Example 35.
\(\ \ \) Show the differential equations are separable using various algebra techniques.
\begin{equation*}
\frac{dy}{dx} = x + xy^2, \qquad
\frac{dP}{dt} = \frac{Pt}{P^2 + 1}, \qquad
\frac{dy}{dx} = 11e^{4x + 3y} \text{.}
\end{equation*}
Solution 1. Factor
Since \(x\) is common in both terms of the sum, we can factor it out as
\begin{equation*}
\frac{dy}{dx} = \underset{not\ separated}{\underbrace{x + xy^2}}
\quad \overset{algebra}{\longrightarrow} \quad
\frac{dy}{dx} = \underset{separated}{\underbrace{
{\color{blue} x } ({\color{green} 1 + y^2 })
}}
\end{equation*}
showing that the differential equation is separable.
Solution 2. Split the Fraction
This example uses the following rule for multiplying fractions:
\begin{equation*}
\frac{A\cdot B}{C\cdot D} = \frac{A}{C} \cdot \frac{B}{D}.
\end{equation*}
First, we split \(P\) and \(t\) in the numerator and denominator, then use the rule above to separate the fraction, like so
\begin{equation*}
\frac{dP}{dt}
= \underset{not\ separated}{\underbrace{\frac{Pt}{P^2 + 1}}}
= \frac{P\cdot t}{(P^2 + 1) \cdot (1) }
= \underset{separated}{\underbrace{
{\color{blue} \frac{P}{P^2 + 1} } \left({\color{green} \frac{t}{1} }\right)
}}
\end{equation*}
and the equation is separable.
Solution 3. Separate the Exponents
Applying the rule
\begin{equation*}
e^{A + B} = e^{A} \cdot e^{B}
\end{equation*}
to this differential equation allows us to show it is separable as follows:
\begin{equation*}
\frac{dy}{dx}
= \underset{not\ separated}{\underbrace{11e^{4x + 3y}}}
= 11(e^{4x} \cdot e^{3y})
= \underset{separated}{\underbrace{
{\color{blue} 11e^{4x} } \cdot {\color{green} e^{3y} }
}}.
\end{equation*}
