Problem 03 | Separation of Variables

$xy \, y' = 1 + y^2$

$xy \dfrac{dy}{dx} = 1 + y^2$

$\dfrac{y \, dy}{1 + y^2} = \dfrac{dx}{x}$

$\displaystyle \frac{1}{2} \int \dfrac{2y \, dy}{1 + y^2} = \int \dfrac{dx}{x}$

$\frac{1}{2} \ln (1 + y^2) = \ln x + \frac{1}{2} \ln c$

$\ln (1 + y^2) = 2\ln x + \ln c$

$\ln (1 + y^2) = \ln x^2 + \ln c$

$\ln (1 + y^2) = \ln cx^2$

$1 + y^2 = cx^2$
 

when   $x = 2$,   $y = 3$
$1 + 3^2 = c(2^2)$

$c = \frac{10}{4} = \frac{5}{2}$
 

then,
$1 + y^2 = \frac{5}{2}x^2$

$y^2 = \frac{5}{2}x^2 - 1$

$y^2 = \dfrac{5x^2 - 2}{2}$

$y = \sqrt{\dfrac{5x^2 - 2}{2}}$

$y = \sqrt{\dfrac{5x^2 - 2}{2} \cdot \dfrac{2}{2}\,}$

$y = \sqrt{\dfrac{10x^2 - 4}{4}}$

$y = \frac{1}{2} \sqrt{10x^2 - 4}$           answer