Homogeneous Differential Equations

Homogeneous Functions and the Degree-Zero Test

Homogeneous functions

Scale both variables — replace $x$ by $\lambda x$ and $y$ by $\lambda y$ — and watch what happens:

1. $F_1(x, y) = y^2 + 2xy$: $F_1(\lambda x, \lambda y) = \lambda^2 F_1(x, y)$ — degree $2$.
2. $F_2(x, y) = 2x - 3y$: scales by $\lambda^1$ — degree $1$.
3. $F_3(x, y) = \cos\frac{y}{x}$: the $\lambda$'s cancel — $\lambda^0 F_3$, degree $0$.
4. $F_4(x, y) = \sin x + \cos y$: scales by no power of $\lambda$ — not homogeneous.

Definition: $F(x, y)$ is a homogeneous function of degree $n$ if $F(\lambda x, \lambda y) = \lambda^n F(x, y)$ for every nonzero constant $\lambda$. Equivalently, $F(x, y) = x^n g\left(\frac{y}{x}\right)$ or $y^n h\left(\frac{x}{y}\right)$.

Homogeneous differential equations

Definition: $\dfrac{dy}{dx} = F(x, y)$ is a homogeneous differential equation if $F$ is homogeneous of degree zero — i.e. the right side can be written entirely in terms of the ratio $\frac{y}{x}$.

The quick recognition test: in a quotient like $\frac{x + 2y}{x - y}$, every term in numerator and denominator has the same total degree (here $1$) — divide top and bottom by $x$ and only $\frac yx$ remains: homogeneous. In $\frac{x^3 + 2y^2}{2xy}$, the degrees mix ($3$ versus $2$) — not homogeneous.

Key Point: Degree-zero homogeneity means the slope field looks the same along every ray through the origin — which is exactly why the ratio $v = \frac yx$ is the natural variable, and why the substitution $y = vx$ (next block) untangles the equation.

[JEE Tip] Terms with constants added — like $(4x + 6y + 5)$ — break homogeneity instantly (the $5$ has degree $0$ while $x, y$ have degree $1$). Scan for stray constants first; it eliminates options in recognition MCQs in seconds.

The Substitution $y = vx$

To solve $\dfrac{dy}{dx} = g\left(\dfrac{y}{x}\right)$:

1. Substitute $y = vx$ (where $v$ is a function of $x$). Differentiating: $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.
2. The equation becomes $v + x\frac{dv}{dx} = g(v)$, i.e. $$x\frac{dv}{dx} = g(v) - v$$
3. Separate (the whole point — $v$ and $x$ now separate): $\dfrac{dv}{g(v) - v} = \dfrac{dx}{x}$.
4. Integrate and back-substitute $v = \frac{y}{x}$: $$\int\frac{dv}{g(v) - v} = \log|x| + C$$

The mirror form $x = vy$

If the equation arrives as $\dfrac{dx}{dy} = h\left(\dfrac{x}{y}\right)$ — degree-zero in the other orientation — substitute $x = vy$, so $\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$, and proceed identically. Equations written in differentials ($2ye^{x/y}dx + (y - 2xe^{x/y})dy = 0$) often signal this form: solve for $\frac{dx}{dy}$ and check for $\frac xy$.

Working habits

1. Show homogeneity first (the exam asks for it): verify $F(\lambda x, \lambda y) = \lambda^0 F(x, y)$, or rewrite the right side as a function of $\frac yx$.
2. Don't forget $-v$: the separable form is $\frac{dv}{g(v) - v}$, not $\frac{dv}{g(v)}$ — omitting the $-v$ is the classic slip.
3. Back-substitute at the end, and simplify constants ($\log$-merging, $\pm C \to C$) exactly as in Section 3.

Key Point: The substitution machine has one purpose: convert to separable. After step 2 you are back in Section 3 territory — every homogeneous problem is a separable problem wearing a disguise.

[JEE Tip] The $v$-integral $\int\frac{dv}{g(v)-v}$ is where Chapter 7 skills return: expect partial fractions ($\frac{2v}{v^2 - 1}$), standard forms ($\frac{v-1}{v^2+v+1}$ needs completing the square), or a plain $\frac{f'}{f}$. Identify the integral type before grinding.

Solved Examples

Example 1: The full showcase

Show that $(x - y)\dfrac{dy}{dx} = x + 2y$ is homogeneous and solve it.

Solution:

1. Homogeneity: $\frac{dy}{dx} = \frac{x + 2y}{x - y}$; replacing $(x, y)$ by $(\lambda x, \lambda y)$ leaves it unchanged ($\lambda^0$) — homogeneous.
2. Substitute $y = vx$: $v + x\frac{dv}{dx} = \frac{1 + 2v}{1 - v}$, so $x\frac{dv}{dx} = \frac{v^2 + v + 1}{1 - v}$.
3. Separate: $\frac{v - 1}{v^2 + v + 1}dv = -\frac{dx}{x}$.
4. Integrate (split the numerator along the derivative $2v + 1$, complete the square for the remainder): $\frac12\log\left(v^2 + v + 1\right) - \sqrt3\tan^{-1}\frac{2v + 1}{\sqrt3} = -\log|x| + C_1$.
5. Back-substitute $v = \frac yx$ and tidy:

Final Answer: $\log\left(x^2 + xy + y^2\right) = 2\sqrt 3\tan^{-1}\left(\dfrac{x + 2y}{\sqrt 3\,x}\right) + C$.

Takeaway: The $v$-integral used both Chapter 7 power-moves: numerator $=$ A(derivative) $+$ B, then the completed-square $\tan^{-1}$ form. Homogeneous problems are integration problems in disguise.

