Homogeneous Differential Equations

Some differential equations can’t be separated as they stand, like $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}$. But if the right side depends only on the ratio $\dfrac{y}{x}$, one clever substitution, $y = vx$, turns the equation into a separable one. This is one of the three exact solving methods in IB Mathematics AA HL (with separation of variables and the integrating factor), so you need to recognize when it applies and carry it through cleanly.

Key ideas

What “homogeneous” means here

A first-order differential equation is homogeneous if it can be written as

$$\frac{dy}{dx} = f\!\left(\frac{y}{x}\right)$$

that is, the right side is a function of $\dfrac{y}{x}$ alone. For example:

Equation	Rewritten in terms of y/x	Homogeneous?
$\dfrac{dy}{dx} = \dfrac{x + y}{x}$	$1 + \dfrac{y}{x}$	Yes
$\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}$	$\dfrac{x}{y} + \dfrac{y}{x}$	Yes
$\dfrac{dy}{dx} = \dfrac{y}{x} + \cos^2\!\left(\dfrac{y}{x}\right)$	already in that form	Yes
$\dfrac{dy}{dx} = \dfrac{x + y^2}{x}$	$1 + \dfrac{y^2}{x}$, which is not a function of y/x	No

A quick way to spot one: if the right side is a fraction whose top and bottom are polynomials in which every term has the same total degree (like $x^2$, $xy$, $y^2$, all degree 2), divide the top and bottom by $x$ to that power. Everything turns into powers of $\dfrac{y}{x}$.

(This use of the word “homogeneous” is different from “homogeneous linear equation” that you may meet in other books. In the IB AA course it means exactly the form above.)

The substitution y = vx

Let $v = \dfrac{y}{x}$, so $y = vx$, where $v$ is a function of $x$. Differentiate with the product rule:

$$\frac{dy}{dx} = v + x\,\frac{dv}{dx}$$

Substitute both into $\dfrac{dy}{dx} = f(v)$:

$$v + x\,\frac{dv}{dx} = f(v) \quad\Rightarrow\quad x\,\frac{dv}{dx} = f(v) - v$$

The new equation in $v$ and $x$ is separable:

$$\int \frac{1}{f(v) - v}\,dv = \int \frac{1}{x}\,dx = \ln|x| + C$$

The method

1. Check that the equation is homogeneous, and write the right side as $f\!\left(\dfrac{y}{x}\right)$.
2. Substitute $y = vx$ and $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. The $v$ terms should simplify, leaving $x\dfrac{dv}{dx} = f(v) - v$.
3. Separate and integrate. You may need partial fractions, a substitution, or a standard integral such as $\displaystyle\int \frac{1}{1 + v^2}\,dv = \arctan v + C$.
4. Back-substitute $v = \dfrac{y}{x}$ to get the general solution in terms of $x$ and $y$.
5. Use the initial condition (if any) to find $C$, giving a particular solution. Make $y$ the subject if the question asks for it, and keep the branch that passes through the given point.

Never forget step 4: an answer containing $v$ is not finished.

Worked examples

Example 1: A first one

Find the general solution of $\dfrac{dy}{dx} = \dfrac{x + y}{x}$ for $x \gt 0$. Then find the particular solution with $y = 2$ when $x = 1$.

Solution. The right side is $1 + \dfrac{y}{x}$, so the equation is homogeneous with $f(v) = 1 + v$. Let $y = vx$:

$$\begin{aligned} v + x\frac{dv}{dx} &= 1 + v \\ x\frac{dv}{dx} &= 1 \\ \int dv &= \int \frac{1}{x}\,dx \\ v &= \ln x + C \end{aligned}$$

Back-substitute $v = \dfrac{y}{x}$ and multiply by $x$:

$$y = x\ln x + Cx$$

For the particular solution, $2 = 1 \cdot \ln 1 + C(1) = C$, so

$$y = x\ln x + 2x$$

Check: $\dfrac{dy}{dx} = \ln x + 1 + 2 = \ln x + 3$, and $\dfrac{x + y}{x} = \dfrac{x + x\ln x + 2x}{x} = \ln x + 3$. ✓

Solutions of $\frac{dy}{dx} = 1 + \frac{y}{x}$: the general solution $y = x\ln x + Cx$ is a family of curves, and the initial condition picks out $C = 2$.

Example 2: A particular solution with a square root

Solve $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}$ for $x \gt 0$, given that $y = 2$ when $x = 1$. Give $y$ in terms of $x$.

