Separable ODEs#

Definition: Separable Ordinary Differential Equation

A first-order ODE \(F(x, y, y')\) with \(F: \mathcal{D}_F \subseteq \mathbb{R} \times \mathbb{R} \times \mathbb{R} \to \mathbb{R}\) is separable if there exist real functions \(M\) and \(N\) such that

\[F(x, y, y') = P(x) + Q(y)y'\]

for all \((x, y, y') \in \mathcal{D}_F\).

More specifically, this means that a real function \(\phi\) is a solution of \(F\) on some subset \(S \subseteq \mathbb{R}\) if and only if

\[P(x) + Q(\phi(x)) \phi'(x) = 0\]

for all \(x \in S\). Many separable ODEs can be solved via direct antidifferentiation.

Theorem: Solving Separable ODEs via Antidifferentiation

Let \(F(x,y,y')\) be a separable ODE expressed as follows:

\[N(y)y' = M(x)\]

A real function \(\phi\) is a solution of \(F\) on some subset \(S \subseteq \mathbb{R}\) if and only if \(\phi\) is differentiable on \(S\) and the antiderivatives of \(M\) and \((N \circ \phi) \cdot \phi'\) exist and are equal on \(S\):

\[\int N(\phi(x))\phi'(x) \, \mathrm{d}x = \int M(x) \, \mathrm{d}x\]

Important

If we can explicitly find \(\int N(\phi(x))\phi'(x) \, \mathrm{d}x\) (e.g. via substitution) and \(\int M(x) \, \mathrm{d}x\), then we can often find an explicit expression for the solutions of the separable ODE.

Example \(y' = yx\)

Consider the following ordinary differential equation:

\[y' = y x\]

We immediately see that \(y = 0\) is a solution on \(\mathbb{R}\).

For \(y \ne 0\), we have a separable ODE:

\[N(y)y' = M(x) \qquad N(y) = \frac{1}{y} \qquad M(x) = x\]

The theorem tells us that \(y\) is a solution on \(\mathbb{R}\) if and only if

\[\int N(y(x))y'(x) \, \mathrm{d}x = \int M(x) \, \mathrm{d}x.\]

In other words:

\[\int \frac{y'(x)}{y(x)} \, \mathrm{d}x = \int M(x) \, \mathrm{d}x\]

\[\ln |y(x)| = \frac{1}{2}x^2 + C\]

\[|y(x)| = \mathrm{e}^{\frac{1}{2}x^2 + C}, \qquad C \in \mathbb{R}\]

\[|y(x)| = \mathrm{e}^C \mathrm{e}^{\frac{1}{2}x^2}\]

\[y(x) = \pm \mathrm{e}^C \mathrm{e}^{\frac{1}{2}x^2}\]

Since \(\pm \mathrm{e}^C\) can take on any value in \(\mathbb{R}_{\ne 0}\) and since \(y = 0\) is also a solution, we can express every solution in the following form:

\[y(x) = C \mathrm{e}^{\frac{1}{2}x^2}, \qquad C \in \mathbb{R}\]

Example: \(y' = -t \mathrm{e}^{y}\)

Consider the following ordinary differential equation:

\[y' = -t \mathrm{e}^{y}\]

It is separable:

\[N(y)y' = M(t) \qquad N(y) = \mathrm{e}^{-y} \qquad M(t) = -t\]

The theorem tells us that \(y\) is a solution on \(\mathbb{R}\) if and only if

\[\int N(y(t))y'(t) \, \mathrm{d}t = \int M(t) \, \mathrm{d}t.\]

In other words:

\[\int \mathrm{e}^{-y(t)} y'(t)\, \mathrm{d}t = \int -t \,\mathrm{d}t\]

\[-\mathrm{e}^{-y(t)} = - \frac{1}{2}t^2 + C, \qquad C \in \mathbb{R}\]

\[y(t) = -\ln \left(\frac{1}{2}t^2 + C\right), \qquad C \in \mathbb{R}\]

Proof

TODO