Solution of First-Order Equations: Separable, Exact & Special Integrating Factors

1. Variables Separable Equations

A first-order differential equation is separable if the derivative factorizes into a product of a function of $x$ alone and a function of $y$ alone:

Integrating both sides directly yields the one-parameter implicit general solution:

Caution: Any constant $y = c$ where $h(c) = 0$ is a potential singular or equilibrium solution that must be checked separately, as division by $h(y)$ assumes $h(y) \neq 0$.

2. Equations Reducible to Separable Form via Substitution $v = ax + by + c$

Equations of the form $\frac{dy}{dx} = f(ax + by + c)$, where $b \neq 0$, are rendered separable by defining the substitution $v = ax + by + c$. Differentiating with respect to $x$:

Integrating both sides yields the solution in terms of $v$, and back-substituting $v = ax + by + c$ gives the final answer.

3. Homogeneous Equations of Degree Zero

A function $f(x, y)$ is homogeneous of degree $n$ if $f(tx, ty) = t^n f(x, y)$ for all $t > 0$. An ODE $M(x, y)dx + N(x, y)dy = 0$ is homogeneous if $M$ and $N$ are homogeneous functions of the same degree $n$. In that case, $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$.

The standard transformation is the Euler substitution: $y = v x$, whence $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting into the ODE:

which is completely separated in the variables $v$ and $x$.

1. The Total Differential & The Exactness Criterion

A first-order differential expression $M(x, y)dx + N(x, y)dy$ is an exact differential in an open simply connected domain $D \subset \mathbb{R}^2$ if there exists a continuously differentiable potential function $\Phi(x, y)$ such that:

If $M(x, y)dx + N(x, y)dy = 0$ is exact, its general solution is immediately given by the level curves:

The differential equation $M(x, y)dx + N(x, y)dy = 0$ is exact if and only if:

2. Construction of the Potential Function $\Phi(x, y)$

To find $\Phi(x, y)$, we integrate $\frac{\partial \Phi}{\partial x} = M(x, y)$ with respect to $x$, treating $y$ as a constant:

where $g(y)$ is an arbitrary function of $y$ acting as the integration constant. Differentiating this expression with respect to $y$ and equating to $N(x, y)$:

By the exactness condition $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the right-hand side is strictly independent of $x$. Integrating $g'(y)$ with respect to $y$ yields $g(y)$ and completes the potential function $\Phi(x, y) = C$.

1. Concept of the Integrating Factor

If the equation $M(x, y)dx + N(x, y)dy = 0$ is not exact ($\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$), it can often be made exact by multiplying by an integrating factor $\mu(x, y) \neq 0$ such that:

2. Standard Single-Variable Integrating Factors

• Integrating factor depending only on $x$ ($\mu = \mu(x)$):
  Here $\frac{\partial \mu}{\partial y} = 0$ and $\frac{\partial \mu}{\partial x} = \frac{d\mu}{dx}$. The condition reduces to: $$\frac{1}{\mu} \frac{d\mu}{dx} = \frac{M_y - N_x}{N}$$ If $\frac{M_y - N_x}{N} = f(x)$ is a function of $x$ alone, then: $$\mu(x) = \exp\left(\int \frac{M_y - N_x}{N} dx\right)$$
• Integrating factor depending only on $y$ ($\mu = \mu(y)$):
  Here $\frac{\partial \mu}{\partial x} = 0$ and $\frac{\partial \mu}{\partial y} = \frac{d\mu}{dy}$. The condition reduces to: $$\frac{1}{\mu} \frac{d\mu}{dy} = \frac{N_x - M_y}{M}$$ If $\frac{N_x - M_y}{M} = g(y)$ is a function of $y$ alone, then: $$\mu(y) = \exp\left(\int \frac{N_x - M_y}{M} dy\right)$$

3. Multi-Variable Integrating Factors & Homogeneous Theorem

• If $\frac{M_y - N_x}{y N - x M} = h(xy)$, then $\mu = \exp\left(\int h(u) du\right)$ where $u = xy$.
• Homogeneous Equation Theorem: If $M dx + N dy = 0$ is a homogeneous equation of degree $n$ and $Mx + Ny \neq 0$, then: $$\mu(x, y) = \frac{1}{Mx + Ny}$$ is always an integrating factor.

1. Standard Linear Form

The general first-order linear differential equation is written in canonical form as:

where $P(x)$ and $Q(x)$ are continuous functions on an interval $I$. Rewriting in differential form: $(P(x)y - Q(x))dx + dy = 0$, where $M = P(x)y - Q(x)$ and $N = 1$.

Testing for exactness: $M_y = P(x)$ and $N_x = 0$. Since $M_y \neq N_x$, the equation is not exact, but:

which depends strictly on $x$ alone!

2. Derivation of the Closed-Form General Solution

The integrating factor is $\mu(x) = \exp\left(\int P(x) dx\right)$. Multiplying the standard form by $\mu(x)$:

By the product rule of differentiation, the left side is the exact derivative of the product $\mu(x) y$:

Integrating both sides directly with respect to $x$:

This closed-form formula provides the complete general solution to any first-order linear ODE.