Solutions to 1st-order ODEs

1. Solution by direct integration¶

When equations are of this form, we can directly integrate:

\begin{align} \frac{dy}{dx} &= y^{\prime} = f(x) \\ \int dy &= \int f(x) dx \\ y(x) &= \int f(x) dx + C \end{align}

(1)

For example:

\begin{align} \frac{dy}{dx} &= x^2 \\ y(x) &= \frac{1}{3} x^3 + C \end{align}

(2)

While these problems look simple, there may not be an obvious closed-form solution to all:

\begin{align} \frac{dy}{dx} &= e^{-x^2} \\ y(x) &= \int e^{-x^2} dx + C \end{align}

(3)

(You may recognize this as leading to the error function, $\text{erf}$ : $\frac{1}{2} \sqrt{\pi} \text{erf}(x) + C$ , so the exact solution to the integral over the range $[0,1]$ is 0.7468.)

2. Solution by separation of variables¶

If the given derivative is a separate function of $x$ and $y$ , then we can solve via separation of variables:

\begin{align} \frac{dy}{dx} &= f(x) g(y) = \frac{h(x)}{j(y)} \\ \int \frac{1}{g(y)} dy &= \int f(x) dx \end{align}

(4)

For example, consider this problem:

y^{\prime} = \frac{dy}{dx} = 1 + y^2 \\

(5)

We can separate this into a problem that looks like $f(y) dy = g(x) dx$ , where $dy = \frac{1}{1+y^2}$ and $g(x) = 1$ .

\begin{align} \int \frac{dy}{1 + y^2} &= \int dx \\ \arctan y &= x + c \\ y(x) &= \tan(x+c) \end{align}

(6)

Unfortunately, not every separable ODE can be integrated:

\begin{align} \frac{dy}{dx} &= \frac{e^x / 2 + 5}{y^2 + \cos y} \\ (y^2 + \cos y) dy &= (e^x / 2 + 5) dx \end{align}

(7)

3. General solution to linear 1st-order ODEs¶

Given a general linear 1st-order ODE of the form

\frac{dy}{dx} + p(x) y = q(x)

(8)

we can solve by integration factor:

y(x) = e^{-\int p(x) dx} \left[ \int e^{\int p(x) dx} q(x) dx + C \right]

(9)

For example, in this equation

y^{\prime} + xy - 5 e^x = 0

(10)

after rearranging to the standard form

y^{\prime} + xy = 5 e^x

(11)

we see that $p(x) = x$ and $q(x) = 5e^x$ .

4. Solution to nonlinear 1st-order ODEs¶

Given a general nonlinear 1st-order ODE

\frac{dy}{dx} + p(x) y = q(x) y^a

(12)

where $a \neq 1$ and $a$ is a constant. This is known as the Bernoulli equation.

We can solve by transforming to a linear equation, by changing the dependent variable from $y$ to $z$ :

\begin{align} \text{let} \quad z &= y^{1-a} \\ \frac{dz}{dx} &= (1-a) y^{-a} \frac{dy}{dx} \end{align}

(13)

Multiply the original equation by $(1-a) y^{-a}$ :

\begin{align} (1-a) y^{-a} \frac{dy}{dx} + (1-a) y^{-a} p(x) y &= (1-a) y^{-a} q(x) y^a \\ \frac{dz}{dx} + p(x) (1-a) z &= q(x) (1-a) \;, \end{align}

(14)

which is now a linear first-order ODE, that looks like

\frac{dz}{dx} + p(x)^{\prime} z = q(x)^{\prime}

(15)

where $p(x)^{\prime} = (1-a) p(x)$ and $q(x)^{\prime} = (1-a)q(x)$ .

We can solve this using the integrating-factor approach discussed above. Then, once we have $z(x)$ , we can find $y(x)$ :

\begin{align} z &= y^{1-a} \\ y &= z^{\frac{1}{1-a}} \end{align}

(16)