Elementary Differential Equations 2.3 Problem 1: Separation of Variables

Find the complete solution of the differential equation \frac{dy}{dx}=\frac{2x+2xy^2}{y+2x^2y}.

SOLUTION:

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

By cross-multiplication,

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

Factor out common monomial factors

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

Simplify

2x\left(1+y^2\right)dx-y\left(1+2x^2\right)dy=0

Combine all terms with x, and combine all terms with y

\frac{2xdx}{\left(1+2x^2\right)}-\frac{ydy}{\left(1+y^2\right)}=0

Since the variables are already separated, we can already apply integration. That is,

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


For \int \frac{2dx}{1+2x^2\:}

Let u=1+2x^2.

=\int \:\frac{\frac{1}{2}du}{u}

=\frac{1}{2}lnu

=\frac{1}{2}ln\:\left(1+2x^2\right)


For \int \:\frac{ydy}{1+y^2}.

Let u=1+y^2

=\int \:\frac{\frac{1}{2}du}{u}

=\frac{1}{2}lnu

=\frac{1}{2}\ln \left|1+y^2\right|


Combining the solutions, we come up with

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

We can simplify the solution further by applying the law of logarithms. That is 

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

\frac{1}{2}ln\left(\frac{1+2x^2}{1+y^2}\right)=C

ln\sqrt{\frac{1+2x^2}{1+y^2}}=C

That is already an acceptable answer, but we can still simplify it further by eliminating the natural logarithm.

e^{ln\:\sqrt{\frac{1+2x^2}{1+y^2}}}=e^C

\sqrt{\frac{1+2x^2}{1+y^2}}=e^C

However, e^C is also a constant so we can name it to another constant, say C_1. So, 

\sqrt{\frac{1+2x^2}{1+y^2}}=C_1

That is also an acceptable answer but we can still simplify the equation by eliminating the square root sign. That is

\left(\sqrt{\frac{1+2x^2}{1+y^2}}\right)^2=\left(C_1\right)^2

\frac{1+2x^2}{1+y^2}=\left(C_1\right)^2

However, \left(C_1\right)^2 is another constant and we can rename it to C_2. That is,

\frac{1+2x^2}{1+y^2}=C_2

Simplifying,

1+2x^2=C_2\left(1+y^2\right)

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