Menu Close

y-cos-x-y-cos-2-x-




Question Number 144356 by Ar Brandon last updated on 24/Jun/21
y′+cos(x)y=cos^2 x
$$\mathrm{y}'+\mathrm{cos}\left(\mathrm{x}\right)\mathrm{y}=\mathrm{cos}^{\mathrm{2}} \mathrm{x} \\ $$
Answered by Olaf_Thorendsen last updated on 24/Jun/21
y′+cos(x)y = cos^2 x   (1)  y = e^(−sinx) u  (1) : e^(−sinx) u′−cos(x)e^(−sinx) u+cos(x)e^(−sinx) u = cos^2 x  e^(−sinx) u′ = cos^2 x  u′ = cos^2 (x)e^(sinx)   u = ∫cos(x)(cos(x)e^(sinx) ) dx  u = cos(x)e^(sinx) +∫sin(x)e^(sinx)  dx  ⇒ y = cos(x)+e^(−sinx) ∫sin(x)e^(sinx)  dx
$${y}'+\mathrm{cos}\left({x}\right){y}\:=\:\mathrm{cos}^{\mathrm{2}} {x}\:\:\:\left(\mathrm{1}\right) \\ $$$${y}\:=\:{e}^{−\mathrm{sin}{x}} {u} \\ $$$$\left(\mathrm{1}\right)\::\:{e}^{−\mathrm{sin}{x}} {u}'−\mathrm{cos}\left({x}\right){e}^{−\mathrm{sin}{x}} {u}+\mathrm{cos}\left({x}\right){e}^{−\mathrm{sin}{x}} {u}\:=\:\mathrm{cos}^{\mathrm{2}} {x} \\ $$$${e}^{−\mathrm{sin}{x}} {u}'\:=\:\mathrm{cos}^{\mathrm{2}} {x} \\ $$$${u}'\:=\:\mathrm{cos}^{\mathrm{2}} \left({x}\right){e}^{\mathrm{sin}{x}} \\ $$$${u}\:=\:\int\mathrm{cos}\left({x}\right)\left(\mathrm{cos}\left({x}\right){e}^{\mathrm{sin}{x}} \right)\:{dx} \\ $$$${u}\:=\:\mathrm{cos}\left({x}\right){e}^{\mathrm{sin}{x}} +\int\mathrm{sin}\left({x}\right){e}^{\mathrm{sin}{x}} \:{dx} \\ $$$$\Rightarrow\:{y}\:=\:\mathrm{cos}\left({x}\right)+{e}^{−\mathrm{sin}{x}} \int\mathrm{sin}\left({x}\right){e}^{\mathrm{sin}{x}} \:{dx} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *