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Question Number 67349 by Joel122 last updated on 26/Aug/19

Solve for y(x)  xy′ = y + 2x^3 sin^2 ((y/x))

$$\mathrm{Solve}\:\mathrm{for}\:{y}\left({x}\right) \\ $$$${xy}'\:=\:{y}\:+\:\mathrm{2}{x}^{\mathrm{3}} \mathrm{sin}^{\mathrm{2}} \left(\frac{{y}}{{x}}\right) \\ $$

Answered by mind is power last updated on 26/Aug/19

let u=(y/x)  ⇒u^′ =((y^′ x−y)/x^2 )=(y^′ /x^ )−(y/x^2 )  xy^, =y+2x^3 sin^2 ((y/x))⇒((y′)/x^ )−(y/x^2 )=2xsin^2 ((y/x))  ⇒u^′ =2xsin^2 (u)  ⇒(u^′ /(2sin^2 (u)))=x  ⇒∫(du/(2sin^2 (u)))=∫xdx  ⇒∫((sin^2 (u)+cos^2 (u))/(2sin^2 (u)))du=(x^2 /2)+c  ∫(1/2)(1+cot^2 (u))du=(x^2 /2)+c  ⇒−(1/2)cot(u)=(x^2 /2)+c  ⇒tg((π/2)−u)=−x^2 +c  ⇒u=(π/2)−tan^(−1) (−x^2 +c)  ⇒(y/x)=(π/2)−tan^(−1) (−x^2 +c)⇒y=x((π/2)−tan^(−1) (−x^2 +c))  ⇒(y/x)

$${let}\:{u}=\frac{{y}}{{x}} \\ $$$$\Rightarrow{u}^{'} =\frac{{y}^{'} {x}−{y}}{{x}^{\mathrm{2}} }=\frac{{y}^{'} }{{x}^{} }−\frac{{y}}{{x}^{\mathrm{2}} } \\ $$$${xy}^{,} ={y}+\mathrm{2}{x}^{\mathrm{3}} {sin}^{\mathrm{2}} \left(\frac{{y}}{{x}}\right)\Rightarrow\frac{{y}'}{{x}^{} }−\frac{{y}}{{x}^{\mathrm{2}} }=\mathrm{2}{xsin}^{\mathrm{2}} \left(\frac{{y}}{{x}}\right) \\ $$$$\Rightarrow{u}^{'} =\mathrm{2}{xsin}^{\mathrm{2}} \left({u}\right) \\ $$$$\Rightarrow\frac{{u}^{'} }{\mathrm{2}{sin}^{\mathrm{2}} \left({u}\right)}={x} \\ $$$$\Rightarrow\int\frac{{du}}{\mathrm{2}{sin}^{\mathrm{2}} \left({u}\right)}=\int{xdx} \\ $$$$\Rightarrow\int\frac{{sin}^{\mathrm{2}} \left({u}\right)+{cos}^{\mathrm{2}} \left({u}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left({u}\right)}{du}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$$$\int\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{cot}^{\mathrm{2}} \left({u}\right)\right){du}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$$$\Rightarrow−\frac{\mathrm{1}}{\mathrm{2}}{cot}\left({u}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$$$\Rightarrow{tg}\left(\frac{\pi}{\mathrm{2}}−{u}\right)=−{x}^{\mathrm{2}} +{c} \\ $$$$\Rightarrow{u}=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left(−{x}^{\mathrm{2}} +{c}\right) \\ $$$$\Rightarrow\frac{{y}}{{x}}=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left(−{x}^{\mathrm{2}} +{c}\right)\Rightarrow{y}={x}\left(\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left(−{x}^{\mathrm{2}} +{c}\right)\right) \\ $$$$\Rightarrow\frac{{y}}{{x}} \\ $$$$ \\ $$

Commented by Joel122 last updated on 26/Aug/19

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Commented by mind is power last updated on 26/Aug/19

y′re welcom

$${y}'{re}\:{welcom} \\ $$

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