# Homogeneous Differential Equations

## Homogeneous Differential Equations

A differential equation of the form $$\frac{dy}{dx}=\frac{{{f}_{1}}\left( x,\,y \right)}{{{f}_{2}}\left( x,\,y \right)}$$, where f₁ (x, y) and f₂ (x, y) are homogeneous functions of x and y of the same degree, is called a homogeneous equation.

For solving such equations, put y = υx and $$\frac{dy}{dx}=\upsilon +x\frac{d\upsilon }{dx}$$,

These substitutions transform the given equation into an equation of the form $$\upsilon +x\frac{d\upsilon }{dx}=f\left( \upsilon \right)$$,

i.e., $$x\frac{d\upsilon }{dx}=f\left( \upsilon \right)-\upsilon$$,

The variables are now separable. Separating the variables and integrating, we get $$\int{\frac{d\upsilon }{f\left( \upsilon \right)-\upsilon }}=\log x+C$$,

Where C is an arbitrary constant.

Now, replacing r by (y/ x) a factor integration, we get the required solution.

Example: Find the solution of the differential equation (x – y) dy – (x + y) dx = 0.

Solution: Given, (x – y) dy – (x + y) dx = 0

(x – y) dy = (x + y) dx

$$\frac{dy}{dx}=\frac{x+y}{x-y}$$ … (i)

Thus, the given differential equation is homogeneous.

So, put y = υx ⇒ $$\frac{dy}{dx}=v+x\frac{dv}{dx}$$,

On putting values of dy/ dx and y in Eq. (i), we get

$$v+x\frac{dv}{dx}=\frac{x+vx}{x-vx}\Rightarrow v+x\frac{dv}{dx}=\frac{1+v}{1-v}$$,

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

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

On integrating both sides, we get

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

Let 1 + v² = t ⇒ $$2v=\frac{dt}{dv}\Rightarrow dv=\frac{dt}{2v}$$,

∴ $${{\tan }^{-1}}v-\int{\frac{v}{t}}\times \frac{dt}{2v}=\log |x|+C$$,

tan⁻¹ v = ½ log |t| = log |x| + C

2 tan⁻¹ v – [log (1 + v²) + 2 log x] = 2 C [put t = 1 + v²]

2 tan⁻¹ v – log [(1 + v²) x²] = 2C

$$2{{\tan }^{-1}}\frac{y}{x}-\log \left[ \left( \frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}} \right){{x}^{2}} \right]=2C$$ (Put v = y/ x)

2 tan⁻¹ (y/ x) – log (x² + y²) = 2 C

tan⁻¹ (y/ x) – ½ log (x² + y²) = C

This is the required solution of the given differential equation.