1.4 Exact equations and integrating factors
Write a first-order equation in the symmetric form \(M(x, y)\,dx + N(x, y)\,dy = 0\), which means \(M + Ny' = 0\) when \(y\) is regarded as a function of \(x\), or \(Mx' + N = 0\) the other way round.
Definition 1.4.1. The equation is exact on a region \(D\) if there is a continuously differentiable \(F\) on \(D\) with \(F_x = M\) and \(F_y = N\). Then \(F\) is a potential, and the solutions lie on the level curves \(F(x, y) = C\).
Indeed, if \(y(x)\) is a solution, \(\frac {d}{dx}F(x, y(x)) = F_x + F_yy' = M + Ny' = 0\), so \(F\) is constant along it.
Theorem 1.4.2. Let \(M\) and \(N\) be continuously differentiable on an open rectangle \(D\). The equation \(M\,dx + N\,dy = 0\) is exact on \(D\) if and only if \(M_y = N_x\) on \(D\).
Proof. If \(F\) exists, \(M_y = F_{xy} = F_{yx} = N_x\), by equality of mixed partial derivatives for \(C^2\) functions. Conversely, suppose \(M_y = N_x\), fix \((x_0, y_0) \in D\), and define \[ F(x, y) = \int _{x_0}^xM(s, y_0)\,ds + \int _{y_0}^yN(x, s)\,ds , \] which is defined because the rectangle contains the path from \((x_0, y_0)\) to \((x, y_0)\) to \((x, y)\). Clearly \(F_y = N(x, y)\). Differentiating under the integral sign, \begin {align*} F_x &= M(x, y_0) + \int _{y_0}^yN_x(x, s)\,ds = M(x, y_0) + \int _{y_0}^yM_y(x, s)\,ds\\ &= M(x, y_0) + M(x, y) - M(x, y_0) = M(x, y) . \end {align*} □
The theorem holds on any simply connected region, but not on every region: on the plane with the origin removed, \(M = -y/(x^2 + y^2)\), \(N = x/(x^2 + y^2)\) satisfy \(M_y = N_x\), yet no single-valued potential exists, since the would-be potential is the polar angle.
An equation that is not exact may become exact after multiplication by a non-vanishing function \(\mu (x, y)\), an integrating factor: the condition is \((\mu M)_y = (\mu N)_x\), a partial differential equation for \(\mu \) that is usually harder than the original problem. Two special cases are easy.
Theorem 1.4.3. If \((M_y - N_x)/N\) is a function \(\rho (x)\) of \(x\) alone, then \(\mu (x) = \exp \int \rho (x)\,dx\) is an integrating factor. If \((N_x - M_y)/M\) is a function \(\sigma (y)\) of \(y\) alone, then \(\mu (y) = \exp \int \sigma (y)\,dy\) is one.
Proof. For \(\mu = \mu (x)\), the condition \((\mu M)_y = (\mu N)_x\) becomes \(\mu M_y = \mu 'N + \mu N_x\), that is \(\mu '/\mu = (M_y - N_x)/N\). If the right side is \(\rho (x)\), the function \(\mu = e^{\int \rho }\) satisfies it, and it is positive. The second case is symmetric. □
The linear equation is the first case: \(y' + p(x)y = q(x)\) is \((py - q)\,dx + dy = 0\), with \((M_y - N_x)/N = p(x)\), and the integrating factor is \(e^{\int p}\), as in Theorem 1.2.1.
Example 1.4.4. The equation \(2xy\,dx + (x^2 + 3y^2)\,dy = 0\) has \(M_y = 2x = N_x\), so it is exact. Integrating \(M\) in \(x\) gives \(x^2y + \psi (y)\); then \(F_y = x^2 + \psi '(y) = x^2 + 3y^2\) gives \(\psi = y^3\). The solutions satisfy \(x^2y + y^3 = C\).
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