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Question Number 83028 by jagoll last updated on 27/Feb/20

$$\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\sec \mathrm{x}=\tan \mathrm{x}$$

Commented by john santu last updated on 27/Feb/20

$$\mathrm{IF}\Rightarrow {\mathrm{e}}^{\int \sec \mathrm{x}\mathrm{dx}}=\mathrm{e}{}^{\ln (\sec \mathrm{x}+\tan \mathrm{x})}$$$$\mathrm{IF}=\sec \mathrm{x}+\tan \mathrm{x}$$$$\Rightarrow \mathrm{y}.(\sec \mathrm{x}+\tan \mathrm{x})=\int \tan \mathrm{x}(\sec \mathrm{x}+\tan \mathrm{x})\mathrm{dx}$$$$\Rightarrow \mathrm{y}.(\sec \mathrm{x}+\tan \mathrm{x})=\int \frac{\sin \mathrm{x}\mathrm{dx}}{\cos {}^2\mathrm{x}}+\int \tan {}^2\mathrm{x}\mathrm{dx}$$$$\Rightarrow \mathrm{y}.(\sec \mathrm{x}+\tan \mathrm{x})=\sec \mathrm{x}+\int (\sec {}^2\mathrm{x}-1)\mathrm{dx}$$$$\Rightarrow \mathrm{y}.(\sec \mathrm{x}+\tan \mathrm{x})=\sec \mathrm{x}+\tan \mathrm{x}-\mathrm{x}+\mathrm{c}$$$$\therefore \mathrm{y}=\frac{\sec \mathrm{x}+\tan \mathrm{x}-\mathrm{x}+\mathrm{c}}{\sec \mathrm{x}+\tan \mathrm{x}}$$

Commented by jagoll last updated on 27/Feb/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by john santu last updated on 27/Feb/20

$$\mathrm{ok}$$

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