Separable and Exact Equations

Outside the linear class, no single method solves every first-order equation. Two important families are solved by direct integration: separable equations, in which the two variables can be separated onto opposite sides, and exact equations, which are the differential of a hidden function. Both produce solutions defined implicitly, by a relation $F(x,y)=c$ rather than a formula $y=\varphi (x)$, and pulling out an explicit solution is often impossible. The independent variable is written $x$ throughout.

Separable equations

Write the general first-order equation $\frac{\mathbf{d}y}{\mathbf{d}x}=f(x,y)$ in the symmetric form

which is always possible by taking $M=-f$ and $N=1$. The equation is separable when $M$ depends on $x$ alone and $N$ on $y$ alone, so it reads $M(x)+N(y)\frac{\mathbf{d}y}{\mathbf{d}x}=0$, or in differential form

The justification is the chain rule. If $H_1$ and $H_2$ are antiderivatives of $M$ and $N$, then $M(x)+N(y)y^{\prime }=H_1^{\prime }(x)+H_2^{\prime }(y)y^{\prime }=\frac{\mathbf{d}}{\mathbf{d}x}[H_1(x)+H_2(y)]$, so the equation says this total derivative is zero and $H_1(x)+H_2(y)=c$. In practice one integrates the $x$-term against $\mathbf{d}x$ and the $y$-term against $\mathbf{d}y$ and adds a single constant.

Constant solutions. A separable equation $\frac{\mathbf{d}y}{\mathbf{d}x}=f(x,y)$ may have a constant solution $y=y_0$ whenever $f(x,y_0)=0$ for all $x$. Separating the variables requires dividing by such a factor, which can drop these solutions; they must be restored by inspection. For $\frac{\mathbf{d}y}{\mathbf{d}x}=\frac{(y-3)\cos x}{1+2y^2}$, the constant $y=3$ is a solution that separation would discard.

Homogeneous equations by substitution

A first-order equation is homogeneous when the right side depends only on the ratio $y/x$:

This is not separable as written, but the substitution $v=y/x$, i.e. $y=xv$, makes it so.¹ Then $\frac{\mathbf{d}y}{\mathbf{d}x}=v+x\frac{\mathbf{d}v}{\mathbf{d}x}$, and the equation becomes

which is separable in $v$ and $x$. Integrate, then replace $v$ by $y/x$ to recover the solution. Because the slope depends only on $y/x$, the direction field is constant along each ray through the origin, and the integral curves are symmetric about it.

The related Bernoulli equation $y^{\prime }+p(t)y=q(t)y^n$ is nonlinear but linearizes under $v=y^{1-n}$ for $n\ne 0,1$, a substitution Leibniz found in 1696; it returns to the integrating-factor method.

Exact equations

Return to the symmetric form $M(x,y)+N(x,y)y^{\prime }=0$ with both coefficients now depending on both variables. Suppose there is a function $\psi (x,y)$ whose partial derivatives are exactly $M$ and $N$:

Then, by the chain rule, $M+N{y}^{\prime }={\psi }_x+{\psi }_y{y}^{\prime }=\frac{\mathbf{d}}{\mathbf{d}x}\psi (x,\varphi (x))$, so the equation reads $\frac{\mathbf{d}}{\mathbf{d}x}\psi (x,y)=0$ and its solutions are the level curves

For $2x+y^2+2xy{y}^{\prime }=0$, the function $\psi =x^2+x{y}^2$ has ${\psi }_x=2x+y^2=M$ and ${\psi }_y=2xy=N$, so the equation is exact and its solutions are $x^2+x{y}^2=c$. The trouble is recognizing $\psi$ when it is less obvious. A symmetry test settles when one exists.

One direction is immediate: if ${\psi }_x=M$ and ${\psi }_y=N$, then $M_y={\psi }_{xy}$ and $N_x={\psi }_{yx}$, and continuity forces the mixed partials to agree. The converse is constructive and gives the solution method.²

Recovering the potential. Once $M_y=N_x$ confirms exactness, integrate ${\psi }_x=M$ in $x$, keeping an arbitrary function of $y$ in place of a constant:

Differentiate in $y$ and set the result equal to $N$, which determines $h^{\prime }(y)$; the exactness condition guarantees $h^{\prime }$ comes out a function of $y$ alone. Integrate $h$ and assemble $\psi$.

For $(y\cos x+2x{e}^y)+(\sin x+x^2{e}^y-1)y^{\prime }=0$, the test gives $M_y=\cos x+2x{e}^y=N_x$, so it is exact. Integrating $M$ in $x$,

which forces $h^{\prime }(y)=-1$, so $h(y)=-y$. The solutions are

Geometrically, the solution curves lie at constant height on the surface $z=\psi (x,y)$; each level $c$ is one integral curve.

Integrating factors that restore exactness

When $M_y\ne N_x$ the equation is not exact, but multiplying by a factor $\mu (x,y)$ may make it so, exactly as with linear equations. The product $\mu M+\mu N{y}^{\prime }=0$ is exact when $(\mu M{)}_y=(\mu N{)}_x$, which expands to the partial differential equation

In general this is as hard as the original equation. It becomes tractable when $\mu$ depends on one variable only. If $\frac{M_y-N_x}{N}$ is a function of $x$ alone, call it $R(x)$, then $\mu$ depends on $x$ alone and satisfies the separable, linear equation

A symmetric test in $y$ handles factors depending only on $y$: if $\frac{N_x-M_y}{M}$ is a function of $y$ alone, an integrating factor $\mu (y)=\exp (\int \frac{N_x-M_y}{M}\mathbf{d}y)$ exists.

The four methods so far cover most first-order equations that yield closed-form solutions.

Summary

• A separable equation splits into $M(x)\mathbf{d}x+N(y)\mathbf{d}y=0$ and integrates side by side into an implicit first integral; watch for constant solutions lost in the split, and read the interval of existence off the solution.
• An exact equation is $\mathbf{d}\psi =0$ for a potential $\psi$, certified by $M_y=N_x$; recover $\psi$ by integrating $M$ in $x$ and matching $N$.
• When exactness fails, an integrating factor in one variable can restore it, found by testing whether $(M_y-N_x)/N$ or $(N_x-M_y)/M$ depends on a single variable.
• The homogeneous substitution $v=y/x$ moves an equation into the separable class.

These direct methods run out quickly; most first-order equations admit no elementary solution at all.