# Homogeneous Differential Equation

## Homogeneous Differential Equation

A differential equation $$\frac{dy}{dx}=\frac{f(x,y)}{g(x,y)}$$ is said to be a homogeneous differential equation in x, y if both f (x, y), g (x, y) are homogeneous function of same degree in x and y.

To find the solution of the Homogeneous Differential Equation put y = v x, then $$\frac{dy}{dx}=v+x\frac{dv}{dx}$$. Substituting these values in given differential equation, then it reduces to variable separable form. Then we find the solution of the differential equation.

Example: Solve the differential equation $$\frac{dy}{dx}=\frac{x-y}{x+y}$$.

Solution: Given that $$\frac{dy}{dx}=\frac{x-y}{x+y}$$ … (1)

Equation (1) is a Homogeneous Differential Equation

Put y = vx

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

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

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

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

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

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

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

$$\frac{-1}{2}\int{\frac{-2(1+v)dv}{1-2v-{{v}^{2}}}}=\int{\frac{dx}{x}}$$ $$\left( \because \int{\frac{f'(x)}{f(x)}}=\log |f(x)|+c \right)$$.

$$\frac{-1}{2}\log |(1-2v-{{v}^{2}})|=\log x+\log c$$,

$$\frac{-1}{2}\log |(1-2v-{{v}^{2}})|=\log xc$$,

$$\frac{-1}{2}\log |(1-\frac{2y}{x}-\frac{{{y}^{2}}}{{{x}^{2}}})|=\log xc$$,

$$\frac{-1}{2}\log |\left( \frac{{{x}^{2}}-2yx-{{y}^{2}}}{{{x}^{2}}} \right)|=\log xc$$,

$$\log |{{\left( \frac{{{x}^{2}}-2yx-{{y}^{2}}}{{{x}^{2}}} \right)}^{\frac{-1}{2}}}|=\log xc$$,

$$\log |{{\left( \frac{{{x}^{2}}}{{{x}^{2}}-2yx-{{y}^{2}}} \right)}^{\frac{1}{2}}}|=\log xc$$,

$${{\left( \frac{{{x}^{2}}}{{{x}^{2}}-2yx-{{y}^{2}}} \right)}^{\frac{1}{2}}}=xc$$,

$$\left( \frac{{{x}^{2}}}{{{x}^{2}}-2yx-{{y}^{2}}} \right)={{x}^{2}}{{c}^{2}}$$,

$${{x}^{2}}-2yx-{{y}^{2}}=\frac{1}{{{c}^{2}}}=k$$,

$${{x}^{2}}-2yx-{{y}^{2}}=k$$.