Homogeneous First-Order Equations and Substitution, differential equations lesson

At every point of the ray $y = 2x$ with $x > 0$ , the equation $y' = 1 + y/x$ gives the same slope $3$ , because the slope depends only on the ratio $y/x$ . Such an equation is usually not separable as written, but taking the ratio as the new unknown makes it separable. This lesson explains the scaling property behind the method, the substitution and its derivative, and the line solutions and half-planes that have to be checked separately.

Key terms

Homogeneous first-order equation

Definition Let $x$ be the independent variable, with $y' = dy/dx$ . A first-order equation $y' = f(x, y)$ is homogeneous, in the sense used for this method, if on the half-plane $x > 0$ , or on the half-plane $x < 0$ , it can be written as

y' = F\!\left(\frac{y}{x}\right)

for a function $F$ of one variable. The same structure comes from a differential form $M\,dx + N\,dy = 0$ whose coefficients are homogeneous functions of the same degree $k$ , meaning

\begin{gathered} M(\lambda x, \lambda y) = \lambda^k M(x, y), \\[0.4em] N(\lambda x, \lambda y) = \lambda^k N(x, y) \end{gathered}

for every $\lambda > 0$ . Where $N \ne 0$ and $x > 0$ , the choice $\lambda = 1/x$ gives $y' = -M(1, y/x)/N(1, y/x)$ , and on $x < 0$ the choice $\lambda = -1/x$ works the same way. This use of the word is different from a homogeneous linear equation, which means an equation with zero forcing, and a homogeneous equation in the present sense is often nonlinear.

In plain words The slope at a point depends on the direction of the point from the origin, not on its distance. Here homogeneous describes how the equation behaves under scaling, not whether it has a forcing term.

Example The equation

(x^2 + y^2)\,dx - xy\,dy = 0

has coefficients that are homogeneous of degree $2$ . Where $xy \ne 0$ it reads $y' = (x^2 + y^2)/(xy)$ , and dividing the numerator and the denominator by $x^2$ gives

y' = \frac{1 + v^2}{v}, \qquad v = \frac{y}{x}.

By contrast, the linear equation $y' = -xy$ is homogeneous in the linear sense, since it has no forcing term, but not in this sense. Along the ray $y = x$ its slope is $-1$ at $(1, 1)$ and $-4$ at $(2, 2)$ .

Scale invariance

Definition A slope function $f(x, y)$ is scale invariant on the half-plane $x > 0$ if

f(\lambda x, \lambda y) = f(x, y)

for every $\lambda > 0$ and every point with $x > 0$ . This holds exactly when $f(x, y) = F(y/x)$ there, with $F(v) = f(1, v)$ , since the choice $\lambda = 1/x$ gives $f(x, y) = f(1, y/x)$ . The same definition applies on $x < 0$ . For an equation $y' = F(y/x)$ the scaling sends solutions to solutions. If $y = \varphi(x)$ is a solution on an interval of positive $x$ , then for every $\lambda > 0$ so is

\psi(x) = \lambda\,\varphi(x/\lambda),

because $\psi'(x) = \varphi'(x/\lambda) = F(\psi(x)/x)$ .

In plain words Stretch the plane away from the origin by any positive factor and the direction field looks exactly the same. So a solution curve stretched in this way is again a solution curve.

Example For $y' = 1 + y/x$ the slope is $2$ at each of the points $(1, 1)$ , $(2, 2)$ and $(3, 3)$ , the same along the whole ray. For $y' = x + y$ the slopes at those points are $2$ , $4$ and $6$ , so that equation is not scale invariant. One solution of the first equation is $y = x\ln x$ on $x > 0$ , since $y' = \ln x + 1 = 1 + y/x$ . Stretching it by a factor $\lambda > 0$ gives $x\ln(x/\lambda) = x(\ln x - \ln\lambda)$ , which is again a solution, because the same check gives $y' = \ln x - \ln\lambda + 1 = 1 + y/x$ .

Ratio substitution

Definition For $y' = F(y/x)$ on an interval of $x$ that does not contain $0$ , the ratio substitution introduces the new unknown

v = \frac{y}{x}, \qquad y = xv.

By the product rule $y' = v + xv'$ , so the equation becomes

xv' = F(v) - v.

A differentiable function $y$ solves the original equation on the interval exactly when $v = y/x$ solves this one. The quotient rule gives the same relation in the form $v' = (xy' - y)/x^2$ .

In plain words Write $y$ as $x$ times the ratio. Both factors change with $x$ , so the derivative of $y$ has two terms. Dropping the term $v$ , or writing $y' = v'$ , is a common error.

Example For $y' = 1 + y/x$ we have $F(v) = 1 + v$ . Substituting gives $v + xv' = 1 + v$ , and the $v$ terms cancel, leaving $xv' = 1$ . Writing $y' = v'$ instead would give $v' = 1 + v$ , a different equation, and none of its solutions $v = Ce^x - 1$ gives a solution $y = x(Ce^x - 1)$ of the original equation.

Reduction to a separable equation

Definition Let $F$ be continuous on an open interval. The equation $xv' = F(v) - v$ is separable, of the form $v' = g(x)h(v)$ with $g(x) = 1/x$ and $h(v) = F(v) - v$ . On an interval of $x$ that does not contain $0$ , a solution whose values $v(x)$ stay where $F(v) \ne v$ satisfies

\int \frac{dv}{F(v) - v} = \ln\lvert x \rvert + C.

