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
\(\frac{dy}{dx}=v+x\frac{dv}{dx}\),
\(v+x\frac{dv}{dx}=\frac{x-vx}{x+vx}\),
\(v+x\frac{dv}{dx}=\frac{1-v}{1+v}\),
\(x\frac{dv}{dx}=\frac{1-v}{1+v}-v\),
\(x\frac{dv}{dx}=\frac{1-2v-{{v}^{2}}}{1+v}\),
\(\int{\frac{(1+v)dv}{1-2v-{{v}^{2}}}}=\int{\frac{dx}{x}}\),
\(\frac{1}{2}\int{\frac{2(1+v)dv}{1-2v-{{v}^{2}}}}=\int{\frac{dx}{x}}\),
\(\frac{-1}{2}\int{\frac{-2(1+v)dv}{1-2v-{{v}^{2}}}}=\int{\frac{dx}{x}}\) \(\left( \because \int{\frac{f'(x)}{f(x)}}=\log |f(x)|+c \right)\).
\(\frac{-1}{2}\log |(1-2v-{{v}^{2}})|=\log x+\log c\),
\(\frac{-1}{2}\log |(1-2v-{{v}^{2}})|=\log xc\),
\(\frac{-1}{2}\log |(1-\frac{2y}{x}-\frac{{{y}^{2}}}{{{x}^{2}}})|=\log xc\),
\(\frac{-1}{2}\log |\left( \frac{{{x}^{2}}-2yx-{{y}^{2}}}{{{x}^{2}}} \right)|=\log xc\),
\(\log |{{\left( \frac{{{x}^{2}}-2yx-{{y}^{2}}}{{{x}^{2}}} \right)}^{\frac{-1}{2}}}|=\log xc\),
\(\log |{{\left( \frac{{{x}^{2}}}{{{x}^{2}}-2yx-{{y}^{2}}} \right)}^{\frac{1}{2}}}|=\log xc\),
\({{\left( \frac{{{x}^{2}}}{{{x}^{2}}-2yx-{{y}^{2}}} \right)}^{\frac{1}{2}}}=xc\),
\(\left( \frac{{{x}^{2}}}{{{x}^{2}}-2yx-{{y}^{2}}} \right)={{x}^{2}}{{c}^{2}}\),
\({{x}^{2}}-2yx-{{y}^{2}}=\frac{1}{{{c}^{2}}}=k\),
\({{x}^{2}}-2yx-{{y}^{2}}=k\).