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

Find the solution of the following differential equation: 

\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

Simplify the answer

\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

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