# differential equations: $y(9x - 2y)dx - x(6x - y)dy = 0$

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Sydney Sales
differential equations: $y(9x - 2y)dx - x(6x - y)dy = 0$

y(9x -2y)dx - x(6x - y) dy = 0

Jhun Vert

$y(9x - 2y) \, dx - x(6x - y) \, dy = 0$

Ley
y = vx
dy = v dx + x dv

$vx(9x - 2vx) \, dx - x(6x - vx)(v \, dx + x \, dv) = 0$

$vx(9x - 2vx) \, dx - vx(6x - vx)\, dx - x^2(6x - vx) \, dv = 0$

$vx(3x - 2vx) \, dx - x^2(6x - vx) \, dv = 0$

$vx^2(3 - 2v) \, dx - x^3(6 - v) \, dv = 0$

$\dfrac{x^2 \, dx}{x^3} - \dfrac{(6 - v) \, dv}{v(3 - 2v)} = 0$

$\dfrac{dx}{x} - \dfrac{(6 - v) \, dv}{v(3 - 2v)} = 0$

For
$\dfrac{6 - v}{v(3 - 2v)} = \dfrac{A}{v} + \dfrac{B}{3 - 2v}$

$6 - v = A(3 - 2v) + Bv$
When v = 0, A = 2
When v = 3/2, B = 3

$\dfrac{dx}{x} - \left( \dfrac{2}{v} + \dfrac{3}{3 - 2v} \right) \, dv = 0$

$\ln x - 2 \ln v + \frac{3}{2} \ln (3 - 2v) = ln c$

$\ln x - \ln v^2 + \ln (3 - 2v)^{3/2} = ln c$

$\ln \dfrac{x(3 - 2v)^{3/2}}{v^2} = ln c$

$\dfrac{x(3 - 2v)^{3/2}}{v^2} = c$

$\dfrac{x\left( 3 - \dfrac{2y}{x} \right)^{3/2}}{\dfrac{y^2}{x^2}} = c$

$\dfrac{x^3\left( \dfrac{3x - 2y}{x} \right)^{3/2}}{y^2} = c$

$\dfrac{x^3\left[ \dfrac{(3x - 2y)^{3/2}}{x^{3/2}} \right]}{y^2} = c$

$\dfrac{x^{3/2}(3x - 2y)^{3/2}}{y^2} = c$

$x^{3/2}(3x - 2y)^{3/2} = cy^2$           answer

Sydney Sales

parang ganito po yung factor sa,
vx (9 x^2 - 2vx)dx - vx ( 6x - vx ) dx

9v x^2dx - 2 v^2 x^2dx - 6vx^2 dx + v^2x^2dx

(3x^2dv - v^2 x^2dx)

Sydney Sales

bakit po naging

vx (3x -2vx) dx

Jhun Vert

Corrections:
$y(9x - 2y) \, dx - x(6x - y) \, dy = 0$

Let
y = vx
dy = v dx + x dv

$vx(9x - 2vx) \, dx - x(6x - vx)(v \, dx + x \, dv) = 0$

$vx(9x - 2vx) \, dx - vx(6x - vx)\, dx - x^2(6x - vx) \, dv = 0$

$vx(3x - vx) \, dx - x^2(6x - vx) \, dv = 0$

$vx^2(3 - v) \, dx - x^3(6 - v) \, dv = 0$

$\dfrac{x^2 \, dx}{x^3} - \dfrac{(6 - v) \, dv}{v(3 - v)} = 0$

$\dfrac{dx}{x} - \dfrac{(6 - v) \, dv}{v(3 - v)} = 0$

For
$\dfrac{6 - v}{v(3 - v)} = \dfrac{A}{v} + \dfrac{B}{3 - v}$

$6 - v = A(3 - v) + Bv$
When v = 0, A = 2
When v = 3, B = 1

$\dfrac{dx}{x} - \left( \dfrac{2}{v} + \dfrac{1}{3 - v} \right) \, dv = 0$

$\ln x - 2 \ln v + \ln (3 - v) = ln c$

$\ln x - \ln v^2 + \ln (3 - v) = ln c$

$\ln \dfrac{x(3 - v)}{v^2} = ln c$

$\dfrac{x(3 - v)}{v^2} = c$

$\dfrac{x\left( 3 - \dfrac{y}{x} \right)}{\dfrac{y^2}{x^2}} = c$

$\dfrac{x^3\left( \dfrac{3x - y}{x} \right)}{y^2} = c$

$\dfrac{x^2(3x - y)}{y^2} = c$

$x^2(3x - y) = cy^2$           answer

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