(3x−y+1) dx +(6x+2y−3) dy = 0


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Question Number 119290 by bobhans last updated on 23/Oct/20

$(3 x - y + 1) d x + (6 x + 2 y - 3) d y = 0$
	Answered by TANMAY PANACEA last updated on 23/Oct/20

	$d y (6 x + 2 y - 3) = (- 3 x + y - 1) d x$ $\frac{d y}{d x} = \frac{- 3 x + y - 1}{6 x + 2 y - 3}$ $x = X + h$ $y = Y + k$ $\frac{d Y}{d X} = \frac{- 3 (X + h) + (Y + k) - 1}{6 (X + h) + 2 (Y + k) - 3}$ $\frac{d Y}{d X} = \frac{- 3 X + Y}{6 X + 2 Y}$ $n o t e - 3 h + k - 1 = 0$ $a n d 6 h + 2 k - 3 = 0$ $- 3 h + k - 1 = 0 \times 2$ $- 6 h + 2 k - 2 = 0$ $6 h + 2 k - 3 = 0$ $s o 4 k = 5 \to k = \frac{5}{4}$ $12 h - 1 = 0 \to h = \frac{1}{12}$ $\frac{d Y}{d X} = \frac{- 3 X + Y}{6 X + 2 Y} = \frac{- 3 + \frac{Y}{X}}{6 + \frac{2 Y}{X}} \to (h o m o g e n e o u s f u n c t i o n)$ $\frac{Y}{X} = v \to \frac{d Y}{d X} = v + X \frac{d v}{d X}$ $v + X \frac{d v}{d X} = \frac{- 3 + v}{6 + 2 v}$ $X \frac{d v}{d X} = \frac{- 3 + v - 6 v - 2 v^{2}}{6 + 2 v}$ $\frac{6 + 2 v}{- 2 v^{2} - 5 v - 3} d v = \frac{d X}{X}$ $\frac{6 + 2 v}{2 v^{2} + 5 v + 3} + \frac{d X}{X} = d C$ $\frac{1}{2} \int \frac{4 v + 5 + 7}{2 v^{2} + 5 v + 3} d v + \int \frac{d X}{X} = \int d C$ $\frac{1}{2} \int \frac{d (2 v^{2} + 5 v + 3)}{2 v^{2} + 5 v + 3} + \frac{7}{4} \int \frac{d v}{v^{2} + \frac{5 v}{2} + \frac{3}{2}} + \int \frac{d X}{X} = \int d C$ $\frac{1}{2} \int \frac{d (2 v^{2} + 5 v + 3)}{2 v^{2} + 5 v + 3} + \frac{7}{4} \int \frac{d v}{v^{2} + 2 \times v \times \frac{5}{4} + \frac{25}{16} + \frac{3}{2} - \frac{25}{16}} + \int \frac{d X}{X} = \int d C$ $\frac{1}{2} \int \frac{d (2 v^{2} + 5 v + 3)}{2 v^{2} + 5 v + 3} + \frac{7}{4} \int \frac{d v}{{(v + \frac{5}{4})}^{2} - {(\frac{1}{4})}^{2}} + \int \frac{d X}{X} = \int d C$ $\frac{1}{2} l n (2 v^{2} + 5 v + 3) + \frac{7}{4} \int p l s u s e f o r m u l a + l n X = C$ $n o w v = \frac{Y}{X} = \frac{y - k}{X - h} = \frac{y - \frac{5}{4}}{x - \frac{1}{12}}$
	Commented by Dwaipayan Shikari last updated on 23/Oct/20

	$\frac{7}{4} \int \frac{d v}{{(v + \frac{5}{4})}^{2} - {(\frac{1}{4})}^{2}} = \frac{7}{2} l o g (\frac{v + \frac{5}{4} - \frac{1}{4}}{v + \frac{6}{4}}) = \frac{7}{2} l o g (\frac{2 v + 2}{2 v + 3})$
	Commented by TANMAY PANACEA last updated on 23/Oct/20

	$t h a n k y o u$
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