Homogeneous Differential Equations

Homogeneous Functions

Before solving homogeneous differential equations, we must understand what a homogeneous function is.

A function $F(x, y)$ is said to be a homogeneous function of degree $n$ if, upon replacing $x$ with $\lambda x$ and $y$ with $\lambda y$ (where $\lambda$ is any non-zero constant), the factor $\lambda$ comes out completely as $\lambda^n$: $$F(\lambda x, \lambda y) = \lambda^n F(x, y)$$

This means that every term of the function scales in the same way.

Examples:

• $F(x, y) = x^2 + 2xy + y^2$ $$F(\lambda x, \lambda y) = (\lambda x)^2 + 2(\lambda x)(\lambda y) + (\lambda y)^2 = \lambda^2(x^2 + 2xy + y^2) = \lambda^2 F(x, y)$$ So this is homogeneous of degree 2.

• $G(x, y) = \sin\left(\frac{y}{x}\right)$ $$G(\lambda x, \lambda y) = \sin\left(\frac{\lambda y}{\lambda x}\right) = \sin\left(\frac{y}{x}\right) = \lambda^0 G(x, y)$$ So this is homogeneous of degree 0.

• $H(x, y) = \sin x + \cos y$ $$H(\lambda x, \lambda y) = \sin(\lambda x) + \cos(\lambda y)$$ Here no single power of $\lambda$ can be factored out. So this is not homogeneous.

Homogeneous Differential Equations

A first-order, first-degree differential equation of the form $$\frac{dy}{dx} = F(x, y)$$ is called a homogeneous differential equation if $F(x, y)$ is a homogeneous function of degree zero.

That means $$F(\lambda x, \lambda y) = F(x, y)$$ for every non-zero constant $\lambda$.

Equivalently, such an equation can be rewritten in the form $$\frac{dy}{dx} = f\left(\frac{y}{x}\right)$$ or, in some cases, more conveniently as $$\frac{dx}{dy} = g\left(\frac{x}{y}\right).$$

This is the key observation that allows us to solve homogeneous equations.

The Solving Algorithm

Homogeneous differential equations are generally not directly separable, but they become separable after a standard substitution.

Method 1: Substituting $y = vx$ This is used when the equation is written as $$\frac{dy}{dx} = f\left(\frac{y}{x}\right).$$

Steps:

1. Put $$y = vx,$$ where $v$ is a function of $x$.
2. Differentiate using the product rule: $$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$
3. Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ into the equation.
4. The equation will reduce to a separable equation in $v$ and $x$.
5. Integrate.
6. Replace $v$ by $\frac{y}{x}$ to get the final answer.

Method 2: Substituting $x = vy$ This is useful when it is easier to write the equation as $$\frac{dx}{dy} = g\left(\frac{x}{y}\right).$$

Steps:

1. Put $$x = vy,$$ where $v$ is a function of $y$.
2. Differentiate with respect to $y$: $$\frac{dx}{dy} = v + y\frac{dv}{dy}.$$
3. Substitute and simplify.
4. Separate variables and integrate.
5. Replace $v$ by $\frac{x}{y}$.

Important idea: The substitution works because a homogeneous function of degree zero depends only on the ratio $\frac{y}{x}$ or $\frac{x}{y}$.

Example 1: Basic Homogeneous Equation

Solve the differential equation: $(x+y)dy + (x-y)dx = 0$.

Solution: Step 1: Write the equation in the form $\frac{dy}{dx} = F(x,y)$. $$(x+y)dy = -(x-y)dx$$ $$\frac{dy}{dx} = -\frac{x-y}{x+y} = \frac{y-x}{x+y}$$ Now the numerator and denominator are both homogeneous expressions of degree 1, so their ratio is homogeneous of degree 0. Hence the equation is homogeneous.

Step 2: Use the substitution $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}.$$ Substitute into the equation: $$v + x\frac{dv}{dx} = \frac{vx - x}{vx + x} = \frac{v-1}{v+1}.$$

Step 3: Isolate the derivative term. $$x\frac{dv}{dx} = \frac{v-1}{v+1} - v$$ $$= \frac{v-1-v(v+1)}{v+1}$$ $$= \frac{v-1-v^2-v}{v+1} = -\frac{1+v^2}{v+1}.$$

Step 4: Separate variables. $$\frac{v+1}{1+v^2}dv = -\frac{dx}{x}.$$

Step 5: Integrate both sides. Split the left side: $$\int \frac{v}{1+v^2}dv + \int \frac{1}{1+v^2}dv = -\int \frac{dx}{x}.$$ So, $$\frac{1}{2}\ln(1+v^2) + \tan^{-1}v = -\ln|x| + C.$$

