathrm{x}}{\sqrt{\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}}}.\mathrm{dx} \\ \Rightarrow \mathrm{y}.\frac{1}{\sqrt{\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}}}=\frac{1}{2}\int \frac{{\sec }^2\mathrm{x}}{\sqrt{{\displaystyle \frac{1}{\sin 2\mathrm{x}}}+{\displaystyle \frac{\cos 2\mathrm{x}}{\sin 2\mathrm{x}}}}}.\mathrm{dx} \\ \Rightarrow \mathrm{y}.\frac{1}{\sqrt{\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}}}=\frac{1}{2}\int \frac{{\sec }^2\mathrm{x}}{\sqrt{{\displaystyle 1+\cos 2\mathrm{x}}}}.\sqrt{\sin 2\mathrm{x}}\mathrm{dx} \\ \Rightarrow \mathrm{y}.\frac{1}{\sqrt{\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}}}=\frac{1}{2}\int \frac{{\sec }^2\mathrm{x}}{\sqrt{{\displaystyle 1+2{\cos }^2\mathrm{x}-1}}}.\sqrt{2\mathrm{sinx}.\mathrm{cosx}}\mathrm{dx} \\ \Rightarrow \mathrm{y}.\frac{1}{\sqrt{\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}}}=\frac{1}{2}\int \frac{{\sec }^2\mathrm{x}}{\sqrt{{\displaystyle 2{\cos }^2\mathrm{x}}}}.\sqrt{2\mathrm{sinx}.\mathrm{cosx}}\mathrm{dx} \\ \Rightarrow \mathrm{y}.\frac{1}{\sqrt{\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}}}=\frac{1}{2}\int \frac{{\sec }^2\mathrm{x}}{\mathrm{cosx}}.\sqrt{\mathrm{sinx}.\mathrm{cosx}}\mathrm{dx} \\ \Rightarrow \mathrm{y}.\frac{1}{\sqrt{\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}}}=\frac{1}{2}\int \frac{{\sec }^2\mathrm{x}}{\sqrt{\mathrm{cosx}}}.\sqrt{\mathrm{sinx}}\mathrm{dx} \\ \Rightarrow \mathrm{y}.\frac{1}{\sqrt{\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}}}=\frac{1}{2}\int {\sec }^2\mathrm{x}.\sqrt{\mathrm{tanx}}\mathrm{dx} \\ \mathrm{We} \mathrm{take} \mathrm{tanx}=\mathrm{t}, \mathrm{then} {\sec }^2\mathrm{x}.\mathrm{dx}=\mathrm{dt} \mathrm{and} \mathrm{hence} \mathrm{we} \mathrm{get} \int \sqrt{\mathrm{t}}.\mathrm{dt}=\frac{{\mathrm{t}}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}=\frac{2}{3}.{\left(\mathrm{tanx}\right)}^{\frac{3}{2}} \\ \mathrm{Hence} \\ \Rightarrow \mathrm{y}.\frac{1}{\sqrt{\mathrm{cosec}2\mathrm{x}+\cot 2\mathrm{x}}}=\frac{1}{2}\frac{2}{3}.{\left(\mathrm{tanx}\right)}^{\frac{3}{2}}+\mathrm{C} \\ \Rightarrow \mathbf{y}\mathbf{.}\frac{\mathbf{1}}{\sqrt{\mathbf{cosec}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{cot}\mathbf{2}\mathbf{x}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{.}{\left(\mathbf{tanx}\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}\mathbf{+}\mathbf{C} \end{gathered}$$

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