Question #866a2

Question

1324 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-20T19:12:27

$y = 3 - 4/x^2$

This is separable:

$1/(3 - y )dy/dx = 2/x$

Integrate wrt x:

$int 1/(3 - y )dy/dx \ dx =int 2/x \ dx$

By chain rule:

$int 1/(3 - y ) \ dy =int 2/x \ dx$

$- ln (3 - y ) = 2 ln x + C$

Apply the $(2,2)$ IV:

$- ln (1 ) = 2 ln 2 + C$

$implies C = - 2 ln 2 = ln (1/4)$

So
$- ln (3 - y ) = ln x^2 + ln(1/ 4)$

$ln (1/(3 - y )) = ln (x^2/4)$

$3 - y = 4/x^2$

$y = 3 - 4/x^2$