Step 6: Put $v = \frac{y}{x}$. $$\frac{1}{2}\ln\left(1+\frac{y^2}{x^2}\right) + \tan^{-1}\left(\frac{y}{x}\right) = -\ln|x| + C.$$ Now, $$\frac{1}{2}\ln\left(\frac{x^2+y^2}{x^2}\right) = \frac{1}{2}\ln(x^2+y^2) - \ln|x|,$$ so the $-\ln|x|$ on the right combines neatly, giving $$\frac{1}{2}\ln(x^2+y^2) + \tan^{-1}\left(\frac{y}{x}\right) = C.$$

Answer: $\frac{1}{2}\ln(x^2 + y^2) + \tan^{-1}\left(\frac{y}{x}\right) = C$

Example 2: Verifying and Solving

Show that the differential equation $x^2 dy + y(x+y) dx = 0$ is homogeneous and solve it.

Solution: Step 1: Rewrite in differential form. $$x^2\frac{dy}{dx} = -y(x+y)$$ $$\frac{dy}{dx} = -\frac{xy+y^2}{x^2}.$$ Let $$F(x,y) = -\frac{xy+y^2}{x^2}.$$ Then, $$F(\lambda x, \lambda y) = -\frac{(\lambda x)(\lambda y) + (\lambda y)^2}{(\lambda x)^2} = -\frac{\lambda^2(xy+y^2)}{\lambda^2x^2} = F(x,y).$$ So $F$ is homogeneous of degree 0. Hence the differential equation is homogeneous.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}.$$ Substitute: $$v + x\frac{dv}{dx} = -(v+v^2).$$

Step 3: Simplify. $$x\frac{dv}{dx} = -v-v^2-v = -(v^2+2v).$$ Thus, $$\frac{dv}{v(v+2)} = -\frac{dx}{x}.$$

Step 4: Integrate. Using partial fractions, $$\frac{1}{v(v+2)} = \frac{1}{2}\left(\frac{1}{v} - \frac{1}{v+2}\right).$$ Therefore, $$\frac{1}{2}\ln\left|\frac{v}{v+2}\right| = -\ln|x| + C.$$ Multiply by 2: $$\ln\left|\frac{v}{v+2}\right| = -2\ln|x| + C_1 = \ln\left|\frac{K}{x^2}\right|.$$ So, $$\frac{v}{v+2} = \frac{K}{x^2}.$$

Step 5: Replace $v=\frac{y}{x}$. $$\frac{y/x}{y/x+2} = \frac{K}{x^2}$$ $$\frac{y}{y+2x} = \frac{K}{x^2}.$$ Cross-multiplying, $$x^2y = K(y+2x).$$

Answer: $x^2 y = K(y + 2x)$

Example 3: When to use $x = vy$

Solve the differential equation: $2ye^{x/y} dx + (y - 2xe^{x/y}) dy = 0$.

Solution: Step 1: Since the term $e^{x/y}$ appears, it is more natural to write the equation in terms of $\frac{dx}{dy}$ and use the substitution $x=vy$.

Rearrange: $$2ye^{x/y}dx = (2xe^{x/y} - y)dy$$ $$\frac{dx}{dy} = \frac{2xe^{x/y} - y}{2ye^{x/y}}.$$

Step 2: Put $$x = vy \implies \frac{dx}{dy} = v + y\frac{dv}{dy}.$$ Substitute: $$v + y\frac{dv}{dy} = \frac{2(vy)e^v - y}{2ye^v} = \frac{2ve^v - 1}{2e^v}.$$

Step 3: Isolate the derivative term. $$y\frac{dv}{dy} = \frac{2ve^v - 1}{2e^v} - v = -\frac{1}{2e^v}.$$

Step 4: Separate variables. $$2e^v dv = -\frac{dy}{y}.$$

Step 5: Integrate. $$\int 2e^v dv = -\int \frac{dy}{y}$$ $$2e^v = -\ln|y| + C.$$ Rearranging, $$2e^v + \ln|y| = C.$$

Step 6: Replace $v = \frac{x}{y}$. $$2e^{x/y} + \ln|y| = C.$$

Answer: $2e^{x/y} + \ln|y| = C$

Example 4: Trigonometric Homogeneous Equation

Solve: $x \frac{dy}{dx} = y + x \tan\left(\frac{y}{x}\right)$.

