Question

Show that the differential equation $\dfrac{{dy}}{{dx}} = \dfrac{{{y^2}}}{{xy - {x^2}}}$ is homogeneous and solve the same.

Answer 1

We know that the differential equation of the first order and of the first degree can be expressed in the form $Mdx + Ndy = 0$, where M and N are both functions of x and y or constants and if M and N are both homogeneous functions of same degree in x and y then the equation is said to be homogeneous equation.

Complete step by step solution:
$\dfrac{{dy}}{{dx}} = \dfrac{{{y^2}}}{{xy - {x^2}}}$ …………………. 1
Let,
$\dfrac{{dy}}{{dx}} = \dfrac{{{y^2}}}{{xy - {x^2}}} = f\left( {x,y} \right)$
Now substitute $x$ by $\lambda x$and $y$ by $\lambda y$, as
$f\left( {\lambda x,\lambda y} \right) = \dfrac{{{\lambda ^2}{y^2}}}{{{\lambda ^2}xy - {\lambda ^2}{x^2}}}$
$f\left( {\lambda x,\lambda y} \right) = \dfrac{{{\lambda ^2}{y^2}}}{{{\lambda ^2}\left( {xy - {x^2}} \right)}}$
Hence, we get
$f\left( {\lambda x,\lambda y} \right) = \dfrac{{{y^2}}}{{xy - {x^2}}} = f\left( {x,y} \right)$ where, $\lambda \ne 0$
Thus, the given differential equation is a Homogeneous differential equation.
Now, to solve it
Let, $y = vx$
i.e.,
$\dfrac{{dy}}{{dx}} = x\dfrac{{dv}}{{dx}} + v$
From equation 1, we get
$x\dfrac{{dv}}{{dx}} + v = \dfrac{{{x^2}{v^2}}}{{{x^2}v - {x^2}}} = \dfrac{{{v^2}}}{{v - 1}}$
$x\dfrac{{dv}}{{dx}} = \dfrac{{{v^2}}}{{v - 1}} - v = \dfrac{{{v^2} - {v^2} + v}}{{v + 1}}$
$x\dfrac{{dv}}{{dx}} = \dfrac{v}{{v - 1}}$or
$x\dfrac{{dv}}{{dx}} = \dfrac{{v - 1}}{v}dv = \dfrac{{dx}}{x}$
Integrating both sides of the equation, we get
$\int {\left( {1 - \dfrac{1}{v}} \right)dv = \int {\dfrac{{dv}}{x}} }$
$v - \log \left| v \right| = \log \left| x \right| + C$
$\dfrac{y}{x} - \log \left| {\dfrac{y}{x}} \right| = \log \left| x \right| + C$ ……………….. 2
Substituting $v = \dfrac{y}{x}$ in equation 2 we get
$\dfrac{y}{x} - \log \left| y \right| + \log \left| x \right| = \log \left| x \right| + C$
We know that,
$\log \left( {\dfrac{m}{n}} \right) = \log m - \log n$
Therefore, the required solution is
$\dfrac{y}{x} - \log \left| y \right| = C$

Additional information:
An equation which involves derivatives of a dependent variable with respect to another independent variable is called a differential equation. If a function has only one independent variable then it is an ordinary differential equation. Differential equation involving a function of several variables of its partial derivatives is called a partial differential equation.
There are 5 methods for solving the differential equation:
Solution by inspection
Variable separable
Homogeneous
Linear differential equation
General

Note: A differential equation is homogeneous if it is a similar function of the anonymous function and its derivatives. For linear differential equations, there are no constant terms. The solutions of any linear ordinary differential equation of any degree or order may be calculated by integration from the solution of the homogeneous equation achieved by eliminating the constant term.