Differential Equations · Lesson 4 of 6

Homogeneous Equations and Ratio Substitutions

“Recognise equations governed by a ratio and reduce them to separable equations.”

Learning Objectives

• Test whether a two-variable function is homogeneous and identify its degree. • Recognise a homogeneous first-order differential equation through its degree-zero right-hand side. • Derive and use the substitution y = vx. • Choose x = vy when x is more conveniently treated as a function of y. • Solve general and particular homogeneous equations, retaining valid straight-line solutions. • Interpret a tangent-slope rule and check the final relation.

A ratio can remove one variable

Some equations resist direct separation because $x$ and $y$ occur together. If they appear only through their ratio, a new variable can capture that ratio. The equation then becomes separable in the new variable and one original variable.

For instance, $(x + 2 y) / (x - y)$ can be divided above and below by $x$ to give $(1 + 2 y / x) / (1 - y / x)$ when $x \neq = 0$ . Multiplying both $x$ and $y$ by the same nonzero number does not change $y / x$ . This scaling observation is the idea behind a homogeneous equation.

Homogeneous functions and their degree

A scaling test gives a precise way to recognise the pattern. Replace $x$ by $λ x$ and $y$ by $λ y$ . If a single fixed power of $λ$ factors out, the function is homogeneous on the domain where both expressions are defined.

Definition

Homogeneous function

A function $F (x, y)$ is homogeneous of degree $n$ if $F (λ x, λ y) = λ^{n} F (x, y)$ for the permitted nonzero scaling values $λ$ . The same degree must work throughout its domain.

Scaling testLaTeX

F (λ x, λ y) = λ^{n} F (x, y)

λ

scales both variables by the same factor. The exponent

n

is the degree of the homogeneous function.

Function	Result after scaling	Conclusion
$y^{2} + 2 x y$	$λ^{2} (y^{2} + 2 x y)$	Homogeneous of degree 2
$2 x - 3 y$	$λ (2 x - 3 y)$	Homogeneous of degree 1
$cos (y / x)$	$cos (y / x)$	Homogeneous of degree 0
$sin x + cos y$	$sin (λ x) + cos (λ y)$	Not homogeneous: no one scaling power works

A homogeneous function can also be expressed using a ratio. For $x \neq = 0$ , $y^{2} + 2 x y = x^{2} [(y / x)^{2} + 2 (y / x)]$ . Similarly, $2 x - 3 y = x [2 - 3 (y / x)]$ . In general the ratio forms are $x^{n} g (y / x)$ or, when $y \neq = 0$ , $y^{n} h (x / y)$ . When $n = 0$ , the prefactor disappears and only a ratio remains.

Two meanings of degree

The degree of a homogeneous function concerns scaling in $x$ and $y$ . The degree of a differential equation concerns polynomial powers of derivatives. These are different classifications. A homogeneous first-order equation normally has a degree-zero right-hand side and first-degree dependence on its derivative.

Definition

Homogeneous differential equation

A first-order equation $d y / d x = F (x, y)$ is homogeneous when $F$ is homogeneous of degree zero. On a suitable domain this can be written $d y / d x = g (y / x)$ . The corresponding form $d x / d y = h (x / y)$ is treated similarly.

Ratio and tangent slopes— Points on the same ray have the same ratio. Here their short tangent segments have the same slope.

The diagram uses $y^{'} = 1 + y / x$ . At $(1, 1)$ , $(2, 2)$ , and $(3, 3)$ the ratio is $1$ , so the prescribed slope is $2$ in each case. The dashed ray identifies equal ratios; it is not itself a solution curve here, because its own slope is $1$ , not $2$ .

Derive the ratio substitution

Let the new variable be the ratio $v = y / x$ . Since $y$ changes with $x$ , this new variable generally changes too. Writing $y = v x$ and differentiating by the product rule is the essential step; treating $v$ as a constant would change the equation.

Derivative after substitutionLaTeX

y = v x, \frac{d y}{d x} = v + x \frac{d v}{d x}

v

is a function of

x

. The term

x d v / d x

records how the ratio changes.

Substituting into $d y / d x = g (y / x)$ gives $v + x d v / d x = g (v)$ , hence $x d v / d x = g (v) - v$ . Now the right side depends only on $v$ , which is exactly the separable structure we need.

