# Separation of Variables| Elementary Differential Equations|dela Fuente, Feliciano, and Uy|Problem 1

#### $\frac{dy}{dx}=\frac{4x+xy^2}{y-x^2y}$

SOLUTION:

$\frac{dy}{dx}=\frac{4x+xy^2}{y-x^2y}$

$\left(y-x^2y\right)dy=\left(4x+xy^2\right)dx$

$y\left(1-x^2\right)dy=x\left(4+y^2\right)dx$

$\frac{ydy}{4+y^2\:}=\frac{xdx}{1-x^2}$

$\frac{ydy}{4+y^2\:}-\frac{xdx}{1-x^2}=0$

$\int \frac{ydy}{4+y^2\:}-\int \frac{xdx}{1-x^2}=\int 0$

There are two integrals, and we will perform them separately.

For $\int \frac{ydy}{4+y^2\:}$

Let $u=4+y^2$, so $du=2ydy$

$\int \:\frac{ydy}{4+y^2}=\int \:\frac{\frac{1}{2}du}{u}=\frac{1}{2}\int \:\frac{du}{u}$

$=\frac{1}{2}ln\left(u\right)+C_1$

$=\frac{1}{2}ln\left(4+y^2\right)+C_1$

For $\int \:\frac{xdx}{1-x^2}$

Let $v=1-x^2$, so $dv=-2xdx$

$\int \:\frac{xdx}{1-x^2}=\int \:\frac{-\frac{1}{2}dv}{v}=-\frac{1}{2}\int \:\frac{dv}{v}$

$=-\frac{1}{2}ln\left(v\right)+C_2$

$=-\frac{1}{2}ln\left(1-x^2\right)+C_2$

Therefore, the solution to the original differential equation is

$\frac{1}{2}ln\left(4+y^2\right)+C_1-\left[-\frac{1}{2}ln\left(1-x^2\right)+C_2\right]=0$

$\frac{1}{2}ln\left(4+y^2\right)+\frac{1}{2}ln\left(1-x^2\right)=C$
$\frac{1}{2}\:ln\left[\left(4+y^2\right)\left(1-x^2\right)\right]=C$
$\:ln\left[\left(4+y^2\right)\left(1-x^2\right)\right]=C$
$\:e^{ln\left[\left(4+y^2\right)\left(1-x^2\right)\right]}=e^C$
$\left(4+y^2\right)\left(1-x^2\right)=C$