(x^5 +3y) dx −x dy = 0


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

$(x^{5} + 3 y) d x - x d y = 0$
	Answered by MAB last updated on 17/Jul/20

	$x^{5} + 3 y - {x y}^{'} = 0$ ${x y}^{'} - 3 y = x^{5}$ $z = \frac{y}{x^{3}} \to z^{'} = \frac{{x y}^{'} - 3 y}{x^{4}}$ $x^{4} z^{'} = x^{5}$ $z^{'} = x$ $z = \frac{1}{2} x^{2} + c (c \in C)$ $y = x^{3} z = \frac{1}{2} x^{5} + {c x}^{3}$
	Answered by john santu last updated on 17/Jul/20
	$x (dy/dx) −3y = x^5 (dy/dx)− (3/x)y = x^4 Integrating factor IF u(x) = e^(∫−(3/x) dx) = e^(ln((1/x^3 ))) u(x) = (1/x^3 ) ⇒y = ((∫(1/x^3 ).x^4 dx+C)/(1/x^3 )) y = x^3 {(1/2)x^2 +C} = (1/2)x^5 +Cx^3 (JS ⊛)$
	$x \frac{d y}{d x} - 3 y = x^{5}$ $\frac{d y}{d x} - \frac{3}{x} y = x^{4}$ $I n t e g r a t i n g f a c t o r$ $I F u (x) = e^{\int - \frac{3}{x} d x} = e^{\ln (\frac{1}{x^{3}})}$ $u (x) = \frac{1}{x^{3}} \Rightarrow y = \frac{\int \frac{1}{x^{3}} . x^{4} d x + C}{\frac{1}{x^{3}}}$ $y = x^{3} {\frac{1}{2} x^{2} + C} = \frac{1}{2} x^{5} + {C x}^{3}$ $(J S ⊛)$
	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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