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Mathematics · Ordinary differential equations · 09

First-Order Homogeneous Equations

If the slope depends only on y/x, what happens when you make that ratio the unknown?

“Homogeneous” here means the ratio form y′=F(y/x). That is a different use of the word from “linear homogeneous.” This page derives the substitution from the product rule and brings the equation back to separation of variables.

Series · 09 / 09

Why is the direction the same all along one ray?

The equation stays y′=1+y/x. Look at one line of constant ratio through the origin, then change C or choose the other half-axis. The gray dashes show the local slope.

Exact solution on the chosen half-axisThe excluded x = 0
Let v=y/x, so y=vx and y′=v+xv′. Then y′=1+y/x becomes xv′=1, which gives y=x(ln|x|+C), valid separately on the positive and the negative half-axis.
One unit may take up a different length on screen along each axis, so read a slope from the change in coordinates. The plot shows only a finite window; leaving it does not mean the solution ends.
Starting value for this constant
Current interval

Two kinds of “homogeneous”

The word “homogeneous” has two meanings in textbooks. This page is about the one where the right side depends only on the ratio y/x.

This page studies first-order equations that can be written as y′=F(y/x), with x≠0. F is a given function of one variable, and its input v=y/x is a ratio. Here x and y are dimensionless; if engineering quantities have different units, choose scales to normalize them first, rather than adding unlike units.

Homogeneous in the ratio sense

y′=1+y/x and y′=(y/x)². The right side looks only at the ratio y/x, so v=y/x can be used. The second one is nonlinear.

Homogeneous in the linear sense

y′+p(x)y=0: the forcing term of the linear standard form is zero. y′=y is of this kind, but it does not depend only on y/x.

For example, this page’s main equation y′=1+y/x, rewritten as y′−y/x=1, is linear nonhomogeneous on either half-axis, and at the same time homogeneous in the ratio sense. The two labels answer different questions, so there is no contradiction.

If you multiply x and y by the same nonzero factor λ, the ratio stays (λy)/(λx)=y/x, so wherever the equation is defined, the slope on the right does not change. On the plot, the direction is the same all along one ray. But the ray itself is not necessarily a solution; you still have to put it into the equation and compare its slope.

v is a function, not a constant

The key to the substitution is that the new v also changes with x, so differentiating it takes the product rule; it can’t be treated as a constant.

  1. Choose a half-axis and define the new unknown

    v(x)=y(x)/x ⇒ y(x)=xv(x)

    The equation already excludes x=0. v usually still changes with x.

  2. Use the product rule

    y′ = v+xv′

    Differentiating x gives 1 and differentiating v gives v′; keep both terms. Dropping xv′ wrongly treats v as a constant.

  3. Substitute into the equation

    v+xv′=F(v) ⇒ xv′=F(v)−v

    The right side no longer mixes the original x and y. That is the whole point of the substitution.

  4. Look for constant-ratio solutions first

    F(c)−c=0 ⇒ v(x)=c ⇒ y=cx

    If they stay within the domain of the equation, these rays may be solutions. Keep them aside so the next division doesn’t lose them.

  5. Separate the remaining branches

    v′/[F(v)−v]=1/x

    Divide only on intervals where F(v)−v≠0. Integrate using the chain rule, not by cancelling symbols unconditionally.

  6. Integrate and change back to y

    ∫ dv/[F(v)−v]=ln|x|+C

    At the end, always put y/x back for v, use the starting value and check the interval. Not every F gives an easy integral.

An equivalent view: use the logarithm as a new position

On either half-axis, set t=ln|x|. Then dt/dx=1/x and dv/dt=xv′, so the equation becomes dv/dt=F(v)−v, an autonomous equation. This only helps explain why separation works. It does not bring x=0 back into the domain, and it does not mix up the direction in which x increases on the negative half-axis with the direction in which t increases.

Worked example: y′=1+y/x, y(1)=2

Work through it once: spot the form, substitute, separate, integrate, change back to y, and use the starting value to fix the constant.

  1. Identify F

    F(v)=1+v; F(v)−v=1

    No constant-ratio root is excluded by division, because 1 can never equal 0. The starting position x=1 calls for the positive half-axis.

  2. Substitute and simplify

    v+xv′=1+v ⇒ xv′=1

    The v on each side cancels, leaving v′=1/x.

  3. Integrate and change back

    v=ln|x|+C ⇒ y=x(ln|x|+C)

    This starting value is on the positive half-axis, so ln x works too.

  4. Put in the starting value

    2=1×(ln1+C) ⇒ C=2

    So y=x(ln x+2) for x>0.

