Question

The solution to the differential equation; $({x^2} - {y^2})dx + 2xydy = 0$, represents which curve-
(A). Line
(B). Circle
(C). Parabola
(D). Ellipse

Answer 1

Hint- First, find out the equation of $\dfrac{{dy}}{{dx}}$, and then simplify it to find out which curve it represents.

Complete step-by-step answer:

We have been given the differential equation, $({x^2} - {y^2})dx + 2xydy = 0 - (1)$
Solving the equation to find $\dfrac{{dy}}{{dx}}$:
$
  ({x^2} - {y^2})dx + 2xydy = 0 \\
   \Rightarrow 2xydy = - ({x^2} - {y^2})dx \\
   \Rightarrow 2xydy = ({y^2} - {x^2})dx \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{({y^2} - {x^2})}}{{2xy}} - (2) \\
$
Now, to find the solution of the given differential equation,
let $y = vx$
Differentiating w.r.t. x, we get-
$
  \dfrac{{dy}}{{dx}} = x\dfrac{{dv}}{{dx}} + v\dfrac{{dx}}{{dx}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = x\dfrac{{dv}}{{dx}} + v \\
 $
Putting value of $\dfrac{{dy}}{{dx}}$ and $y = vx$ in equation (2), we get-
\[
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{({y^2} - {x^2})}}{{2xy}} \\
   \Rightarrow x\dfrac{{dv}}{{dx}} + v = \dfrac{{{{(xv)}^2} - {x^2}}}{{2x(xv)}} \\
   \Rightarrow x\dfrac{{dv}}{{dx}} = \dfrac{{{x^2}{v^2} - {x^2}}}{{2{x^2}v}} - v \\
   \Rightarrow x\dfrac{{dv}}{{dx}} = \dfrac{{{x^2}{v^2} - {x^2} - 2{x^2}{v^2}}}{{2{x^2}v}} \\
   \Rightarrow x\dfrac{{dv}}{{dx}} = \dfrac{{ - {x^2}{v^2} - {x^2}}}{{2{x^2}v}} \\
   \Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{1}{x}\left( {\dfrac{{ - {x^2}({v^2} + 1)}}{{2{x^2}v}}} \right) \\
   \Rightarrow \dfrac{{dv}}{{dx}} = - \dfrac{1}{x}\left( {\dfrac{{{v^2} + 1}}{{2v}}} \right) \\
   \Rightarrow \dfrac{{2vdv}}{{{v^2} + 1}} = - \dfrac{{dx}}{x} \\
 \]
Integrating both sides, we get
\[
  \int {\dfrac{{2vdv}}{{{v^2} + 1}}} = - \int {\dfrac{{dx}}{x}} \\
  \int {\dfrac{{2vdv}}{{{v^2} + 1}}} = - \log |x| + C - (2) \\
 \]
Now, let us put $t = {v^2} + 1$
Differentiating w.r.t v, we get-
$
  \dfrac{{d({v^2} + 1)}}{{dv}} = \dfrac{{dt}}{{dv}} \\
  2v = \dfrac{{dt}}{{dv}} \\
   \Rightarrow dv = \dfrac{{dt}}{{2v}} \\
$
Now, from equation (2),
$
  \int {\left( {\dfrac{{2vdt}}{{t.2v}}} \right)} = - \log |x| + c \\
  \int {\dfrac{{dt}}{t} = - \log |x| + c} \\
   \Rightarrow \log |t| = - \log |x| + c \\
  \because t = {v^2} + 1 \\
  \therefore \log |{v^2} + 1| = - \log |x| + c \\
   \Rightarrow \log |{v^2} + 1| + \log |x| = c \\
   \Rightarrow \log |x({v^2} + 1)| = c \\
 $
Now, put $v = \dfrac{y}{x}$, we get-
$
  \log |\dfrac{{{y^2} + {x^2}}}{x}| = \log c \\
   \Rightarrow {y^2} + {x^2} = cx \\
   \Rightarrow {x^2} + {y^2} = cx \\
   \Rightarrow {x^2} + {y^2} - cx = 0 \\
 $
Hence, the solution to the differential equation represents a circle with centre at c.
So, the correct option is (B).

Note – Whenever such types of questions appear, always try to convert the given equation in the form of dy/dx, and then solve it by assuming y = vx. So that, while integrating it won’t be so difficult. Once, done with the integration, check the equation is of which curve.