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Answer 1 · 2018-01-26T16:31:19

This is Linear Differential Equation.
$x y = ln (\frac{c}{y})$ $xy= ln(c/y)$

Instead of $\frac{d y}{d x}$ $dy/dx$ we form $\frac{d x}{d y}$ $dx/dy$ because of linear $x$ $x$ in Numerator.

$y^{2} d x + (x y + 1) d y = 0$ $y^2dx+(xy+1)dy=0$

Dividing by $d y$ $dy$

$y^{2} \frac{d x}{d y} + (x y + 1) = 0$ $y^2dx/dy+(xy+1)=0$

Dividing by $y^{2}$ $y^2$

$\frac{d x}{d y} + (\frac{x}{y} + \frac{1}{y^{2}}) = 0$ $dx/dy+(x/y+1/y^2)=0$

$\frac{d x}{d y} + \frac{x}{y} = - \frac{1}{y^{2}}$ $dx/dy +x/y = -1/y^2$

Comparing with:
$\frac{d x}{d y} + P (y) . x = Q (y)$ $dx/dy + P(y).x=Q(y)$

$P (y) = \frac{1}{y}$ $P(y) = 1/y$ and $Q (y) = - \frac{1}{y^{2}}$ $Q(y) = -1/y^2$

Integrating Factor:
$e^{\int \frac{1}{y} d y}$ $e^(int 1/y dy)$

$e^{ln y}$ $e^lny$

i.e $y$ $y$

Now,
$x . I F = \int I F . (- \frac{1}{y^{2}}) d y$ $x . IF = int IF . (-1/y^2) dy$

$x y = \int y (- \frac{1}{y^{2}}) d y$ $x y = int y (-1/y^2) dy$

$x y = \int - \frac{1}{y} d y$ $x y = int -1/y dy$

$x y = - ln y + ln c$ $xy = -ln y + ln c$

$x y = ln (\frac{c}{y})$ $xy = ln(c/y)$