Solve in the order of finding the integrsting stroke: 1. ydx−(x^3 y+x)−xdy=0 2. (xy^2 +y)dx−xdy=0


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Question Number 124763 by Mammadli last updated on 05/Dec/20

$S o l v e i n t h e o r d e r o f f i n d i n g t h e i n t e g r s t i n g s t r o k e :$ $1 . y d x - (x^{3} y + x) - x d y = 0$ $2 . ({x y}^{2} + y) d x - x d y = 0$
Commented by Mammadli last updated on 05/Dec/20
Sorry dear ser.. 1. ydx-(x³y+x)dy=0
	Answered by benjo_mathlover last updated on 05/Dec/20

	$2 . \frac{d y}{d x} = \frac{{x y}^{2} + y}{x} = y^{2} + \frac{y}{x}$ $p u t v = \frac{y}{x} \Rightarrow y = v x a n d \frac{d y}{d x} = v + x \frac{d v}{d x}$ $\Leftrightarrow v + x \frac{d v}{d x} = v^{2} x^{2} + v$ $\Rightarrow x \frac{d v}{d x} = v^{2} x^{2} \Rightarrow \frac{d v}{v^{2}} = x d x$ $\int \frac{d v}{v^{2}} = \int x d x \Rightarrow - \frac{1}{v} = \frac{1}{2} x^{2} + c$ $- \frac{1}{v} = \frac{x^{2} + C}{2}; - \frac{x}{y} = \frac{x^{2} + C}{2}$ $y = - \frac{2 x}{x^{2} + C}$
	Commented by Mammadli last updated on 05/Dec/20
	thk you dear ser, please: ydx-(x³y+x)dy=0
	Answered by liberty last updated on 06/Dec/20

	$(1) y d x - (x^{3} y + x) d y = 0$ $l e t x = y v \Rightarrow d x = v d y + y d v$ $\Leftrightarrow y (v d y + y d v) - (y^{3} v^{3} + y v) d y = 0$ $\Leftrightarrow y^{2} d v - y^{3} v^{3} d y = 0$ $\Leftrightarrow d v = y v^{3} d y; \frac{d v}{v^{3}} = y d y$ $\int \frac{d v}{v^{3}} = \int y d y; - \frac{1}{2 v^{2}} = \frac{1}{2} y^{2} + c$ $\frac{y^{2}}{x^{2}} = - y^{2} + C; [- 2 c = C]$ $y^{2} (\frac{1}{x^{2}} + 1) = C; y^{2} = \frac{{C x}^{2}}{x^{2} + 1}$ $y = \pm \frac{λ x}{\sqrt{x^{2} + 1}}; [λ = \sqrt{C}]$
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