Question

(2+2y^2)dx+3xy^2dy=0

Answer 1

Answer:

$\dfrac {3}{2y}ln|1+y^2| + ln|\dfrac {A}{x^2}| - 3y = 0$

Solution

$(2+2y^2)dx + 3xy^2dy = 0$

Separating terms gives:

$3xy^2dy = -2(1+y^2)dx$

dividing both sides by $x(1+y^2)$ gives:

$\dfrac {3xy^2}{x(1+y^2)}dy = \dfrac {-2(1+y^2)}{x(1+y^2)}dx$

$=> \dfrac {3y^2}{1+y^2}dy = \dfrac {-2}{x}dx$

Reducing $\dfrac {y^2}{1+y^2}$ will give:

$\dfrac {y^2}{1+y^2} = 1 - \dfrac{1}{1+y^2}$

$=> 3(1-\dfrac {1}{1+y^2})dy = \dfrac{-2}{x}dx$

Integrating both sides gives:

$\intop3(1-\dfrac {1}{1+y^2})dy = \intop\dfrac{-2}{x}dx$

$=> 3\intop(1-\dfrac {1}{1+y^2})dy = -2\intop\dfrac{1}{x}dx$

$=> 3(y-\dfrac{1}{2y}ln|1+y^2|=-2ln|x|+ln(A)$

$=> 3y - \dfrac{3}{2y}ln|1+y^2| = ln|x^{-2}| + ln(A)$

$=> 3y = \dfrac {3}{2y}ln|1+y^2| + ln|\dfrac {A}{x^2}|$

$\dfrac {3}{2y}ln|1+y^2| + ln|\dfrac {A}{x^2}| - 3y = 0$