Separated homogeneous equationLaTeX

\int \frac{d v}{g ( v ) - v} = ln ∣ x ∣ + C

This step assumes

x \neq = 0

and

g (v) - v \neq = 0

. Integrate first, then replace

v

y / x

If $g (v_{0}) = v_{0}$ , the constant ratio $v = v_{0}$ gives the straight line $y = v_{0} x$ , provided the original equation is defined on it. Division by $g (v) - v$ would exclude it. Check these possibilities separately or verify that they are restored by a constant value in the final family.

Example — A rational homogeneous equation

Problem
Show that $(x - y) y^{'} = x + 2 y$ is homogeneous and solve it.

1.Where $x - y \neq = 0$ , $y^{'} = (x + 2 y) / (x - y)$ . Scaling numerator and denominator by $λ$ cancels the factor, so the right-hand side is homogeneous of degree zero.
2.Use $y = v x$ . Then $v + x v^{'} = (1 + 2 v) / (1 - v)$ , so $x v^{'} = (v^{2} + v + 1) / (1 - v)$ .
3.Separate as $\frac{v - 1}{v ^{2} + v + 1} d v = - d x / x$ . The numerator is $\frac{1}{2} (2 v + 1) - \frac{3}{2}$ .
4.For the first part, integrate the derivative of the denominator to get $\frac{1}{2} ln (v^{2} + v + 1)$ . For the second, complete the square: $v^{2} + v + 1 = (v + 1/2)^{2} + 3/4$ .
5.Therefore $\frac{1}{2} ln (v^{2} + v + 1) - 3 tan^{- 1} (\frac{2 v + 1}{3}) = - ln ∣ x ∣ + C_{1}$ .
6.Move the logarithm of $x$ to the left and use $x^{2} (v^{2} + v + 1) = x^{2} + x y + y^{2}$ . Doubling and renaming the constant gives $ln (x^{2} + x y + y^{2}) = 23 tan^{- 1} (\frac{x + 2 y}{3 x}) + C$ .
7.The ratio expression applies locally with $x \neq = 0$ , and the slope form requires $x \neq = y$ . The inverse-tangent expression is read on the chosen solution interval; its branch cannot be changed midway without adjusting the constant.

Example — A trigonometric function of a ratio

Problem
Solve $x cos (y / x) y^{'} = y cos (y / x) + x$ .

1.The expression requires $x \neq = 0$ . Where a differentiable solution exists, $cos (y / x) \neq = 0$ , because otherwise the equation would force $x = 0$ .
2.Divide to get $y^{'} = y / x + sec (y / x)$ . This depends only on $y / x$ , so it is homogeneous.
3.Set $y = v x$ . Then $v + x v^{'} = v + sec v$ , leaving $x v^{'} = sec v$ .
4.Separate: $cos v d v = d x / x$ . Integrate: $sin v = ln ∣ x ∣ + C$ .
5.Return to the original variables: $sin (y / x) = ln ∣ x ∣ + C$ , on intervals respecting the original restrictions.

Reverse the independent variable when useful

A differential relation involving $d x$ and $d y$ need not be forced into $d y / d x$ . Sometimes solving for $d x / d y$ produces a simpler ratio $x / y$ . In that case treat $x$ as a function of $y$ and use the product rule with respect to $y$ .

Reverse-variable substitutionLaTeX

x = v y, \frac{d x}{d y} = v + y \frac{d v}{d y}

Here

v = x / y

depends on

y

. The derivative is with respect to

y

, not

x

Example — Choose the more convenient ratio

Problem
Solve $2 y e^{x / y} d x + (y - 2 x e^{x / y}) d y = 0$ , with $x = 0$ when $y = 1$ .

1.The equation requires $y \neq = 0$ . Solve for $d x / d y$ : $d x / d y = x / y - \frac{1}{2} e^{- x / y}$ . The right side depends only on $x / y$ .
2.Let $x = v y$ . Substitution gives $v + y d v / d y = v - \frac{1}{2} e^{- v}$ , so $2 e^{v} d v = - d y / y$ .
3.Integrate: $2 e^{v} = - ln ∣ y ∣ + C$ . Hence $2 e^{x / y} + ln ∣ y ∣ = C$ .
4.At $(x, y) = (0, 1)$ , $C = 2$ . The particular solution is $2 e^{x / y} + ln ∣ y ∣ = 2$ , on a suitable interval with $y > 0$ .

Connect the substitution to curve geometry

A homogeneous slope rule produces a family of curves. The algebraic form of that family can be simpler than the equation used to obtain it. The derivative of the final relation provides a quick check and reveals which points have a defined slope.

