Question

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

Answer 1

In this question first order differential equation is given $\dfrac{{dy}}{{dx}} = f(x,y)$, which is called homogeneous equation. If the right side satisfies the condition $f(tx,ty) = f(x,y)$for all $f$. In other words 1st order differential equation in the form $\dfrac{{dy}}{{dx}} = f(x,y)$ is homogeneous, if it does not depend on $x$ and $y$ separately but only the ratio $\dfrac{x}{y}$or $\dfrac{y}{x}$. Homogeneous equations are in the form $\dfrac{{dy}}{{dx}} = F(\dfrac{y}{x})$.

Complete step by step answer:
$\dfrac{{dy}}{{dx}} = \dfrac{{{y^2}}}{{xy - {x^2}}}$
Now, we will try to show $\dfrac{{dy}}{{dx}} = f(x,y)$ which does not depend on the $x$ and $y$separately but only on the ratio $\dfrac{x}{y}$ or $\dfrac{y}{x}$.
As given in the differential equation $\dfrac{{dy}}{{dx}} = \dfrac{{{y^2}}}{{xy - {x^2}}}$.
So, $\dfrac{{dy}}{{dx}} = \dfrac{{{y^2}}}{{xy - {x^2}}}$
Dividing left hand side numerator and denominator by ${x^2}$, we have,
$\dfrac{{dy}}{{dx}} = \dfrac{{{{(\dfrac{y}{x})}^2}}}{{(\dfrac{y}{x}) - 1}}$
Hence, the expression shows that the 1st order differential equation does not depend on $x$ and $y$ but only in the form of $\dfrac{x}{y}$ or $\dfrac{y}{x}$. So, it is homogeneous.
Now, we are going to solve the equation so, assume that
$\dfrac{y}{x} = v$ so,
$y = vx$
Now, differentiating both sides with respect to$dx$ , we have

$$\because \dfrac{{dy}}{{dx}} = v\dfrac{{dx}}{{dx}} + x\dfrac{{dv}}{{dx}} \to equation(1) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}} \\\ \because \dfrac{{dy}}{{dx}} = \dfrac{{{{(\dfrac{y}{x})}^2}}}{{(\dfrac{y}{x}) - 1}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{v^2}}}{{v - 1}} \to equation(2) \\\$$

Now, putting the value of equation (2) in equation (1)

$$\Rightarrow x\dfrac{{dv}}{{dx}} = \dfrac{{{v^2}}}{{v - 1}} - v \\\ \Rightarrow x\dfrac{{dv}}{{dx}} = \dfrac{{{v^2} - {v^2} + v}}{{v - 1}} \\\ \Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{v}{{v - 1}} \times \dfrac{1}{x} \\\ \Rightarrow \smallint \tfrac{{v - 1}}{v}dv = \smallint \tfrac{{dx}}{x} \\\ \Rightarrow \smallint (1 - \tfrac{1}{v})dv = \log \left| x \right| \\\ \Rightarrow v - \log \left| v \right| = \log \left| x \right| + c \\\ \Rightarrow v - \log \left| {v + x} \right| + c = 0 \\\$$

Now, putting the value of v , $[\because y = vx] \Rightarrow v = \dfrac{y}{x}$
$\therefore \dfrac{y}{x} - \log \left| {\dfrac{y}{x} + x} \right| + c = 0$

Note:
In this question, first check if the equation is homogeneous or not. If yes then go ahead, if not then first homogenize the equation and then solve. Be careful while putting the value of v when the equation is solved