First-order first-degree differential equation

Definition

In implicit form

A first-order first-degree differential equation is a differential equation that is both a first-order differential equation and a first-degree differential equation. Explicitly, it has the form:

$P(x,y) \frac{dy}{dx} = Q(x,y)$

where $P,Q$ are known functions. Here, $x$ is the independent variable and $y$ is the dependent variable.

In explicit form

Any first-order first-degree differential equation can be converted to an (almost) equivalent first-order explicit differential equation, i.e., a differential equation of the form:

$\frac{dy}{dx} = G(x,y)$

Conversion between the forms

A first-order first-degree differential equation can be converted to explicit form as follows. Start with:

$P(x,y) \frac{dy}{dx} = Q(x,y)$

Now, divide both sides by $P(x,y)$ and set $G(x,y) := Q(x,y)/P(x,y)$, giving the explicit form.

Note that the process may involve some slight change in the set of solutions. In particular, any solution that identically satisfies both $P(x,y) = 0$ and $Q(x,y) = 0$ may be lost when we divide. In most cases, there are no such solutions, and there are usually at most finitely many such solutions.

Existence and uniqueness of solutions

• Peano existence theorem guarantees the existence of a local solution to any initial value problem for an explicit first-order first-degree differential equation $\frac{dy}{dx} = G(x,y)$ with initial value point $(x_0,y_0)$ provided that $G$ is continuous.
• Picard-Lindelof theorem establishes existence and uniqueness under somewhat stronger continuity and differentiability assumptions.

Solution strategies in particular cases

Below are some formats of equations for which general strategies are known. Note that the letter $f$ is no longer used for the solution function but may be used for other functions:

Equation type	Quick summary of solution strategy
first-order linear differential equation which in simplified form looks like $y' + p(x)y = q(x)$	Use the integrating factor $e^{H(x)}$ where $H'=p$. The general solution is $y = Ce^{-H(x)} + e^{-H(x)}\int p(x)e^{H(x)} \, dx$
separable differential equation which is of the form $y' = f(x)g(y)$ (any first-order first-degree autonomous differential equation is separable, though there are separable differential equations that aren't autonomous)	Separate and solve as $\int \frac{dy}{g(y)} = \int f(x) \, dx$. Also find solutions corresponding to $y = k$ where $g(k) = 0$.
first-order exact differential equation $F(x,y,y') = 0$	Try to find a relation $R(x,y)$ such that $F(x,y,y') = \frac{d}{dx}[R(x,y)]$ using implicit differentiation. Finding the $R$, even if it does exist, can be tricky.
Bernoulli differential equation $y' + p(x)y = q(x)y^n$ ($n \ne 0,1$)	Divide both sides by $y^n$ (set aside possible stationary solution $y = 0$), then substitute $w = 1/y^{n-1}$ to get a first-order linear differential equation with dependent variable $w$ and independent variable $x$.