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Question Number 97799 by abdomathmax last updated on 09/Jun/20

$$\mathrm{solve}{\mathrm{y}}^{{}^{\prime }}\mathrm{cosx}+\mathrm{y}\mathrm{sinx}=\mathrm{cosx}+\mathrm{sinx}$$

Commented by john santu last updated on 10/Jun/20

$$\frac{dy}{dx}+\tan x.y=1+\tan x$$$$\mathrm{IF}u\left(x\right)=e^{\int \tan xdx}=e^{\ln \left(\sec x\right)}=\sec x$$$$\iff \sec x\frac{dy}{dx}+\tan x\sec x.y=\sec x(1+\tan x)$$$$\int \frac{d}{dx}(y.\sec x)=\int \sec x(1+\tan x)dx$$$$y.\sec x=\ln \mid \sec x+\tan x\mid +\sec x+c$$$$y=\cos x\ln \mid \sec x+\tan x\mid +\mathrm{c}.\cos x+1$$

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