Solution: Step 1: Divide by $x$. $$\frac{dy}{dx} = \frac{y}{x} + \tan\left(\frac{y}{x}\right).$$ This is of the form $f(y/x)$, so it is homogeneous.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}.$$ Substitute: $$v + x\frac{dv}{dx} = v + \tan v.$$

Step 3: Simplify. $$x\frac{dv}{dx} = \tan v$$ $$\cot v \, dv = \frac{dx}{x}.$$

Step 4: Integrate. $$\int \cot v \, dv = \int \frac{dx}{x}$$ $$\ln|\sin v| = \ln|x| + C = \ln|Cx|.$$ So, $$\sin v = Cx.$$

Step 5: Replace $v = \frac{y}{x}$. $$\sin\left(\frac{y}{x}\right) = Cx.$$

Answer: $\sin\left(\frac{y}{x}\right) = Cx$

Example 5: Initial Value Problem (IVP)

Solve the initial value problem: $x \frac{dy}{dx} - y + x \sin\left(\frac{y}{x}\right) = 0$, given that $y = \pi$ when $x = 2$.

Solution: Step 1: Rewrite the equation. $$x\frac{dy}{dx} = y - x\sin\left(\frac{y}{x}\right)$$ $$\frac{dy}{dx} = \frac{y}{x} - \sin\left(\frac{y}{x}\right).$$

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}.$$ Substitute: $$v + x\frac{dv}{dx} = v - \sin v.$$

Step 3: Simplify and separate. $$x\frac{dv}{dx} = -\sin v$$ $$\csc v \, dv = -\frac{dx}{x}.$$

Step 4: Integrate. Using $\int \csc v \, dv = \ln|\csc v - \cot v|$, $$\ln|\csc v - \cot v| = -\ln|x| + C = \ln\left|\frac{C}{x}\right|.$$ Therefore, $$\csc v - \cot v = \frac{C}{x}.$$

Step 5: Replace $v = \frac{y}{x}$. $$\csc\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right) = \frac{C}{x}.$$

Step 6: Apply the condition $x=2$, $y=\pi$. Then $\frac{y}{x} = \frac{\pi}{2}$, so $$\csc\left(\frac{\pi}{2}\right) - \cot\left(\frac{\pi}{2}\right) = \frac{C}{2}$$ $$1 - 0 = \frac{C}{2}$$ $$C = 2.$$

Hence the particular solution is $$\csc\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right) = \frac{2}{x}.$$

Answer: $\csc\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right) = \frac{2}{x}$

Example 6: Radical Homogeneous Equation

Solve: $x \frac{dy}{dx} = y + \sqrt{x^2 + y^2}$.

Solution: Step 1: Divide by $x$: $$\frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{x^2+y^2}}{x}.$$ Now, $$\frac{\sqrt{x^2+y^2}}{x} = \sqrt{1+\left(\frac{y}{x}\right)^2}$$ when working in the standard homogeneous setting. Thus, $$\frac{dy}{dx} = \frac{y}{x} + \sqrt{1+\left(\frac{y}{x}\right)^2}.$$

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}.$$ Substitute: $$v + x\frac{dv}{dx} = v + \sqrt{1+v^2}.$$

Step 3: Simplify. $$x\frac{dv}{dx} = \sqrt{1+v^2}.$$ So, $$\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}.$$

Step 4: Integrate. Using $$\int \frac{dv}{\sqrt{1+v^2}} = \ln|v+\sqrt{1+v^2}|,$$ we get $$\ln|v+\sqrt{1+v^2}| = \ln|x| + C = \ln|Cx|.$$ Therefore, $$v+\sqrt{1+v^2} = Cx.$$

Step 5: Replace $v = \frac{y}{x}$. $$\frac{y}{x} + \sqrt{1+\frac{y^2}{x^2}} = Cx.$$ Multiplying through by $x$ gives $$y + \sqrt{x^2+y^2} = Cx^2.$$

Answer: $y + \sqrt{x^2 + y^2} = Cx^2$

Example 7: Implicit General Solution

Solve: $(x^2+y^2)dx - 2xy \, dy = 0$.

Solution: Step 1: Rearrange. $$2xy \, dy = (x^2+y^2)dx$$ $$\frac{dy}{dx} = \frac{x^2+y^2}{2xy}.$$ This is homogeneous because numerator and denominator are both of degree 2.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}.$$ Substitute: $$v + x\frac{dv}{dx} = \frac{1+v^2}{2v}.$$

Step 3: Simplify. $$x\frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1-v^2}{2v}.$$ Thus, $$\frac{2v}{1-v^2}dv = \frac{dx}{x}.$$

Step 4: Integrate. Let $u = 1-v^2$, so $du = -2v \, dv$. Then $$\int \frac{2v}{1-v^2}dv = -\int \frac{du}{u} = -\ln|1-v^2|.$$ Therefore, $$-\ln|1-v^2| = \ln|x| + C.$$ So, $$\ln\left|\frac{1}{1-v^2}\right| = \ln|Cx|$$ $$\frac{1}{1-v^2} = Cx.$$

Step 5: Replace $v = \frac{y}{x}$. $$\frac{1}{1-y^2/x^2} = Cx$$ $$\frac{x^2}{x^2-y^2} = Cx.$$ This simplifies to $$x = C(x^2-y^2).$$ Equivalently, $$x^2-y^2 = Kx,$$ where $K = \frac{1}{C}$.

