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Question Number 133889 by EDWIN88 last updated on 25/Feb/21

y′−y.tan x = cos x−2x sin x y((π/6)) = 0 ⇒y(x)=?

$$\:\mathrm{y}'−\mathrm{y}.\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{cos}\:\mathrm{x}−\mathrm{2x}\:\mathrm{sin}\:\mathrm{x} \\ $$$$\mathrm{y}\left(\frac{\pi}{\mathrm{6}}\right)\:=\:\mathrm{0}\:\Rightarrow\mathrm{y}\left(\mathrm{x}\right)=? \\ $$

Answered by bemath last updated on 25/Feb/21

Answered by Ñï= last updated on 25/Feb/21

y′−ytanx=cosx−2xsinx ⇒y′cosx−ysinx=cos^2 x−2xsinxcosx ⇒(ycosx)′=cos^2 x−xsin2x y=xcosx+(C/(cosx)) y(π/6)=(√3)π/12+((2C)/( (√3)))=0 C=−(π/8) y=xcosx−(π/(8cosx))

$${y}'−{ytanx}={cosx}−\mathrm{2}{xsinx} \\ $$$$\Rightarrow{y}'{cosx}−{ysinx}={cos}^{\mathrm{2}} {x}−\mathrm{2}{xsinxcosx} \\ $$$$\Rightarrow\left({ycosx}\right)'={cos}^{\mathrm{2}} {x}−{xsin}\mathrm{2}{x} \\ $$$${y}={xcosx}+\frac{{C}}{{cosx}} \\ $$$${y}\left(\pi/\mathrm{6}\right)=\sqrt{\mathrm{3}}\pi/\mathrm{12}+\frac{\mathrm{2}{C}}{\:\sqrt{\mathrm{3}}}=\mathrm{0} \\ $$$${C}=−\frac{\pi}{\mathrm{8}} \\ $$$${y}={xcosx}−\frac{\pi}{\mathrm{8}{cosx}} \\ $$

Answered by mathmax by abdo last updated on 25/Feb/21

h→y^′ −ytanx =0 ⇒(y^′ /y)=tanx ⇒ln∣y∣=∫ ((sinx)/(cosx))dx =−ln∣cosx∣ +c ⇒ y =k.(1/(∣cosx∣)) solution on {x ∣cosx>0} ⇒y =(k/(cosx)) ⇒ y^′ =(k^′ /(cosx))+((ksinx)/(cos^2 x)) mvc method →(k^′ /(cosx))+((ksinx)/(cos^2 x))−(k/(cosx))((sinx)/(cosx)) =cosx−2xsinx ⇒k^′ =cos^2 x−2xsinx cosx ⇒ k =∫ cos^2 x dx−∫ xsin(2x)dx but ∫ cos^2 x dx =∫ ((1+cos(2x))/2)dx =(x/2)+(1/4)sin(2x) +c_0 ∫ xsin(2x)dx =_(by parts) −(x/2)cos(2x)+(1/2)∫ cos(2x)dx =−(x/2)cos(2x)+(1/4)sin(2x) +c_1 ⇒ k =(x/2)+(1/4)sin(2x)+(x/2)cos(2x)−(1/4)sin(2x) +λ ⇒ y(x)=(1/(cosx)){(x/2)+(x/2)cos(2x)+λ} y((π/6))=0 ⇒(2/( (√3)))((π/(12))+(π/(12))×(1/2)+λ)=0 ⇒λ=((3π)/(12))=(π/4) ⇒ y(x)=(1/(cosx))((x/2)+(x/2)cos(2x)+(π/4))

$$\mathrm{h}\rightarrow\mathrm{y}^{'} −\mathrm{ytanx}\:=\mathrm{0}\:\Rightarrow\frac{\mathrm{y}^{'} }{\mathrm{y}}=\mathrm{tanx}\:\Rightarrow\mathrm{ln}\mid\mathrm{y}\mid=\int\:\frac{\mathrm{sinx}}{\mathrm{cosx}}\mathrm{dx}\:=−\mathrm{ln}\mid\mathrm{cosx}\mid\:+\mathrm{c}\:\Rightarrow \\ $$$$\mathrm{y}\:=\mathrm{k}.\frac{\mathrm{1}}{\mid\mathrm{cosx}\mid}\:\:\mathrm{solution}\:\mathrm{on}\:\left\{\mathrm{x}\:\mid\mathrm{cosx}>\mathrm{0}\right\}\:\Rightarrow\mathrm{y}\:=\frac{\mathrm{k}}{\mathrm{cosx}}\:\Rightarrow \\ $$$$\mathrm{y}^{'} \:=\frac{\mathrm{k}^{'} }{\mathrm{cosx}}+\frac{\mathrm{ksinx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:\:\mathrm{mvc}\:\mathrm{method}\:\rightarrow\frac{\mathrm{k}^{'} }{\mathrm{cosx}}+\frac{\mathrm{ksinx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}−\frac{\mathrm{k}}{\mathrm{cosx}}\frac{\mathrm{sinx}}{\mathrm{cosx}} \\ $$$$=\mathrm{cosx}−\mathrm{2xsinx}\:\Rightarrow\mathrm{k}^{'} \:=\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{2xsinx}\:\mathrm{cosx}\:\Rightarrow \\ $$$$\mathrm{k}\:=\int\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}−\int\:\mathrm{xsin}\left(\mathrm{2x}\right)\mathrm{dx}\:\:\mathrm{but} \\ $$$$\int\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}\:=\int\:\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}\mathrm{dx}\:=\frac{\mathrm{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2x}\right)\:+\mathrm{c}_{\mathrm{0}} \\ $$$$\int\:\mathrm{xsin}\left(\mathrm{2x}\right)\mathrm{dx}\:=_{\mathrm{by}\:\mathrm{parts}} \:\:−\frac{\mathrm{x}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{2x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int\:\mathrm{cos}\left(\mathrm{2x}\right)\mathrm{dx} \\ $$$$=−\frac{\mathrm{x}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{2x}\right)+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2x}\right)\:+\mathrm{c}_{\mathrm{1}} \:\Rightarrow \\ $$$$\mathrm{k}\:=\frac{\mathrm{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2x}\right)+\frac{\mathrm{x}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{2x}\right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2x}\right)\:+\lambda\:\Rightarrow \\ $$$$\mathrm{y}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{cosx}}\left\{\frac{\mathrm{x}}{\mathrm{2}}+\frac{\mathrm{x}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{2x}\right)+\lambda\right\} \\ $$$$\mathrm{y}\left(\frac{\pi}{\mathrm{6}}\right)=\mathrm{0}\:\Rightarrow\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left(\frac{\pi}{\mathrm{12}}+\frac{\pi}{\mathrm{12}}×\frac{\mathrm{1}}{\mathrm{2}}+\lambda\right)=\mathrm{0}\:\Rightarrow\lambda=\frac{\mathrm{3}\pi}{\mathrm{12}}=\frac{\pi}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{y}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{cosx}}\left(\frac{\mathrm{x}}{\mathrm{2}}+\frac{\mathrm{x}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{2x}\right)+\frac{\pi}{\mathrm{4}}\right) \\ $$

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