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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\({{x}^{2}}-2yx-{{y}^{2}}=k\).