Example 2: A trig ratio

Show that $x\cos\left(\dfrac{y}{x}\right)\dfrac{dy}{dx} = y\cos\left(\dfrac{y}{x}\right) + x$ is homogeneous and solve it.

Solution:

1. Homogeneity: $\frac{dy}{dx} = \frac{y\cos(y/x) + x}{x\cos(y/x)}$ — scaling cancels ($\lambda^0$): homogeneous.
2. Substitute $y = vx$: $v + x\frac{dv}{dx} = \frac{v\cos v + 1}{\cos v}$, so $x\frac{dv}{dx} = \frac{1}{\cos v}$.
3. Separate and integrate: $\cos v\,dv = \frac{dx}{x}$ gives $\sin v = \log|x| + \log|C|$.
4. Back-substitute:

Final Answer: $\sin\left(\dfrac{y}{x}\right) = \log|Cx|$.

Example 3: The $x = vy$ variant with a condition

Show that $2ye^{x/y}dx + \left(y - 2xe^{x/y}\right)dy = 0$ is homogeneous and find the particular solution with $x = 0$ when $y = 1$.

Solution:

1. Rearrange: $\frac{dx}{dy} = \frac{2xe^{x/y} - y}{2ye^{x/y}}$ — a degree-zero function of $\frac xy$: homogeneous, mirror form.
2. Substitute $x = vy$: $v + y\frac{dv}{dy} = \frac{2ve^v - 1}{2e^v}$, so $y\frac{dv}{dy} = -\frac{1}{2e^v}$.
3. Separate and integrate: $2e^v dv = -\frac{dy}{y}$ gives $2e^v = -\log|y| + C$.
4. Back-substitute and apply the condition: $2e^{x/y} + \log|y| = C$; at $(0, 1)$: $2 + 0 = C$.

Final Answer: $2e^{x/y} + \log|y| = 2$.

Example 4: A family of curves

Show that the family of curves whose tangent slope at any point $(x, y)$ is $\dfrac{x^2 + y^2}{2xy}$ is $x^2 - y^2 = Cx$.

Solution:

1. Set up: $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy} = \frac{1 + v^2}{2v}$ with $v = \frac yx$ — homogeneous.
2. Substitute: $x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 - v^2}{2v}$.
3. Separate: $\frac{2v}{v^2 - 1}dv = -\frac{dx}{x}$; integrate: $\log\left|v^2 - 1\right| = -\log|x| + \log|C_1|$.
4. Exponentiate and back-substitute: $\left(v^2 - 1\right)x = \pm C_1$, i.e. $y^2 - x^2 = \pm C_1 x$.

Final Answer: $x^2 - y^2 = Cx$.

Example 5: A short one

Solve $y' = \dfrac{x + y}{x}$.

Solution:

1. Recognise: $\frac{x+y}{x} = 1 + \frac yx$ — homogeneous.
2. Substitute $y = vx$: $v + x\frac{dv}{dx} = 1 + v$, so $x\frac{dv}{dx} = 1$.
3. Separate and integrate: $dv = \frac{dx}{x}$ gives $v = \log|x| + C$.
4. Back-substitute: $\frac yx = \log|x| + C$.

Final Answer: $y = x\log|x| + Cx$.

Example 6: Particular solution with an arctangent

Solve $(x + y)\,dy + (x - y)\,dx = 0$, given $y = 1$ when $x = 1$.

Solution:

1. Rearrange: $\frac{dy}{dx} = \frac{y - x}{y + x}$ — homogeneous.
2. Substitute $y = vx$: $x\frac{dv}{dx} = \frac{v - 1}{v + 1} - v = -\frac{v^2 + 1}{v + 1}$.
3. Separate: $\frac{v + 1}{v^2 + 1}dv = -\frac{dx}{x}$; integrate term-wise: $\frac12\log\left(v^2 + 1\right) + \tan^{-1}v = -\log|x| + C_1$.
4. Back-substitute and tidy: $\log\left(x^2 + y^2\right) + 2\tan^{-1}\frac yx = C$.
5. Apply $(1, 1)$: $\log 2 + 2\cdot\frac{\pi}{4} = C$.

Final Answer: $\log\left(x^2 + y^2\right) + 2\tan^{-1}\dfrac{y}{x} = \log 2 + \dfrac{\pi}{2}$.

Example 7: Partial fractions in the $v$-integral

Solve $x^2dy + \left(xy + y^2\right)dx = 0$, given $y = 1$ when $x = 1$.

Solution:

1. Rearrange: $\frac{dy}{dx} = -\frac{xy + y^2}{x^2} = -\left(v + v^2\right)$ with $v = \frac yx$ — homogeneous.
2. Substitute: $x\frac{dv}{dx} = -v^2 - 2v$.
3. Separate and split: $\frac{dv}{v(v + 2)} = -\frac{dx}{x}$; partial fractions: $\frac12\log\left|\frac{v}{v+2}\right| = -\log|x| + C_1$.
4. Exponentiate and back-substitute: $\frac{v}{v + 2} = \frac{C}{x^2}$ gives $\frac{y}{y + 2x} = \frac{C}{x^2}$, i.e. $x^2y = C(y + 2x)$.
5. Apply $(1, 1)$: $1 = 3C$, so $C = \frac13$.

Final Answer: $3x^2y = y + 2x$, i.e. $y + 2x = 3x^2y$.

Takeaway: Steps 1-2 are always the same three lines; all the variety lives in the $v$-integral. Bring your Chapter 7 toolkit — here, a two-factor partial fraction decomposition.