cos x (dy/dx) + y.sin x = 2cos^3 x.sin x


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Question Number 121378 by liberty last updated on 07/Nov/20

$\cos x \frac{dy}{dx} + y . \sin x = 2 \cos^{3} x . \sin x$
	Answered by Ar Brandon last updated on 07/Nov/20

	$cosx \cdot \frac{dy}{dx} + y \cdot sinx = {2 \cos}^{3} x \cdot sinx$ $secx \frac{dy}{dx} + ysecxtanx = 2 cosxsinx$ $\frac{d (ysecx)}{dx} = \sin 2 x \Rightarrow d (ysecx) = \sin 2 xdx$ $ysecx = \int \sin 2 xdx = - \frac{\cos 2 x}{2} + C$
	Answered by TANMAY PANACEA last updated on 07/Nov/20

	$\frac{d y}{d x} + (t a n x) y = 2 {c o s}^{2} x s i n x$ $I . F e^{\int t a n x d x} = e^{l n s e c x} = s e c x$ $s e c x \frac{d y}{d x} + (s e c x t a n x) y = s e c x \times 2 {c o s}^{2} x s i n x$ $s e c x d y + (s e c x t a n x) y d x = s i n 2 x d x$ $d (y s e c x) = - d (\frac{c o s 2 x}{2}) + d C$ $i n t r e g a t i n g$ $y s e c x = \frac{- c o s 2 x}{2} + C$
	Answered by Ar Brandon last updated on 07/Nov/20

	$secx \frac{dy}{dx} + ysecxtanx = 2 cosxsinx$ $\frac{d (ysecx)}{dx} = 2 sinxcosx$ $ysecx = \int 2 sinxcosxdx$ $= \sin^{2} x + K$ $\Rightarrow y = \sin^{2} xcosx + K cosx$
	Answered by john santu last updated on 07/Nov/20

	Answered by Dwaipayan Shikari last updated on 07/Nov/20

	$\frac{d y}{d x} + y t a n x = 2 {c o s}^{2} x s i n x$ $I . F = e^{\int t a n x} = e^{\int l o g (s e c x)} = s e c x$ $y . s e c x = \int 2 {c o s}^{2} x . s i n x . s e c x$ $y . s e c x = \int s i n 2 x$ $y = - \frac{c o s 2 x c o s x}{2} + C c o s x$
	Answered by mathmax by abdo last updated on 07/Nov/20
	$cosx y^′ +sinx y =2cos^3 x sinx h→ cosxy^′ +sinx y =0 ⇒cosx y^′ =−sinx y ⇒(y^′ /y)=−((sinx)/(cosx)) ⇒ ln∣y∣ =ln∣cosx∣ +c ⇒y =k ∣cosx∣ let determine solution on w={x /cosx>0} ⇒y=kcosx mvc method y^′ =k^′ cosx−ksinx e ⇒k^′ cos^2 x−ksinx cosx +k cosx sinx =2cos^3 x sinx ⇒ k^′ =2cosx sinx ⇒k^′ =sin(2x) ⇒k =−(1/2)cos(2x) +λ ⇒ y(x)=(−(1/2)cos(2x)+λ)cosx =−(1/2)cosx cos(2x)+λcosx$
	$cosx y^{^{'}} + sinx y = {2 \cos}^{3} x sinx$ $h \to {cosxy}^{^{'}} + sinx y = 0 \Rightarrow cosx y^{^{'}} = - sinx y \Rightarrow \frac{y^{^{'}}}{y} = - \frac{sinx}{cosx} \Rightarrow$ $\ln ∣ y ∣ = \ln ∣ cosx ∣ + c \Rightarrow y = k ∣ cosx ∣ let determine solution on$ $w = {x / cosx > 0} \Rightarrow y = kcosx mvc method$ $y^{^{'}} = k^{^{'}} cosx - ksinx$ $e \Rightarrow k^{^{'}} \cos^{2} x - ksinx cosx + k cosx sinx = {2 \cos}^{3} x sinx \Rightarrow$ $k^{^{'}} = 2 cosx sinx \Rightarrow k^{^{'}} = \sin (2 x) \Rightarrow k = - \frac{1}{2} \cos (2 x) + λ \Rightarrow$ $y (x) = (- \frac{1}{2} \cos (2 x) + λ) cosx = - \frac{1}{2} cosx \cos (2 x) + λ cosx$
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