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Question Number 103759 by bemath last updated on 17/Jul/20

(x^5 +3y) dx −x dy = 0

$$\left({x}^{\mathrm{5}} +\mathrm{3}{y}\right)\:{dx}\:−{x}\:{dy}\:=\:\mathrm{0}\: \\ $$

Answered by MAB last updated on 17/Jul/20

x^5 +3y−xy′=0  xy′−3y=x^5   z=(y/x^3 )→z′=((xy′−3y)/x^4 )  x^4 z′=x^5   z′=x  z=(1/2)x^2 +c          (c∈C)  y=x^3 z=(1/2)x^5 +cx^3

$${x}^{\mathrm{5}} +\mathrm{3}{y}−{xy}'=\mathrm{0} \\ $$$${xy}'−\mathrm{3}{y}={x}^{\mathrm{5}} \\ $$$${z}=\frac{{y}}{{x}^{\mathrm{3}} }\rightarrow{z}'=\frac{{xy}'−\mathrm{3}{y}}{{x}^{\mathrm{4}} } \\ $$$${x}^{\mathrm{4}} {z}'={x}^{\mathrm{5}} \\ $$$${z}'={x} \\ $$$${z}=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +{c}\:\:\:\:\:\:\:\:\:\:\left({c}\in\mathbb{C}\right) \\ $$$${y}={x}^{\mathrm{3}} {z}=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{5}} +{cx}^{\mathrm{3}} \\ $$

Answered by john santu last updated on 17/Jul/20

x (dy/dx) −3y = x^5    (dy/dx)− (3/x)y = x^4    Integrating factor   IF u(x) = e^(∫−(3/x) dx) = e^(ln((1/x^3 )))   u(x) = (1/x^3 ) ⇒y = ((∫(1/x^3 ).x^4  dx+C)/(1/x^3 ))  y = x^3 {(1/2)x^2 +C} = (1/2)x^5 +Cx^3   (JS ⊛)

$${x}\:\frac{{dy}}{{dx}}\:−\mathrm{3}{y}\:=\:{x}^{\mathrm{5}} \: \\ $$$$\frac{{dy}}{{dx}}−\:\frac{\mathrm{3}}{{x}}{y}\:=\:{x}^{\mathrm{4}} \: \\ $$$${Integrating}\:{factor}\: \\ $$$${IF}\:{u}\left({x}\right)\:=\:{e}^{\int−\frac{\mathrm{3}}{{x}}\:{dx}} =\:{e}^{\mathrm{ln}\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)} \\ $$$${u}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:\Rightarrow{y}\:=\:\frac{\int\frac{\mathrm{1}}{{x}^{\mathrm{3}} }.{x}^{\mathrm{4}} \:{dx}+{C}}{\frac{\mathrm{1}}{{x}^{\mathrm{3}} }} \\ $$$${y}\:=\:{x}^{\mathrm{3}} \left\{\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +{C}\right\}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{5}} +{Cx}^{\mathrm{3}} \\ $$$$\left({JS}\:\circledast\right)\: \\ $$

Commented by Ar Brandon last updated on 17/Jul/20

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Answered by bemath last updated on 17/Jul/20

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