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


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

$y^{'} - y . \tan x = \cos x - 2 x \sin x$ $y (\frac{π}{6}) = 0 \Rightarrow y (x) = ?$
	Answered by bemath last updated on 25/Feb/21

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

	$y^{'} - y t a n x = c o s x - 2 x s i n x$ $\Rightarrow y^{'} c o s x - y s i n x = {c o s}^{2} x - 2 x s i n x c o s x$ $\Rightarrow {(y c o s x)}^{'} = {c o s}^{2} x - x s i n 2 x$ $y = x c o s x + \frac{C}{c o s x}$ $y (π / 6) = \sqrt{3} π / 12 + \frac{2 C}{\sqrt{3}} = 0$ $C = - \frac{π}{8}$ $y = x c o s x - \frac{π}{8 c o s x}$
	Answered by mathmax by abdo last updated on 25/Feb/21
	$h→y^′ −ytanx =0 ⇒(y^′ /y)=tanx ⇒ln∣y∣=∫ ((sinx)/(cosx))dx =−ln∣cosx∣ +c ⇒ y =k.(1/(∣cosx∣)) solution on {x ∣cosx>0} ⇒y =(k/(cosx)) ⇒ y^′ =(k^′ /(cosx))+((ksinx)/(cos^2 x)) mvc method →(k^′ /(cosx))+((ksinx)/(cos^2 x))−(k/(cosx))((sinx)/(cosx)) =cosx−2xsinx ⇒k^′ =cos^2 x−2xsinx cosx ⇒ k =∫ cos^2 x dx−∫ xsin(2x)dx but ∫ cos^2 x dx =∫ ((1+cos(2x))/2)dx =(x/2)+(1/4)sin(2x) +c_0 ∫ xsin(2x)dx =_(by parts) −(x/2)cos(2x)+(1/2)∫ cos(2x)dx =−(x/2)cos(2x)+(1/4)sin(2x) +c_1 ⇒ k =(x/2)+(1/4)sin(2x)+(x/2)cos(2x)−(1/4)sin(2x) +λ ⇒ y(x)=(1/(cosx)){(x/2)+(x/2)cos(2x)+λ} y((π/6))=0 ⇒(2/( (√3)))((π/(12))+(π/(12))×(1/2)+λ)=0 ⇒λ=((3π)/(12))=(π/4) ⇒ y(x)=(1/(cosx))((x/2)+(x/2)cos(2x)+(π/4))$
	$h \to y^{^{'}} - ytanx = 0 \Rightarrow \frac{y^{^{'}}}{y} = tanx \Rightarrow \ln ∣ y ∣= \int \frac{sinx}{cosx} dx = - \ln ∣ cosx ∣ + c \Rightarrow$ $y = k . \frac{1}{∣ cosx ∣} solution on {x ∣ cosx > 0} \Rightarrow y = \frac{k}{cosx} \Rightarrow$ $y^{^{'}} = \frac{k^{^{'}}}{cosx} + \frac{ksinx}{\cos^{2} x} mvc method \to \frac{k^{^{'}}}{cosx} + \frac{ksinx}{\cos^{2} x} - \frac{k}{cosx} \frac{sinx}{cosx}$ $= cosx - 2 xsinx \Rightarrow k^{^{'}} = \cos^{2} x - 2 xsinx cosx \Rightarrow$ $k = \int \cos^{2} x dx - \int xsin (2 x) dx but$ $\int \cos^{2} x dx = \int \frac{1 + \cos (2 x)}{2} dx = \frac{x}{2} + \frac{1}{4} \sin (2 x) + c_{0}$ $\int xsin (2 x) dx =_{by parts} - \frac{x}{2} \cos (2 x) + \frac{1}{2} \int \cos (2 x) dx$ $= - \frac{x}{2} \cos (2 x) + \frac{1}{4} \sin (2 x) + c_{1} \Rightarrow$ $k = \frac{x}{2} + \frac{1}{4} \sin (2 x) + \frac{x}{2} \cos (2 x) - \frac{1}{4} \sin (2 x) + λ \Rightarrow$ $y (x) = \frac{1}{cosx} {\frac{x}{2} + \frac{x}{2} \cos (2 x) + λ}$ $y (\frac{π}{6}) = 0 \Rightarrow \frac{2}{\sqrt{3}} (\frac{π}{12} + \frac{π}{12} \times \frac{1}{2} + λ) = 0 \Rightarrow λ = \frac{3 π}{12} = \frac{π}{4} \Rightarrow$ $y (x) = \frac{1}{cosx} (\frac{x}{2} + \frac{x}{2} \cos (2 x) + \frac{π}{4})$
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