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Question Number 37344 by math khazana by abdo last updated on 12/Jun/18

solve the d.e. y^′ −xy =cosx .

$${solve}\:{the}\:{d}.{e}.\:{y}^{'} \:−{xy}\:\:={cosx}\:. \\ $$

Commented by math khazana by abdo last updated on 12/Jun/18

he⇒y^′ −xy=0 ⇒(y^′ /y)=x ⇒ln∣y∣ = (x^2 /2) +c ⇒ y =k e^(x^2 /2) mvc method give y^′ =k^′ e^(x^2 /2) +kx e^(x^2 /2) (e) ⇒k^′ e^(x^2 /2) +kx e^(x^2 /2) −xk e^(x^2 /2) =cosx ⇒ k^′ =e^(−(x^2 /2)) cosx ⇒ k(x) = ∫_. ^x e^(−(t^2 /2)) cost +λ ⇒ y(x)=( ∫_. ^x e^(−(t^2 /2)) cost dt +λ)e^(x^2 /2) ⇒ y(x) = e^(x^2 /2) ∫_. ^x e^(−(t^2 /2)) cost dt +λ e^(x^2 /2) .

$${he}\Rightarrow{y}^{'} −{xy}=\mathrm{0}\:\Rightarrow\frac{{y}^{'} }{{y}}={x}\:\Rightarrow{ln}\mid{y}\mid\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{c}\:\Rightarrow \\ $$$${y}\:={k}\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\:\:\:{mvc}\:{method}\:{give}\: \\ $$$${y}^{'} ={k}^{'} \:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\:+{kx}\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\:\left({e}\right)\:\Rightarrow{k}^{'} {e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\:+{kx}\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} −{xk}\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} ={cosx} \\ $$$$\Rightarrow\:{k}^{'} \:={e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:{cosx}\:\Rightarrow\:{k}\left({x}\right)\:=\:\int_{.} ^{{x}} \:\:{e}^{−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} {cost}\:+\lambda\:\Rightarrow \\ $$$${y}\left({x}\right)=\left(\:\int_{.} ^{{x}} \:\:{e}^{−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} {cost}\:{dt}\:+\lambda\right){e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\Rightarrow \\ $$$${y}\left({x}\right)\:=\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\:\int_{.} ^{{x}} \:\:{e}^{−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} \:{cost}\:{dt}\:\:+\lambda\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\:. \\ $$

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