Exact Equations and Integrating Factor

Where Exact Equations Come From

If the function $F(x, y)$ is differentiable, then its total differential is $dF = (\partial F/\partial x),dx + (\partial F/\partial y),dy$. The equation $dF = 0$ means that $F$ is constant along the integral curves: $F(x, y) = C$. This is the implicit solution—the family of level lines of the function $F$.

But what if we are given the equation $P(x,y),dx + Q(x,y),dy = 0$, and do not know if it is the differential of some function? The criterion for exactness and the method for finding $F$—this is the theory of exact equations.

Exact ODEs: Definition and Criterion

The equation $P(x,y),dx + Q(x,y),dy = 0$ is called exact if there exists a function $F(x,y)$ such that $\partial F/\partial x = P$ and $\partial F/\partial y = Q$. Then the equation takes the form $dF = 0$, and its general solution is $F(x, y) = C$.

Criterion of exactness: If $P$ and $Q$ have continuous partial derivatives in a simply connected domain, then the equation is exact if and only if $\partial P/\partial y = \partial Q/\partial x$.

Meaning: this is the condition of equality of mixed partial derivatives of the function $F$ (Schwarz's theorem). If $P = F_x$ and $Q = F_y$, then $P_y = F_{xy} = F_{yx} = Q_x$.

Finding the Function F: Step-by-step Algorithm

Step 1: Check the criterion $\partial P/\partial y = \partial Q/\partial x$.

Step 2: Integrate with respect to $x$: $F(x, y) = \int P(x, y) dx + \varphi(y)$, where $\varphi(y)$ is an unknown function of $y$ only (analogous to the constant of integration).

Step 3: From the condition $\partial F/\partial y = Q$, find $\varphi'(y)$ and integrate.

Expanded Example: Solve $(2xy + y^2),dx + (x^2 + 2xy),dy = 0$.

Check: $\partial P/\partial y = 2x + 2y$; $\partial Q/\partial x = 2x + 2y$. Equal—the equation is exact. ✓

Find $F$: $F = \int (2xy + y^2),dx = x^2 y + x y^2 + \varphi(y)$.

From $\partial F/\partial y = Q$: $x^2 + 2xy + \varphi'(y) = x^2 + 2xy \rightarrow \varphi'(y) = 0 \rightarrow \varphi = \text{const}$.

General solution: $x^2 y + x y^2 = C$, or $xy(x + y) = C$.

Geometrically: the family of curves $xy(x + y) = C$ covers the plane. Through each point (not lying on the coordinate axes) passes exactly one such curve.

Integrating Factor

What if the equation is not exact ($\partial P/\partial y \neq \partial Q/\partial x$)? Sometimes it is possible to find a function $\mu(x, y)$ such that the equation $\mu P,dx + \mu Q,dy = 0$ becomes exact. Such $\mu$ is called an integrating factor.

The condition for exactness of the new equation: $\partial (\mu P)/\partial y = \partial (\mu Q)/\partial x$. This is a partial differential equation for $\mu$—in general it is not simpler than the original. But in particular cases it is solvable.

If $\mu$ depends only on $x$: Then the ratio $(\partial P/\partial y - \partial Q/\partial x)/Q = g(x)$ should depend only on $x$. Then $\mu = e^{\int g(x),dx}$.

If $\mu$ depends only on $y$: Then the ratio $(\partial Q/\partial x - \partial P/\partial y)/P = h(y)$ should depend only on $y$. Then $\mu = e^{\int h(y),dy}$.

Example: $y,dx - x,dy = 0$. Here $P = y$, $Q = -x$. $\partial P/\partial y = 1 \neq \partial Q/\partial x = -1$. Not exact.

$(\partial P/\partial y - \partial Q/\partial x)/Q = (1 - (-1))/(-x) = -2/x = g(x)$. Depends only on $x$! $\mu = e^{\int -2/x,dx} = e^{-2\ln x} = 1/x^2$.

Multiply: $(y/x^2),dx - (1/x),dy = 0$. Check: $\partial (y/x^2)/\partial y = 1/x^2$, $\partial (-1/x)/\partial x = 1/x^2$. Exact! Find $F$: $F = \int y/x^2,dx = -y/x + \varphi(y)$. From $\partial F/\partial y = -1/x$: $-1/x + \varphi' = -1/x \rightarrow \varphi = \text{const}$. Answer: $y/x = C$, or $y = Cx$.

Bernoulli Equation

$y' + p(x)y = q(x)y^n$ ($n \neq 0, 1$)—a nonlinear equation reducible to linear.

Substitute $z = y^{1-n}$: $z' = (1-n)y^{-n} y'$. Divide the equation by $y^n$ and multiply by $(1-n)$:

$z' + (1-n)p(x)z = (1-n)q(x)$.

This is a first-order linear equation for $z$—solved by the standard method.

Application: Logistic growth equation: $N' = kN(1 - N/K)$—a saturation growth model, where $K$ is the carrying capacity. This is Bernoulli's equation with $n = 2$. Substitute $z = 1/N$ leads to the linear equation $z' - k z = -k/K$. Solution: $N(t) = K / \left(1 + \frac{K - N_0}{N_0}e^{-kt}\right)$. This S-shaped curve—the "logistic curve"—describes real populations, epidemic spread, and market penetration for new technologies more accurately than Malthus's model.

Clairaut Equation and Its Special Solutions

$y = x y' + f(y')$—Clairaut's equation. Differentiate with respect to $x$: $y' = y' + x y'' + f'(y')y''$, hence $y''[x + f'(y')] = 0$.

Case 1: $y'' = 0$, that is $y' = C$. Substitute into the original: $y = Cx + f(C)$. This is a family of straight lines—general solution.

Case 2: $x + f'(y') = 0$. Together with the original equation this gives the parametric equation of the envelope—the special solution, which is tangent to every straight line in the family.

Example: $y = x y' + (y')^2$. General solution: $y = Cx + C^2$. Particular solution: $x + 2C = 0 \rightarrow C = -x/2$, $y = -x^2/4$. This is a parabola, which is tangent to every straight line in the family $y = Cx + C^2$. A beautiful geometric fact: envelope-parabola corresponds to a "parabolic reflector"—the principle of concentrating reflected rays at the focus.

Reflection question: The integrating factor allows us to solve an equation that is "in itself" not exact. Do you think an integrating factor always exists for any first-order ODE?