Solution. Divide the top and bottom by $x^2$: $\dfrac{dy}{dx} = \dfrac{1 + (y/x)^2}{y/x}$, so $f(v) = \dfrac{1 + v^2}{v}$. Then

$$f(v) - v = \frac{1 + v^2}{v} - \frac{v^2}{v} = \frac{1}{v}$$

So $x\dfrac{dv}{dx} = \dfrac{1}{v}$. Separate and integrate:

$$\int v\,dv = \int \frac{1}{x}\,dx \quad\Rightarrow\quad \frac{v^2}{2} = \ln x + C$$

Back-substitute and multiply by $2x^2$ (renaming $2C$ as $C$):

$$y^2 = 2x^2\ln x + Cx^2$$

At $(1, 2)$: $4 = 0 + C$, so $C = 4$ and $y^2 = x^2(2\ln x + 4)$. Since $y = 2 \gt 0$ at the start, take the positive root:

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

This is defined while $2\ln x + 4 \gt 0$, that is, for $x \gt e^{-2} \approx 0.135$.

Example 3: Partial fractions inside

Solve $x^2\dfrac{dy}{dx} = y^2 + 2xy$, given that $y = 1$ when $x = 1$.

Solution. Divide by $x^2$: $\dfrac{dy}{dx} = \left(\dfrac{y}{x}\right)^2 + 2\left(\dfrac{y}{x}\right)$, so $f(v) = v^2 + 2v$. With $y = vx$:

$$x\frac{dv}{dx} = v^2 + 2v - v = v^2 + v = v(v + 1)$$

Separate, and split with partial fractions: $\dfrac{1}{v(v + 1)} = \dfrac{1}{v} - \dfrac{1}{v + 1}$.

$$\begin{aligned} \int \left(\frac{1}{v} - \frac{1}{v + 1}\right) dv &= \int \frac{1}{x}\,dx \\ \ln|v| - \ln|v + 1| &= \ln|x| + C \\ \frac{v}{v + 1} &= Ax && \text{exponentiate; } A = \pm e^C \end{aligned}$$

Back-substitute: $\dfrac{v}{v + 1} = \dfrac{y/x}{y/x + 1} = \dfrac{y}{y + x}$. At $(1, 1)$: $\dfrac{1}{2} = A$. So

$$\frac{y}{y + x} = \frac{x}{2} \quad\Rightarrow\quad 2y = xy + x^2 \quad\Rightarrow\quad y(2 - x) = x^2 \quad\Rightarrow\quad y = \frac{x^2}{2 - x}$$

Check at $x = 1$: $y = \dfrac{1}{1} = 1$. ✓ The solution has a vertical asymptote at $x = 2$, so the solution through $(1, 1)$ is valid for $x \lt 2$ (the side of the asymptote that contains $x = 1$).

Example 4: A trigonometric integral

Find the particular solution of $\dfrac{dy}{dx} = \dfrac{y}{x} + \cos^2\!\left(\dfrac{y}{x}\right)$, $x \gt 0$, with $y = \dfrac{\pi}{4}$ when $x = 1$.

Solution. Here $f(v) = v + \cos^2 v$, so $x\dfrac{dv}{dx} = \cos^2 v$. Separate, using $\dfrac{1}{\cos^2 v} = \sec^2 v$:

$$\int \sec^2 v\,dv = \int \frac{1}{x}\,dx \quad\Rightarrow\quad \tan v = \ln x + C$$

Back-substitute: $\tan\!\left(\dfrac{y}{x}\right) = \ln x + C$. At $x = 1$, $y = \dfrac{\pi}{4}$: $\tan\dfrac{\pi}{4} = 1 = 0 + C$, so $C = 1$. Then

$$\frac{y}{x} = \arctan(\ln x + 1) \quad\Rightarrow\quad y = x\arctan(\ln x + 1)$$

(Angles are in radians, as always in calculus.)

Common mistakes

Writing dy/dx = x dv/dx. If $y = vx$, both $v$ and $x$ change, so you need the product rule: $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. Dropping the $v$ means the $v$ terms won’t cancel and the equation won’t separate.

Forgetting to subtract v. After substituting you get $x\dfrac{dv}{dx} = f(v) - v$, not $f(v)$. In Example 2 that subtraction is what made $\dfrac{1 + v^2}{v}$ collapse to $\dfrac{1}{v}$.

Leaving the answer in terms of v. The question is about $x$ and $y$. Always replace $v$ with $\dfrac{y}{x}$ before you finish, and before you use the initial condition (or substitute $v = \dfrac{y_0}{x_0}$ if you prefer to find $C$ earlier).

Using the initial condition before integrating. The condition is about the solution curve, so it can only find $C$ after you have integrated. Plugging numbers into the differential equation itself tells you a slope, not the constant.

Mishandling the constant. After $\ln|v| - \ln|v + 1| = \ln|x| + C$, exponentiating gives $\dfrac{v}{v + 1} = Ax$, not $\dfrac{v}{v + 1} = x + A$. Combine the logs first, then exponentiate.

