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Question Number 144356 by Ar Brandon last updated on 24/Jun/21

$${\mathrm{y}}^{\prime }+\cos \left(\mathrm{x}\right)\mathrm{y}={\cos }^2\mathrm{x}$$

Answered by Olaf_Thorendsen last updated on 24/Jun/21

$$y^{\prime }+\cos \left(x\right)y={\cos }^{2}x\left(1\right)$$$$y=e^{-\sin x}u$$$$\left(1\right):e^{-\sin x}{u}^{\prime }-\cos \left(x\right)e^{-\sin x}u+\cos \left(x\right)e^{-\sin x}u={\cos }^{2}x$$$$e^{-\sin x}{u}^{\prime }={\cos }^{2}x$$$$u^{\prime }={\cos }^2\left(x\right)e^{\sin x}$$$$u=\int \cos \left(x\right)\left(\cos \left(x\right)e^{\sin x}\right)dx$$$$u=\cos \left(x\right)e^{\sin x}+\int \sin \left(x\right)e^{\sin x}dx$$$$\Rightarrow y=\cos \left(x\right)+e^{-\sin x}\int \sin \left(x\right)e^{\sin x}dx$$

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