  5. Check it in full

    y′=ln x+3; 1+y/x=1+ln x+2

    The two sides match, and y(1)=2.

  6. State the largest interval

    I=(0,∞)

    The formula is finite for every x>0, and the equation is undefined at x=0. x ln x does tend to 0, but filling in a value does not remove the denominator problem in the original equation.

With the starting value y(−1)=2 instead, 2=−C, so C=−2 and the solution is y=x(ln|x|−2) on (−∞,0). Here d ln|x|/dx is still 1/x, so the check is exactly the same. Each half-axis gets its own constant; the two are not joined across the origin into one solution of the initial value problem.

A second example: don’t lose the constant-ratio solutions

Division can quietly throw solutions away. Go back and check every fixed ratio that a division removed, one by one.

y′=(y/x)²,   y(1)=2This equation is nonlinear, but it still has this page’s ratio form.

  1. Substitute, then look for zero factors

    v+xv′=v² ⇒ xv′=v(v−1)

    v=0 and v=1 both make the right side zero, so list y=0 and y=x first. Substituting gives 0=0 and 1=1. Neither meets y(1)=2, but both are solutions of the equation.

  2. Separate the other branches

    v′/[v(v−1)]=1/x

    Here v≠0, 1 and x≠0.

  3. Partial fractions, then integrate

    −ln|v|+ln|v−1|=ln|x|+C

    Because 1/[v(v−1)]=−1/v+1/(v−1).

  4. Absorb the fixed sign into the constant

    (v−1)/v=Ax ⇒ v=1/(1−Ax)

    The sign is handled on one connected stretch of solution and one half-axis, with A≠0. Changing back gives y=x/(1−Ax).

  5. Use the starting value to choose A

    2=1/(1−A) ⇒ A=1/2

    This gives y=x/(1−x/2).

  6. Check it and the interval

    y′=1/(1−x/2)² = (y/x)²

    The starting value is right. The equation excludes x=0, and the formula blows up at x=2, so the largest interval containing the starting point x=1 is (0,2).

A=0 does not come from this division by nonzero factors, though putting it into the final formula does give back y=x; it can be included only after its own check. y=0 cannot be reached by any finite A in this formula and has to be listed separately. These are exactly the branches you miss if you only memorize the general formula.

For a general F that fails the uniqueness conditions, there may also be other continuations near an equilibrium ratio. Here F(v)=v² is smooth enough that constant-ratio and non-equilibrium solutions cannot be spliced together at any x≠0. Don’t take the completeness of one example as a guarantee for every F.

Spot, substitute, choose a side, check

Spot → substitute → choose a half-axis → check. Skip choosing the half-axis, and you can’t say clearly where the answer holds.

1. In the main example, with C=0, is y=x ln|x| a solution? Is y=x?

See the full reasoning

The first is a solution on either half-axis: its derivative, ln|x|+1, equals 1+y/x. The second has derivative 1 while the right side is 2, so it is not a solution. That the direction on the plot is the same along a ray does not mean the ray itself follows that direction.

2. Main example, y′=1+y/x with y(−1)=−1: what are the constant and the interval?

See the full reasoning

−1=−(ln1+C)=−C, so C=1. The solution is y=x(ln|x|+1) on I=(−∞,0). Its derivative is ln|x|+2, the right side is ln|x|+2 as well, and the starting value is right.

3. For y′=(y/x)² with y(1)=1, can you divide by v−1 and carry on?

See the full reasoning

The starting ratio v(1)=1 is exactly the root that division excludes. Try the constant-ratio solution y=x first: its derivative, 1, equals (y/x)²=1, and it meets the starting value. The largest interval is (0,∞); the function x can cross 0, but the equation cannot. Since the right side is smooth for x>0, this solution is unique.

Next tool: the product rule in reverse

The next tool runs the product rule backward: make the whole left side become the derivative of a product.

This stretch has connected classification, conditions for solutions, separation of variables, intervals, the behavior of autonomous equations and the ratio substitution. First-order linear equations come next. They ask whether multiplying by some function can turn the whole left side into the derivative of a product; that is where the integrating factor begins.

The page on first-order linear equations will rewrite this same example as y′−y/x=1 and use the product rule to find a second route to the solution.

Sources and scope

Definitions and the conditions of theorems were checked against the authors and university materials below. The explanations, worked examples and interactive plots here are written independently. Decimals on the plots are rounded displays; the mathematical conclusions rest on derivation and substitution. This page does not prove the general existence and uniqueness theorems it cites.