Taking the wrong square root. When you solve $y^2 = \ldots$, choose the sign that matches the initial condition (positive $y$ in Example 2) and say where the solution is defined.

Practice

1. (Warm-up) Decide whether each differential equation is homogeneous. If it is, write the right side as a function of $v = \dfrac{y}{x}$.

• (a) $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{2xy}$
• (b) $\dfrac{dy}{dx} = \dfrac{x + y^2}{x}$
• (c) $\dfrac{dy}{dx} = \dfrac{3y - x}{x + y}$

Solution

(a) Every term on top and bottom has degree 2. Divide by $x^2$: $\dfrac{1 + v^2}{2v}$. Homogeneous.

(b) The top has terms of degree 1 and 2, so dividing can’t make everything a function of $v$: $\dfrac{x + y^2}{x} = 1 + \dfrac{y^2}{x} = 1 + v^2 x$, which still contains $x$. Not homogeneous.

(c) Every term has degree 1. Divide by $x$: $\dfrac{3v - 1}{1 + v}$. Homogeneous.

2. (Warm-up) Show that the substitution $y = vx$ turns $\dfrac{dy}{dx} = \dfrac{x + 2y}{x}$ into $x\dfrac{dv}{dx} = 1 + v$. Hence find the general solution, for $x \gt 0$.

Solution

The right side is $1 + 2v$. With $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$:

$$v + x\frac{dv}{dx} = 1 + 2v \quad\Rightarrow\quad x\frac{dv}{dx} = 1 + v$$

Separate and integrate:

$$\int \frac{1}{1 + v}\,dv = \int \frac{1}{x}\,dx \quad\Rightarrow\quad \ln|1 + v| = \ln x + C \quad\Rightarrow\quad 1 + v = Ax$$

Back-substitute: $1 + \dfrac{y}{x} = Ax$, so $y = Ax^2 - x$.

Check: $\dfrac{dy}{dx} = 2Ax - 1$ and $\dfrac{x + 2(Ax^2 - x)}{x} = 2Ax - 1$. ✓

3. (Warm-up) Find the general solution of $\dfrac{dy}{dx} = \dfrac{xy + y^2}{x^2}$, $x \gt 0$.

Solution

$f(v) = v + v^2$, so $x\dfrac{dv}{dx} = v^2$. Then

$$\int v^{-2}\,dv = \int \frac{1}{x}\,dx \quad\Rightarrow\quad -\frac{1}{v} = \ln x + C \quad\Rightarrow\quad v = -\frac{1}{\ln x + C}$$

So $y = -\dfrac{x}{\ln x + C}$.

4. (Core) Solve $x\dfrac{dy}{dx} = y + 2x$, $x \gt 0$, given that $y = 3$ when $x = 1$.

Solution

Divide by $x$: $\dfrac{dy}{dx} = \dfrac{y}{x} + 2$, so $f(v) = v + 2$ and $x\dfrac{dv}{dx} = 2$.

$$\int dv = \int \frac{2}{x}\,dx \quad\Rightarrow\quad v = 2\ln x + C \quad\Rightarrow\quad y = 2x\ln x + Cx$$

At $(1, 3)$: $3 = 0 + C$, so $y = 2x\ln x + 3x$.

5. (Core) Solve $\dfrac{dy}{dx} = \dfrac{x^2 + 3y^2}{2xy}$, $x \gt 0$, given that $y = 1$ when $x = 1$. Give $y$ in terms of $x$, and state where the solution is defined.

Solution

$f(v) = \dfrac{1 + 3v^2}{2v}$, so

$$x\frac{dv}{dx} = \frac{1 + 3v^2}{2v} - v = \frac{1 + 3v^2 - 2v^2}{2v} = \frac{1 + v^2}{2v}$$

Separate. The top of $\dfrac{2v}{1 + v^2}$ is the derivative of the bottom:

$$\int \frac{2v}{1 + v^2}\,dv = \int \frac{1}{x}\,dx \quad\Rightarrow\quad \ln(1 + v^2) = \ln x + C \quad\Rightarrow\quad 1 + v^2 = Ax$$

Back-substitute and multiply by $x^2$: $x^2 + y^2 = Ax^3$. At $(1, 1)$: $2 = A$. So $y^2 = 2x^3 - x^2 = x^2(2x - 1)$, and with $y \gt 0$,

$$y = x\sqrt{2x - 1}$$

defined for $x \gt \dfrac{1}{2}$ (the solution must be differentiable on an interval containing $x = 1$).

6. (Core) Consider $\dfrac{dy}{dx} = \dfrac{y^2 - x^2}{2xy}$.

• (a) Show that the general solution can be written $x^2 + y^2 = Ax$, and describe these curves.
• (b) Find the solution curve that passes through $(1, 1)$.

