### Definition 34.

A first-order differential equation is separable if it can be written in the separable form:
\begin{equation}
\frac{dy}{dx} = f(x) \cdot g(y)\tag{5}
\end{equation}

The separation of variables method works when we can separate the dependent and independent variables by multiplication. Let’s define this property formally.

\begin{equation}
\frac{dy}{dx} = f(x) \cdot g(y)\tag{5}
\end{equation}

The notation \(f(x)\) and \(g(y)\) are place-holders for expressions that only contain the variable \(x\) and the variable \(y\text{,}\) respectively. Equivalently, the definition could be written as

\begin{equation*}
\frac{dy}{dx} = \left(
\begin{array}{c}
\text{ONLY } x \text{ terms} \\
\text{ and/or constants}\\
\end{array} \right) \cdot \left(
\begin{array}{c}
\text{ONLY } y \text{ terms} \\
\text{and/or constants}\\
\end{array} \right),
\end{equation*}

but it is much easier to just write \(f(x) \cdot g(y)\text{.}\)

For example, the following differential equations are separable:

\begin{equation*}
\frac{dy}{dx} = 6xy, \quad \frac{dy}{dx} = (x^2 + 1)(y - 6)^3, \quad \frac{dy}{dx} = \sin(y)\cos(x)\text{.}
\end{equation*}

Whereas the following differential equations are not separable:

\begin{equation*}
\frac{dy}{dx} = 6x + y, \quad \frac{dy}{dx} = (x + y)(y - 6)^3, \quad \frac{dy}{dx} = \cos(x - y)\text{.}
\end{equation*}

A differential equation can still be separable even if one or both of the variables, \(x\) or \(y\text{,}\) do not appear in the equation. For example, the differential equation,

\begin{equation*}
\frac{dy}{dx} = y^2
\end{equation*}

is still separable since it can be written as

\begin{equation*}
\frac{dy}{dx} = (\underset{f(x)}{\underset{\uparrow}{ \vphantom{|} 1 }}) \cdot (\underset{g(y)}{\underset{\uparrow}{ \vphantom{|} y^2 }}) \text{.}
\end{equation*}

Similarly, the following equations are also separable:

\begin{equation*}
\frac{dy}{dx} = 6x + 5, \quad \frac{dy}{dx} = (y - 6)^3, \quad \frac{dy}{dx} = 15\text{.}
\end{equation*}

Understanding the separable form of differential equations is crucial for effectively applying the separation of variables method. By recognizing when an equation can be expressed as a product of functions involving only the independent and dependent variables, you unlock a straightforward path to finding solutions.

- multiplication
- Correct!
- addition
- Read the first sentence of this page.
- integration
- Read the first sentence of this page.
- gender
- Really?

- \(\ds\frac{dy}{dx} = \sin(x)\cos(y)\)
- Incorrect. This equation is separable because it can be expressed as a product of functions involving only \(x\) and \(y\text{.}\)
- \(\ds\frac{dy}{dx} = e^x \cdot y^2\)
- Incorrect. This equation is separable as the variables are already separated by multiplication.
- \(\ds \frac{dy}{dx} = x + y\)
- Correct! This equation is not separable because the terms involving \(x\) and \(y\) are added together, not multiplied.
- \(\ds\frac{dy}{dx} = \frac{x}{y}\)
- Incorrect. This equation is separable because the variables are divided, which can still be separated into a product of functions.

True.

- We can show it is separable by rewriting it as\begin{equation*} \frac{dz}{dt} = \underset{f(t)}{\underbrace{(\ \pi + 5 \ )}} \cdot \underset{g(t)}{\underbrace{(\ 1 \ )}} \qquad \text{ or } \qquad \frac{dP}{dt} = \underset{f(z)}{\underbrace{(\ \cos^2 z \ )}} \cdot \underset{g(t)}{\underbrace{(\ \pi + 5 \ )}} \end{equation*}
False.

- We can show it is separable by rewriting it as\begin{equation*} \frac{dz}{dt} = \underset{f(t)}{\underbrace{(\ \pi + 5 \ )}} \cdot \underset{g(t)}{\underbrace{(\ 1 \ )}} \qquad \text{ or } \qquad \frac{dP}{dt} = \underset{f(z)}{\underbrace{(\ \cos^2 z \ )}} \cdot \underset{g(t)}{\underbrace{(\ \pi + 5 \ )}} \end{equation*}

- \(\ds\frac{dy}{dx} = y\left(\frac{1}{x} + \frac{1}{y}\right)\)
- Incorrect. This rewriting does not separate the variables effectively.
- \(\ds\frac{dy}{dx} = \frac{1}{x}\left(y + 1\right)\)
- Incorrect. This form still combines \(x\) and \(y\) terms in a way that is not separable.
- \(\ds\frac{dy}{dx} = \frac{1}{x}\cdot y - \frac{1}{x}\)
- Correct! This form separates the variables, making it possible to identify whether the equation is separable.
- None of the above.
- Correct! This equation is not separable, so it is impossible to write it in separable form.

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