Answer: $x^2 - y^2 = Kx$

Example 8: Reverse Substitution Case ($x = vy$)

Solve: $2xy \, dx + (x^2 + 2y^2) dy = 0$.

Solution: Step 1: It is easier to isolate $\frac{dx}{dy}$. $$2xy \, dx = -(x^2+2y^2)dy$$ $$\frac{dx}{dy} = -\frac{x^2+2y^2}{2xy}.$$ This is homogeneous, so use $x=vy$.

Step 2: Put $$x = vy \implies \frac{dx}{dy} = v + y\frac{dv}{dy}.$$ Substitute: $$v + y\frac{dv}{dy} = -\frac{v^2+2}{2v}.$$

Step 3: Simplify. $$y\frac{dv}{dy} = -\frac{v^2+2}{2v} - v = -\frac{3v^2+2}{2v}.$$ So, $$\frac{2v}{3v^2+2}dv = -\frac{dy}{y}.$$

Step 4: Integrate. Let $u = 3v^2+2$, then $du = 6v \, dv$, so $2v \, dv = \frac{du}{3}$. Hence, $$\frac{1}{3}\int \frac{du}{u} = -\int \frac{dy}{y}$$ $$\frac{1}{3}\ln|3v^2+2| = -\ln|y| + C.$$ Multiply by 3: $$\ln|3v^2+2| = -3\ln|y| + C_1 = \ln\left|\frac{K}{y^3}\right|.$$ Therefore, $$y^3(3v^2+2) = K.$$

Step 5: Replace $v = \frac{x}{y}$. $$y^3\left(3\frac{x^2}{y^2} + 2\right) = K$$ $$3x^2y + 2y^3 = K.$$

Answer: $3x^2 y + 2y^3 = K$

Example 9: Complex Logarithmic Homogeneous Equation

Solve: $x \frac{dy}{dx} = y(\ln y - \ln x + 1)$.

Solution: Step 1: Use log properties. $$\ln y - \ln x = \ln\left(\frac{y}{x}\right).$$ So the equation becomes $$x\frac{dy}{dx} = y\left(\ln\left(\frac{y}{x}\right)+1\right)$$ $$\frac{dy}{dx} = \frac{y}{x}\left(\ln\left(\frac{y}{x}\right)+1\right).$$ This is homogeneous.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}.$$ Substitute: $$v + x\frac{dv}{dx} = v(\ln v + 1) = v\ln v + v.$$

Step 3: Simplify. $$x\frac{dv}{dx} = v\ln v.$$ Thus, $$\frac{dv}{v\ln v} = \frac{dx}{x}.$$

Step 4: Integrate. Let $u = \ln v$, then $du = \frac{1}{v}dv$. So, $$\int \frac{du}{u} = \int \frac{dx}{x}$$ $$\ln|u| = \ln|x| + C = \ln|Cx|.$$ Hence, $$u = Cx$$ that is, $$\ln v = Cx.$$

Step 5: Replace $v = \frac{y}{x}$. $$\ln\left(\frac{y}{x}\right) = Cx.$$

Answer: $\ln\left(\frac{y}{x}\right) = Cx$

Example 10: Special Form

Solve: $\frac{dy}{dx} = \frac{y^2}{xy - x^2}$.

Solution: Step 1: Rewrite in homogeneous form by dividing numerator and denominator by $x^2$. $$\frac{dy}{dx} = \frac{(y/x)^2}{(y/x)-1}.$$ So the equation is homogeneous.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}.$$ Substitute: $$v + x\frac{dv}{dx} = \frac{v^2}{v-1}.$$

Step 3: Simplify. $$x\frac{dv}{dx} = \frac{v^2}{v-1} - v$$ $$= \frac{v^2 - v(v-1)}{v-1} = \frac{v}{v-1}.$$ So, $$\frac{v-1}{v}dv = \frac{dx}{x}.$$

Step 4: Integrate. $$\int \left(1 - \frac{1}{v}\right) dv = \int \frac{dx}{x}$$ $$v - \ln|v| = \ln|x| + C.$$ Now, $$\ln|x| + \ln|v| = \ln|vx|,$$ so we may write $$v = \ln|vx| + C.$$

Step 5: Replace $v = \frac{y}{x}$. Since $vx = y$, $$\frac{y}{x} = \ln|y| + C.$$

Answer: $\frac{y}{x} = \ln|y| + C$