Solution

(a) $f(v) = \dfrac{v^2 - 1}{2v}$, so

$$x\frac{dv}{dx} = \frac{v^2 - 1}{2v} - v = \frac{v^2 - 1 - 2v^2}{2v} = -\frac{1 + v^2}{2v}$$$$\int \frac{2v}{1 + v^2}\,dv = -\int \frac{1}{x}\,dx \quad\Rightarrow\quad \ln(1 + v^2) = -\ln|x| + C \quad\Rightarrow\quad 1 + v^2 = \frac{A}{x}$$

Multiply by $x^2$: $x^2 + y^2 = Ax$. Completing the square, $\left(x - \dfrac{A}{2}\right)^2 + y^2 = \dfrac{A^2}{4}$: circles with centre $\left(\dfrac{A}{2}, 0\right)$ on the $x$-axis, all passing through the origin.

(b) At $(1, 1)$: $2 = A$, so $x^2 + y^2 = 2x$, the circle $(x - 1)^2 + y^2 = 1$. As a function, the solution through $(1, 1)$ is the upper half, $y = \sqrt{2x - x^2}$ for $0 \lt x \lt 2$.

7. (Core) Solve $\dfrac{dy}{dx} = \dfrac{y}{x} + \tan\!\left(\dfrac{y}{x}\right)$, $x \gt 0$, given that $y = \dfrac{\pi}{6}$ when $x = 1$.

Solution

$x\dfrac{dv}{dx} = \tan v$. Separate, using $\dfrac{1}{\tan v} = \dfrac{\cos v}{\sin v}$:

$$\int \frac{\cos v}{\sin v}\,dv = \int \frac{1}{x}\,dx \quad\Rightarrow\quad \ln|\sin v| = \ln x + C \quad\Rightarrow\quad \sin v = Ax$$

So $\sin\!\left(\dfrac{y}{x}\right) = Ax$. At $(1, \tfrac{\pi}{6})$: $\sin\dfrac{\pi}{6} = \dfrac{1}{2} = A$. Then

$$\sin\!\left(\frac{y}{x}\right) = \frac{x}{2} \quad\Rightarrow\quad y = x\arcsin\!\left(\frac{x}{2}\right)$$

valid for $0 \lt x \lt 2$.

8. (Challenge) Show that the solution of $\dfrac{dy}{dx} = \dfrac{y - x}{y + x}$ through the point $(1, 0)$ satisfies

$$\frac{1}{2}\ln\left(x^2 + y^2\right) + \arctan\!\left(\frac{y}{x}\right) = 0$$

Solution

$f(v) = \dfrac{v - 1}{v + 1}$, so

$$x\frac{dv}{dx} = \frac{v - 1}{v + 1} - v = \frac{v - 1 - v^2 - v}{v + 1} = -\frac{v^2 + 1}{v + 1}$$

Separate, and split the integrand into two standard pieces:

$$\int \left(\frac{v}{v^2 + 1} + \frac{1}{v^2 + 1}\right) dv = -\int \frac{1}{x}\,dx$$$$\frac{1}{2}\ln(v^2 + 1) + \arctan v = -\ln|x| + C$$

Move $\ln|x| = \dfrac{1}{2}\ln x^2$ to the left and combine:

$$\frac{1}{2}\ln\big(x^2(v^2 + 1)\big) + \arctan v = C \quad\Rightarrow\quad \frac{1}{2}\ln(x^2 + y^2) + \arctan\!\left(\frac{y}{x}\right) = C$$

since $x^2 v^2 = y^2$. At $(1, 0)$: $\dfrac{1}{2}\ln 1 + \arctan 0 = 0 = C$. ✓ (This solution can’t be made explicit, so leaving it in this implicit form is fine.)

9. (Challenge) Solve $x^2\dfrac{dy}{dx} = x^2 + xy + y^2$, $x \gt 0$, given that $y = 0$ when $x = 1$. Find the largest interval containing $x = 1$ on which the solution is defined, giving the endpoints exactly.

Solution

Divide by $x^2$: $f(v) = 1 + v + v^2$, so $x\dfrac{dv}{dx} = 1 + v^2$.

$$\int \frac{1}{1 + v^2}\,dv = \int \frac{1}{x}\,dx \quad\Rightarrow\quad \arctan v = \ln x + C$$

At $(1, 0)$, $v = 0$: $\arctan 0 = 0 = C$. So $\arctan\!\left(\dfrac{y}{x}\right) = \ln x$, and

$$y = x\tan(\ln x)$$

$\tan$ is defined on $-\dfrac{\pi}{2} \lt \ln x \lt \dfrac{\pi}{2}$ around $\ln 1 = 0$, so the solution is defined for

$$e^{-\pi/2} \lt x \lt e^{\pi/2}$$

that is, about $0.208 \lt x \lt 4.81$ (to 3 s.f.). At either end, $y$ has a vertical asymptote.