Homogeneous equations

What you will be able to do

• decide whether a function or an equation is homogeneous;
• solve a homogeneous equation with $y = vx$, or with $x = vy$ when that is shorter;
• return to $x$ and $y$ and fix the constant from a point.

The same slope along every ray

Look at $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{2xy}$. Replace $x$ by $3x$ and $y$ by $3y$. The top becomes $9(x^2 + y^2)$ and the bottom becomes $9 \times 2xy$. The nine cancels, and the slope is the same. So the slope at $(3, 6)$ equals the slope at $(1, 2)$. Both points lie on the ray $y = 2x$.

This is the idea. In a homogeneous equation the slope is the same at every point of a ray from the origin. The simulation shows it.

Homogeneous functions

A function $f(x, y)$ is homogeneous of degree $n$ if

for every $\lambda > 0$ in the domain. For example, $x^2 + 3xy$ has degree $2$, $\dfrac{x + y}{x - y}$ has degree $0$, and $x^2 + y$ is not homogeneous.

An equation $\dfrac{dy}{dx} = \dfrac{P(x, y)}{Q(x, y)}$ is homogeneous when $P$ and $Q$ are homogeneous of the same degree. Then the right side is a function of $\dfrac{y}{x}$ alone, which is $F\!\left(\dfrac{y}{x}\right)$, because we can take $\lambda = \dfrac{1}{x}$.

A warning about the word. In NCERT and JEE Main this is a "homogeneous differential equation". In A-Level Further Maths, "homogeneous" means a linear equation with zero on the right side, such as $y'' + 3y' + 2y = 0$. Those are different ideas. Here we use the Indian meaning.

The steps

1. Check. Write the equation as $\dfrac{dy}{dx} = F\!\left(\dfrac{y}{x}\right)$. Test the degrees, or replace $x, y$ by $\lambda x, \lambda y$.
2. Substitute. Put $y = vx$. By the product rule, $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.
3. Separate. The equation becomes $x\dfrac{dv}{dx} = F(v) - v$. Then $\dfrac{dv}{F(v) - v} = \dfrac{dx}{x}$.
4. Integrate both sides.
5. Return. Replace $v$ by $\dfrac{y}{x}$. Then use the point to fix $C$.

If $F(v) - v$ is zero for some $v = v_1$, then $y = v_1 x$ is a straight-line solution that the division loses.

Worked examples

Example 1. Solve $(x^2 + y^2)\,dx = 2xy\,dy$.

1. Write $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{2xy}$. This is homogeneous of degree $0$.
2. With $y = vx$: $v + x\dfrac{dv}{dx} = \dfrac{1 + v^2}{2v}$. So $x\dfrac{dv}{dx} = \dfrac{1 - v^2}{2v}$.
3. Separate: $\dfrac{2v\,dv}{1 - v^2} = \dfrac{dx}{x}$. Integrate: $-\ln|1 - v^2| = \ln|x| + c$.
4. So $x(1 - v^2) = C$. Put $v = \dfrac{y}{x}$: $x - \dfrac{y^2}{x} = C$. Multiply by $x$: $x^2 - y^2 = Cx$.

Example 2. Solve $x\dfrac{dy}{dx} = y + x\sin^2\dfrac{y}{x}$ with $y = \dfrac{\pi}{4}$ at $x = 1$.

1. Divide by $x$: $\dfrac{dy}{dx} = \dfrac{y}{x} + \sin^2\dfrac{y}{x}$. With $y = vx$: $v + x\dfrac{dv}{dx} = v + \sin^2 v$.
2. So $x\dfrac{dv}{dx} = \sin^2 v$. Separate: $\operatorname{cosec}^2 v\,dv = \dfrac{dx}{x}$.
3. Integrate: $-\cot v = \ln x + c$ for $x > 0$. So $\cot\dfrac{y}{x} = C - \ln x$.
4. At $(1, \dfrac{\pi}{4})$: $\cot\dfrac{\pi}{4} = 1 = C - 0$. So $\cot\dfrac{y}{x} = 1 - \ln x$.

Example 3. Solve $\dfrac{dy}{dx} = \dfrac{y}{x + y}$. Here $x = vy$ is shorter, because the equation is easier to write for $\dfrac{dx}{dy}$.

1. Turn it over: $\dfrac{dx}{dy} = \dfrac{x + y}{y} = \dfrac{x}{y} + 1$.
2. Put $x = vy$, so $\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$. Then $v + y\dfrac{dv}{dy} = v + 1$.
3. So $\dfrac{dv}{dy} = \dfrac{1}{y}$ and $v = \ln|y| + C$.
4. Return: $\dfrac{x}{y} = \ln|y| + C$, or $x = y\ln|y| + Cy$.

With $y = vx$ the same equation takes more steps. Choose the substitution that gives the simpler right side.

A non-example. Try $\dfrac{dy}{dx} = \dfrac{x + y + 1}{x - y}$. The top has a constant $1$, so replacing $x, y$ by $\lambda x, \lambda y$ does not give a power of $\lambda$. The equation is not homogeneous, and $y = vx$ does not separate it. It needs a shift of origin, which the Standard course of this chapter explains.

Practise with new numbers

Try it in Python

The first part prints the slope of $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{2xy}$ at three points on each of three rays. The second part compares the slope of a solution $x^2 - y^2 = 3x$ with the equation.

Common mistakes

• Forgetting that $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. Treating $v$ as a constant loses the second term.
• Mixing $x$ and $v$ after the substitution. Separate completely before you integrate.
• Not returning to $y$. The answer must be in $x$ and $y$, not in $v$.
• Using the substitution on a function that is not homogeneous. Check the degrees first.
• Confusing the words. A-Level "homogeneous" is a different idea, as explained above.

Remember these

• $f(\lambda x, \lambda y) = \lambda^n f(x, y)$ means that $f$ is homogeneous of degree $n$.
• A homogeneous equation has $\dfrac{dy}{dx} = F\!\left(\dfrac{y}{x}\right)$. Put $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.
• After separating, integrate and replace $v$ by $\dfrac{y}{x}$.
• Use $x = vy$ when the equation is simpler for $\dfrac{dx}{dy}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Is $f(x, y) = \dfrac{x^3 + y^3}{xy}$ homogeneous? If so, of what degree?

2. Use $y = vx$ to solve $\dfrac{dy}{dx} = \dfrac{2x + y}{x}$.

3. Is $x\dfrac{dy}{dx} = y + x^2$ homogeneous?

4. Solve $\dfrac{dy}{dx} = \dfrac{x^2 + 3y^2}{2xy}$.

5. Solve $\dfrac{dy}{dx} = \dfrac{xy}{x^2 + y^2}$ for $y > 0$, using $x = vy$.

References

• OpenStax, Calculus Volume 2, Chapter 4, section 4.3 Separable Differential Equations (for the separable form that the substitution reaches).
• NCERT, Mathematics Part II for Class XII, Chapter 9, the section on homogeneous differential equations.