Example — Recover a family from its tangent slopes

Problem
Find the family whose tangent slope at $(x, y)$ is $(x^{2} + y^{2}) / (2 x y)$ .

1.The equation is $y^{'} = (x^{2} + y^{2}) / (2 x y)$ , requiring $x y \neq = 0$ . Divide numerator and denominator by $x^{2}$ to reveal $(1 + v^{2}) / (2 v)$ with $v = y / x$ .
2.With $y = v x$ , $v + x v^{'} = (1 + v^{2}) / (2 v)$ , so $x v^{'} = (1 - v^{2}) / (2 v)$ .
3.For $v^{2} \neq = 1$ , separate: $\frac{2 v}{v ^{2} - 1} d v = - d x / x$ . Integrate: $ln ∣ v^{2} - 1∣ = - ln ∣ x ∣ + C_{1}$ .
4.Combine the logarithms and absorb sign into a constant: $(v^{2} - 1) x = C_{2}$ . Substituting $v = y / x$ gives $y^{2} - x^{2} = C_{2} x$ .
5.Rename the constant to write $x^{2} - y^{2} = C x$ . The excluded ratios $v = 1$ and $v = - 1$ give $y = x$ and $y = - x$ ; both are recovered when $C = 0$ .
6.Differentiate the final family: $2 x - 2 y y^{'} = C$ . Substituting $C = (x^{2} - y^{2}) / x$ gives $y^{'} = (x^{2} + y^{2}) / (2 x y)$ wherever $x y \neq = 0$ .

Example — Handle a square root on a specified interval

Problem
Solve $x y^{'} - y = x^{2} + y^{2}$ on $x > 0$ .

1.With a square root, do not replace $x^{2}$ by x on an arbitrary real interval. Here $x > 0$ , so $x^{2} + y^{2} = x 1 + (y / x)^{2}$ . Positive simultaneous scaling preserves the ratio expression on this domain.
2.Put $y = v x$ . The left side is $x^{2} v^{'}$ , so the equation reduces to $x v^{'} = 1 + v^{2}$ .
3.Separate and integrate: $\int d v / 1 + v^{2} = ln x + C$ , hence $ln (v + 1 + v^{2}) = ln x + C$ .
4.Exponentiation gives $v + 1 + v^{2} = K x$ with $K > 0$ . Replacing v yields $y + x^{2} + y^{2} = K x^{2}$ .
5.An explicit form is $y = (K x^{2} - K^{- 1}) /2$ . Its derivative gives $x y^{'} - y = (K x^{2} + K^{- 1}) /2 = x^{2} + y^{2}$ , so the result checks with the positive square root.

Example — Apply an initial value after integration

Problem
Solve $(x + y) d y + (x - y) d x = 0$ with $y (1) = 1$ .

1.Write $y^{'} = (y - x) / (x + y)$ . Set $y = v x$ to get $v + x v^{'} = (v - 1) / (1 + v)$ .
2.After subtraction, $x v^{'} = - (1 + v^{2}) / (1 + v)$ . Thus $\frac{1 + v}{1 + v ^{2}} d v = - d x / x$ .
3.Integration gives $tan^{- 1} v + \frac{1}{2} ln (1 + v^{2}) = - ln ∣ x ∣ + C$ .
4.Combine logarithms and replace $v$ : $tan^{- 1} (y / x) + \frac{1}{2} ln (x^{2} + y^{2}) = C$ .
5.At $(1, 1)$ , $C = π /4 + \frac{1}{2} ln 2$ . The equation with this constant is the required local solution around the specified point.

Organise a longer homogeneous equation

Factors involving differentials can obscure the ratio structure. Collect the $d x$ and $d y$ terms first, then divide numerator and denominator by a common power. The substitution and separation steps are unchanged, but excluded factors need particular care.

Example — A ratio hidden inside a differential relation

Problem
Solve $(x d y - y d x) y sin (y / x) = (y d x + x d y) x cos (y / x)$ .