After solving for $v$ , the solution of the original equation is $y = xv$ , the whole expression for $v$ multiplied by $x$ . Its interval must avoid $x = 0$ , where the original slope is undefined, and every point where the formula breaks down.

In plain words After the substitution, the dependence on $x$ and the dependence on $v$ come apart, so the method for separable equations applies. Then multiply by $x$ to return to $y$ .

Example Solve $y' = 1 + y/x$ with $y(1) = 0$ on $x > 0$ . The substitution gave $xv' = 1$ , so $v' = 1/x$ and $v = \ln x + C$ . Then $y = x(\ln x + C)$ , and $y(1) = C = 0$ , so

y = x\ln x.

The check is direct, since $y' = \ln x + 1$ and $1 + y/x = 1 + \ln x$ . The maximal interval is $(0, \infty)$ . As $x \to 0^+$ the solution tends to $0$ while its slope tends to $-\infty$ , and because the equation itself is undefined at $x = 0$ , no solution passes through that line.

Excluded solution branches

Definition An excluded solution branch is a solution of the original equation that the separated formula does not produce. Each root $c$ of $F(c) = c$ gives the constant ratio $v = c$ and hence the line solution $y = cx$ on each half-plane, because then $y' = c = F(c)$ . Separating divides by $F(v) - v$ , so these roots must be found before dividing, and their lines added to the answer when the formula does not contain them. In addition, the line $x = 0$ is excluded from the start, the half-planes $x > 0$ and $x < 0$ are handled separately, with $\ln\lvert x \rvert$ in place of $\ln x$ , and the explicit formula for $y$ holds only on intervals that avoid its own singular points.

In plain words Before dividing by $F(v) - v$ , ask which constant ratios make it zero. Each one gives a straight-line solution that the final formula may leave out.

Example For

y' = \frac{y}{x} + \left(\frac{y}{x}\right)^2

we have $F(v) = v + v^2$ and $xv' = v^2$ . The root $v = 0$ gives $y = 0$ , which clearly solves the equation. For $v \ne 0$ , separating gives

\begin{gathered} -\frac{1}{v} = \ln\lvert x \rvert + C, \\[0.4em] y = -\frac{x}{\ln\lvert x \rvert + C}. \end{gathered}

No value of $C$ gives $y = 0$ , so the zero solution has to be added separately. Each nonzero branch lives on an interval that avoids both $x = 0$ and the pole where $\lvert x \rvert = e^{-C}$ . With $C = 1$ , for instance, the branch through $(1, -1)$ is $y = -x/(\ln x + 1)$ on $(e^{-1}, \infty)$ .

Check yourself

Decide which of the equations $y' = (x^2 + 3y^2)/(2xy)$ , $y' = (y + 1)/x$ and $y' = -xy$ have the form $y' = F(y/x)$ .

Show the answer

The first one does. Dividing the numerator and the denominator by $x^2$ gives $F(v) = (1 + 3v^2)/(2v)$ with $v = y/x$ , valid where $xy \ne 0$ . The second does not, since along the ray $y = x$ its slope is $2$ at $(1, 1)$ but $3/2$ at $(2, 2)$ . The third does not either. It is a homogeneous linear equation, because it has no forcing term, but its slope is $-1$ at $(1, 1)$ and $-4$ at $(2, 2)$ , so it is not scale invariant. The two meanings of homogeneous are different properties.

Find all solutions of

y' = \frac{y}{x} + \frac{(y - x)^2}{x^2}

on the half-plane $x > 0$ .

Show the answer

With $v = y/x$ the right side is $F(v) = v + (v - 1)^2$ , so the substitution gives $xv' = (v - 1)^2$ . The root $v = 1$ gives the line $y = x$ , and indeed its slope $1$ equals $1 + 0$ . For $v \ne 1$ , separating gives

-\frac{1}{v - 1} = \ln x + C,

so $v = 1 - 1/(\ln x + C)$ and

y = x - \frac{x}{\ln x + C}.

No value of $C$ gives $y = x$ , so the complete answer is this family together with the line $y = x$ . Each curve of the family lives on an interval of $x > 0$ that avoids $x = e^{-C}$ , that is on $(0, e^{-C})$ or on $(e^{-C}, \infty)$ .

Solve the initial-value problem below and find the maximal interval of its solution.

y' = \frac{x^2 + y^2}{xy}, \qquad y(1) = 2

Show the answer

Here $F(v) = v + 1/v$ , so the substitution gives $xv' = 1/v$ . Since $F(v) - v = 1/v$ is never zero, no line solution is lost. Separating gives $v\,dv = dx/x$ , so on $x > 0$

\frac{v^2}{2} = \ln x + C.

At $x = 1$ the ratio is $v = 2$ , so $C = 2$ and $v^2 = 2\ln x + 4$ . Since $y(1) = 2 > 0$ , the positive root is the right one, and

y = x\sqrt{2\ln x + 4}.

The equation divides by $y$ , so the solution must stay positive, which requires $2\ln x + 4 > 0$ , that is $x > e^{-2}$ . The maximal interval is $(e^{-2}, \infty)$ , where $e^{-2} \approx 0.135$ . As $x \to e^{-2}$ from the right, $y \to 0$ and the slope grows without bound.

Homogeneous First-Order Equations and Substitution

Key terms

Homogeneous first-order equation

Scale invariance

Ratio substitution

Reduction to a separable equation

Excluded solution branches

Check yourself

How this lesson connects

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