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Question Number 121960 by bemath last updated on 13/Nov/20

$$\frac{dy}{dx}+3y\tan x=\frac{1}{2}\sin 2x$$

Answered by liberty last updated on 13/Nov/20

$${\mathrm{y}}^{\prime }+3\mathrm{y}\tan \mathrm{x}=\sin \mathrm{x}\cos \mathrm{x}$$$$\mathrm{IF}\mathrm{u}={\mathrm{e}}^{\int 3\tan \mathrm{x}\mathrm{dx}}={\mathrm{e}}^{-3\int \frac{\mathrm{d}\left(\cos \mathrm{x}\right)}{\cos \mathrm{x}}}={\mathrm{e}}^{\ln \left(\sec {}^3\mathrm{x}\right)}$$$$\mathrm{u}=\sec {}^3\mathrm{x}$$$$\mathrm{General}\mathrm{solution}$$$$\mathrm{y}=\frac{\int \sin \mathrm{x}\cos \mathrm{x}\sec {}^3\mathrm{x}\mathrm{dx}+\mathrm{C}}{\sec {}^3\mathrm{x}}$$$$\mathrm{y}=\mathrm{C}.\cos {}^3\mathrm{x}+\cos {}^3\mathrm{x}\int \frac{\sin \mathrm{x}}{\cos {}^2\mathrm{x}}\mathrm{dx}$$$$\mathrm{y}=\mathrm{C}.\cos {}^3\mathrm{x}-\cos {}^3\mathrm{x}\int \frac{\mathrm{d}\left(\cos \mathrm{x}\right)}{\cos {}^2\mathrm{x}}$$$$\mathrm{y}=\mathrm{C}.\cos {}^3\mathrm{x}+\cos {}^2\mathrm{x}$$

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