y′ + y = x (y^2 )^(1/(3 ))


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Question Number 96065 by john santu last updated on 29/May/20

$y^{'} + y = x \sqrt[3]{y^{2}}$
	Answered by bobhans last updated on 29/May/20

	$Bernoulli eq$ $v = y^{1 - \frac{2}{3}} = y^{\frac{1}{3}} \Rightarrow \frac{dv}{dx} = \frac{1}{3} y^{- \frac{2}{3}} . \frac{dy}{dx}$ $\frac{dy}{dx} = {3 y}^{\frac{2}{3}} \frac{dv}{dx}$ $\Rightarrow {3 y}^{\frac{2}{3}} v^{'} + y = {xy}^{\frac{2}{3}}$ $v^{'} + \frac{1}{3} v = \frac{1}{3} x .$ $integrating factor u (x) = e^{\int \frac{1}{3} dx}$ $u (x) = e^{\frac{1}{3} x} \Rightarrow e^{\frac{1}{3} x} v = \int \frac{1}{3} {xe}^{\frac{1}{3} x} dx$ ${3 e}^{\frac{1}{3} x} y^{\frac{1}{3}} = {xe}^{\frac{1}{3} x} - {3 e}^{\frac{1}{3} x} + C$ $y^{\frac{1}{3}} = \frac{1}{3} x - 1 + {Ce}^{- \frac{1}{3} x} ∙ .$
	Commented by john santu last updated on 29/May/20
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	Answered by abdomathmax last updated on 29/May/20
	$(e) ⇒y^′ +y =x y^(2/3) ⇒(y^′ /y^(2/3) ) +(y/y^(2/3) ) =x let z =y^(1−(2/3)) =y^(1/(3 )) ⇒z^3 =y ⇒y^′ =3z^2 z^′ (e) ⇒3z^2 z^′ ×(z^3 )^(−(2/3)) +z =x ⇒ 3z^′ +z =x (he)→3z^′ =−z ⇒(z^′ /z)=−(1/3) ⇒ln∣z∣ =−(1/3)x +c ⇒ z =k e^(−(x/3)) lagrange method →z^′ =k^′ e^(−(x/3)) −(1/3)k e^(−(x/3)) (e)⇒3k^′ e^(−(x/3)) −ke^(−(x/3)) +k e^(−(x/3)) =x ⇒ 3k^′ =x e^(x/3) ⇒k^′ =(1/3)x e^(x/3) ⇒ k =∫ x((1/3)e^(x/3) )dx =xe^(x/3) −∫e^(x/3) dx =xe^(x/3) −3e^(x/3) =(x−3)e^(x/3) +c ⇒ z(x) ={(x−3)e^(x/3) +c}e^(−(x/3)) =x−3 +ce^(−(x/3)) y =z^3 ⇒y =(x−3 +ce^(−(x/3)) )^3$
	$(e) \Rightarrow y^{^{'}} + y = x y^{\frac{2}{3}} \Rightarrow \frac{y^{^{'}}}{y^{\frac{2}{3}}} + \frac{y}{y^{\frac{2}{3}}} = x let$ $z = y^{1 - \frac{2}{3}} = y^{\frac{1}{3}} \Rightarrow z^{3} = y \Rightarrow y^{^{'}} = {3 z}^{2} z^{^{'}}$ $(e) \Rightarrow {3 z}^{2} z^{^{'}} \times {(z^{3})}^{- \frac{2}{3}} + z = x \Rightarrow$ ${3 z}^{^{'}} + z = x$ $(he) \to {3 z}^{^{'}} = - z \Rightarrow \frac{z^{^{'}}}{z} = - \frac{1}{3} \Rightarrow \ln ∣ z ∣ = - \frac{1}{3} x + c \Rightarrow$ $z = k e^{- \frac{x}{3}} lagrange method \to z^{^{'}} = k^{^{'}} e^{- \frac{x}{3}}$ $- \frac{1}{3} k e^{- \frac{x}{3}}$ $(e) \Rightarrow {3 k}^{^{'}} e^{- \frac{x}{3}} - {ke}^{- \frac{x}{3}} + k e^{- \frac{x}{3}} = x \Rightarrow$ ${3 k}^{^{'}} = x e^{\frac{x}{3}} \Rightarrow k^{^{'}} = \frac{1}{3} x e^{\frac{x}{3}} \Rightarrow$ $k = \int x (\frac{1}{3} e^{\frac{x}{3}}) dx = {xe}^{\frac{x}{3}} - \int e^{\frac{x}{3}} dx$ $= {xe}^{\frac{x}{3}} - {3 e}^{\frac{x}{3}} = (x - 3) e^{\frac{x}{3}} + c \Rightarrow$ $z (x) = {(x - 3) e^{\frac{x}{3}} + c} e^{- \frac{x}{3}} = x - 3 + {ce}^{- \frac{x}{3}}$ $y = z^{3} \Rightarrow y = {(x - 3 + {ce}^{- \frac{x}{3}})}^{3}$
	Answered by Sourav mridha last updated on 29/May/20

	$\Rightarrow (y^{\frac{1}{3}} - x) dx + \frac{1}{y^{\frac{2}{3}}} dy = 0$ $\frac{\partial}{\partial y} (y^{\frac{1}{3}} - x) = \frac{1}{3} y^{- \frac{2}{3}} \neq \frac{\partial}{\partial x} (y^{- \frac{2}{3}}) = 0$ $so this is not exact differential$ $now : [\frac{1}{y^{- \frac{2}{3}}} (\frac{1}{3} y^{- \frac{2}{3}} - 0) = \frac{1}{3}]$ $s o I . F = e^{\int \frac{d x}{3}} = e^{\frac{x}{3}} m u l t i : withthis w e$ $g e t : e^{\frac{x}{3}} [y^{\frac{1}{3}} - x] d x + e^{\frac{x}{3}} . y^{- \frac{2}{3}} d y = 0$ $now : \int_{(k e e p i n g y as constant)} e^{\frac{x}{3}} [y^{\frac{1}{3}} - x] d x$ $= 3 e^{\frac{x}{3}} . y^{\frac{1}{3}} - 3 {x e}^{\frac{x}{3}} + 9 e^{\frac{x}{3}}$ $and : \int_{(which dos n^{'} t contain x)} (0) dy = 0$ $so the solution is :$ $e^{\frac{x}{3}} [y^{\frac{1}{3}} - x + 3] = c$
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