Question

The curve amongst the family of curves, represented by the differential equation, $\left( {{x}^{2}}-{{y}^{2}} \right)dx+2xydy=0$ which passes through the point (1,1) is
A. A circle with center on the y-axis
B. A circle with center on the x-axis
C. An ellipse with major axis along the y-axis
D. A hyperbola with transverse axis along the x-axis

Answer 1

Hint: In this question, the equation of the curve is given in differential form i.e. involving dx and dy. Therefore, we should first try to solve the equation to obtain the equation of the curve involving only x and y. Then, we can match it with the curves given in the options and choose the correct answer.

Complete step-by-step answer:
The given equation of the curve is
$\begin{align}
  & \left( {{x}^{2}}-{{y}^{2}} \right)dx+2xydy=0 \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{{{y}^{2}}-{{x}^{2}}}{2xy}...............(1.1) \\
\end{align}$
We find that in the RHS of equation (1.1), the degree of the terms i.e. the total power of x and y of each term in the numerator and the denominator is 2. Therefore, this is the case of a homogeneous differential equation.
Therefore, we can take $y=vx$ for some function v. Therefore, taking derivative on both sides of $y=vx$, we obtain
$\dfrac{dy}{dx}=\dfrac{d(vx)}{dx}=v\dfrac{dx}{dx}+x\dfrac{dv}{dx}=v+x\dfrac{dv}{dx}..........(1.2)$
And putting $y=vx$ in the RHS of equation (1.1), we obtain
$\dfrac{{{y}^{2}}-{{x}^{2}}}{2xy}=\dfrac{{{\left( vx \right)}^{2}}-{{x}^{2}}}{2x\times vx}=\dfrac{{{x}^{2}}\left( {{v}^{2}}-1 \right)}{{{x}^{2}}\times 2v}=\dfrac{\left( {{v}^{2}}-1 \right)}{2v}..............(1.3)$
Therefore, using equations (1.2) and (1.3), we can rewrite equation (1.1) as
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{{{y}^{2}}-{{x}^{2}}}{2xy} \\
 & \Rightarrow v+x\dfrac{dv}{dx}=\dfrac{{{v}^{2}}-1}{2v} \\
\end{align}$
Now, we can separate the terms involving v and x in the last line of the above equation to write it as
$\begin{align}
  & x\dfrac{dv}{dx}=\dfrac{{{v}^{2}}-1}{2v}-v=\dfrac{{{v}^{2}}-1-2{{v}^{2}}}{2v}=\dfrac{-\left( {{v}^{2}}+1 \right)}{2v} \\
 & \Rightarrow \dfrac{2vdv}{\left( {{v}^{2}}+1 \right)}=\dfrac{-dx}{x}.................(1.4) \\
\end{align}$
Now, we know that the integral of $\dfrac{1}{x}$ is given by
$\int{\dfrac{1}{x}dx}=\log x+C............(1.5)$
Where C is an arbitrary constant.
Again, if we take
 $\begin{align}
  & {{v}^{2}}+1=s \\
 & \Rightarrow 2vdv=ds..............(1.6) \\
\end{align}$
in LHS of equation (1.4), we can write it as
$\dfrac{2vdv}{\left( {{v}^{2}}+1 \right)}=\dfrac{ds}{s}.................(1.7)$
Therefore, we can rewrite equation (1.4) using (1.7) as
$\dfrac{ds}{s}=-\dfrac{dx}{x}$
Taking integral on both sides and using equation (1.5), we obtain
$\begin{align}
  & \int{\dfrac{ds}{s}}=-\int{\dfrac{dx}{x}} \\
 & \Rightarrow \log s+{{C}_{1}}=-\log x+{{C}_{2}} \\
 & \Rightarrow \log s=-\log x+{{C}_{3}} \\
\end{align}$
Where ${{C}_{1}}$ and ${{C}_{2}}$ are arbitrary constants and ${{C}_{3}}={{C}_{2}}-{{C}_{1}}$ which is also a constant. Taking exponentials on both sides and using the fact that $-\log x=\log \left( \dfrac{1}{x} \right)$ and ${{e}^{\log x}}=x$ for any value of x, we have
$\begin{align}
  & \log s=-\log x+{{C}_{3}}=\log \left( \dfrac{1}{x} \right)+{{C}_{3}} \\
 & \Rightarrow {{e}^{\log s}}={{e}^{\log \left( \dfrac{1}{x} \right)+{{C}_{3}}}}={{e}^{\log \left( \dfrac{1}{x} \right)}}\times {{e}^{{{C}_{3}}}} \\
 & \Rightarrow s=\dfrac{1}{x}\times {{C}_{4}}...................(1.8) \\
\end{align}$
Where ${{C}_{4}}={{e}^{{{C}_{3}}}}$ is another constant. Now, rewriting the value of s as $s={{v}^{2}}+1$ and $v=\dfrac{y}{x}$, we can rewrite equation (1.8) as
$\begin{align}
  & s=\dfrac{1}{x}\times {{C}_{4}} \\
 & \Rightarrow {{\left( \dfrac{y}{x} \right)}^{2}}+1=\dfrac{{{C}_{4}}}{x}...................(1.9) \\
\end{align}$
Which is the equation of the curve. As it passes through (1,1), x=1 and y=1 should satisfy equation (1.9), therefore, we should have
$\begin{align}
  & {{\left( \dfrac{1}{1} \right)}^{2}}+1=\dfrac{{{C}_{4}}}{1} \\
 & \Rightarrow {{C}_{4}}=2 \\
\end{align}$
Therefore, we can use this value of ${{C}_{4}}$ in equation (1.9) to obtain
$\begin{align}
  & {{\left( \dfrac{y}{x} \right)}^{2}}+1=\dfrac{2}{x} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-2x=0...................(1.10) \\
\end{align}$
Now, we see that the degree and the coefficient of the terms with highest degree of both x and y in (1.10) is same. Therefore, it represents a circle. Comparing (1.10) with the general equation of a circle with center at (-g,-f) is given by
${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$
Comparing it with equation (1.10), we find that g=-1 and f=0. Therefore, the center of the circle represented by (1.10) should lie at (-1,0) and thus on the x-axis as the y-coordinate is 0.
Therefore, option (b) is the correct answer.

Note: We should note that in equation (1.8), we put all the constant terms on one side and renamed it as ${{C}_{4}}$. We could do it because as ${{C}_{1}}$, ${{C}_{2}}$ and ${{C}_{3}}$ were arbitrary constants and their values were not determined and in the final equations just ${{C}_{4}}$ appeared, therefore, we could find the equation of the curve using ${{C}_{4}}$ only and not ${{C}_{1}},{{C}_{2}}$ or ${{C}_{3}}$.