1.For $x \neq = 0$ , collect terms to obtain $\frac{d y}{d x} = \frac{x y c o s ( y / x ) + y ^{2} s i n ( y / x )}{x y s i n ( y / x ) - x ^{2} c o s ( y / x )}$ wherever this quotient is defined.
2.Put $y = v x$ . Division by $x^{2}$ and subtraction of $v$ give $x v^{'} = \frac{2 v c o s v}{v s i n v - c o s v}$ .
3.For nonzero factors, separate: $(tan v - 1/ v) d v = 2 d x / x$ .
4.Integrate to obtain $- ln ∣ cos v ∣ - ln ∣ v ∣ = 2 ln ∣ x ∣ + C_{1}$ . Hence $x^{2} v cos v = C$ .
5.The result is $x y cos (y / x) = C$ . When $y cos (y / x) \neq = 0$ , it can also be written $sec (y / x) = K x y$ by changing the constant.
6.The product form retains $C = 0$ . Check the straight lines $y = 0$ and $y = v_{0} x$ with $cos v_{0} = 0$ directly in the original relation; they satisfy it on intervals with $x \neq = 0$ . A reciprocal form alone would miss them.

Do not drop the product-rule term

From $y = v x$ , the derivative is $v + x v^{'}$ , not just $x v^{'}$ and not just $v$ . Write the product rule explicitly each time until it becomes automatic. After solving, substitute $v = y / x$ back into the result.

Check your understanding

A homogeneous test and a substitution choice come before integration. Use these questions to distinguish the degree-zero scaling property from the unrelated degree of a differential equation.

Quick check

What is the degree of the homogeneous function $y^{2} + 2 x y$ ?

Quick check

Which function is homogeneous of degree zero?

Quick check

If $y = v x$ and v depends on x, what is $d y / d x$ ?

Quick check

For $d x / d y = h (x / y)$ , which substitution is appropriate?

Quick check

After $y = v x$ , what does $y^{'} = 1 + y / x$ become?

Quick check

Which function is not homogeneous?

Quick check

What should be checked before dividing by $g (v) - v$ ?

Quick check

For $y^{'} = 1 + y / x$ , what slope is prescribed at $(2, 2)$ ?

Practice Problems

Show homogeneity and solve $(x^{2} + x y) d y = (x^{2} + y^{2}) d x$ .
Solve $y^{'} = (x + y) / x$ on an interval with $x \neq = 0$ .
Solve $(x - y) d y - (x + y) d x = 0$ .
Solve $(x^{2} - y^{2}) d x + 2 x y d y = 0$ .
Solve $x^{2} y^{'} = x^{2} - 2 y^{2} + x y$ by setting $y = v x$ .
Solve $x y^{'} - y + x sin (y / x) = 0$ . Check constant ratios excluded during separation.
Solve $y d x + x ln (y / x) d y - 2 x d y = 0$ where $y / x > 0$ . Try $x = v y$ .
Solve $(1 + e^{x / y}) d x + e^{x / y} (1 - x / y) d y = 0$ where $y \neq = 0$ .
Solve $[x sin^{2} (y / x) - y] d x + x d y = 0$ with $y (1) = π /4$ .
Solve $y^{'} - y / x + csc (y / x) = 0$ with $y (1) = 0$ , or explain why the request is impossible.
Solve $2 x y + y^{2} - 2 x^{2} y^{'} = 0$ with $y (1) = 2$ .
Classify $y^{2} d x + (x^{2} - x y - y^{2}) d y = 0$ as homogeneous or not, and justify by scaling.
Solve $x^{2} d y + (x y + y^{2}) d x = 0$ with $y (1) = 1$ .
Collect the differentials in $(x cos (y / x) + y sin (y / x)) y d x = (y sin (y / x) - x cos (y / x)) x d y$ and relate it to the longer worked example.

With $v = y / x$ , $x v^{'} = (1 - v) / (1 + v)$ . Integrating gives $- v - 2 ln ∣1 - v ∣ = ln ∣ x ∣ + C$ for $v \neq = 1$ . The line $y = x$ must be added separately.

Key Takeaways

• Homogeneous functions have a fixed scaling degree; homogeneous first-order equations have a degree-zero right-hand side. • The substitution $y = v x$ produces $d y / d x = v + x d v / d x$ and reduces the equation to separation. • Use $x = v y$ when the equation is simpler with $x$ depending on $y$ . • Replace the ratio variable after integration and apply any given point. • Check constant-ratio solutions and ensure the original equation is defined at a proposed initial point.

Previous · Lesson 3

Separation of Variables and Growth Models

Next · Lesson 5

Linear Equations and Integrating Factors

A ratio can remove one variable

Homogeneous functions and their degree

Derive the ratio substitution

Reverse the independent variable when useful

Connect the substitution to curve geometry

Organise a longer homogeneous equation

Check your understanding

Quiz

Practice Problems

Key Takeaways