2.3 Homogeneous differential equation

  Differential Equations

A differential equation of the form $$\frac{{dy}}{{dx}} = \frac{{f(x,y)}}{{g(x,y)}}$$ where $f$ and $g$ are homogeneous functions of same degree in $x$ and $y$, is the homogeneous differential equation.

The solution of homogeneous differential equation can be find out by putting $y=vx$ or $x=vy$ in the given differential equation where $v$ is some unknown function.

This substitution is done to transform the given equation to a variable separable form and solution can be found out by separating the variables and then integrating it.

Question 7. Solve $$({x^2} - {y^2})dx + 2xydy = 0$$ Given that $y=1$ when $x=1$.

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

Put $y=vx$, we get $$\frac{{dy}}{{dx}} = v + \frac{{dv}}{{dx}}$$$$\Rightarrow v + x\frac{{dv}}{{dx}} = - \frac{{1 - {v^2}}}{{2v}}$$$$\Rightarrow \int {\frac{{2v}}{{1 + {v^2}}}dv = - \int {\frac{{dx}}{x}} }$$$$\Rightarrow \ln (1 + {v^2}) = - \ln x + c $$

At $x=1$, $y=1$. Put in $y=vx$, we get $v=1$. Therefore, $$\ln (1 + {1^2}) = - \ln 1 + c$$$$c = \ln 2$$ $$\therefore \ln \left\{ {\left( {1 + \frac{{{y^2}}}{{{x^2}}}} \right) - x} \right\} = \ln 2$$$$\Rightarrow {x^2} + {y^2} = 2x$$