Question

The solution of $\frac{dy}{dx} = \frac{x^{2} + y^{2} + 1}{2xy}$, satisfying $ y\left(1\right) = 0$ is given by

• A. hyperbola
• B. circle
• C. ellipse
• D. parabola

Answer 1

The Correct Option is A

Given differential equation is
$\frac{dy}{dx} = \frac{x^{2} + y^{2} + 1}{2xy} $
$\Rightarrow 2xy dy = \left(x^{2} + 1\right)dx +y^{2} dx$
$\Rightarrow \frac{xd\left(y^{2}\right) -y^{2} dx}{x^{2}}$
$= \left(\frac{x^{2} + 1 }{x^{2}} \right) dx$
$\Rightarrow \int d\left(\frac{y^{2}}{x}\right) = \int\left(1+ \frac{1}{x^{2}}\right)dx$
$\Rightarrow \frac{y^{2}}{x} = x - \frac{1}{x}C $
$\Rightarrow y^{2} = \left(x^{2} - 1 + Cx\right) $
When $x = 1, y = 0$
Then, $0 = 1 - 1 + C$
$\Rightarrow C= 0$
$\therefore$ The solution is $x^{2}-y^{2}=1$ i.e., hyperbola.

Answer 2

The given differential equation is:
\(\frac{dy}{dx} = \frac{(x^2 + y^2 + 1)}{(2xy)}\)

This is a non-linear differential equation and cannot be solved by simple integration techniques. However, we can make use of the fact that the given differential equation is a homogeneous equation.

To solve this differential equation, we can make the substitution \(y=vx\). Then, we have:
\(\frac{dy}{dx} = v + x\frac{dv}{dx}\)

Substituting this in the given differential equation, we get:
\(v + x\frac{dv}{dx} = \frac{(x^2 + v^2x^2 + 1)}{(2x^2v)}\)

Multiplying both sides by \(\frac{2v}{x}\) and simplifying, we get:
\(\frac{2v}{(v^2 - 1)} dv= \frac{(x^2 + 1)}{x^2} dx\)

Integrating both sides, we get:
\(ln|v - 1| - ln|v + 1| = ln|x| + C\)

where C is the constant of integration. Rearranging and using the initial condition \(y(1) = 0\) (which implies \(v(1) = 0\), we get:
\(v = \frac{y}{x} = \frac{(1 - x^2)}{(1 + x^2)}\)

Substituting this value of v in the equation \(y = vx\), we get:
\(y = x \times \frac{(1 - x^2)}{(1 + x^2)}\)

This is the solution of the given differential equation, satisfying \(y(1) = 0\).

Now, we can rewrite the equation in terms of y and x as:
\(y(1 + x^2) = x(1 - x^2)\)

This can be simplified to:
\(x^2 + y^2 = x\)

This is the equation of a circle with center at \((\frac{1}{2}, 0)\) and radius \(\frac{1}{2}\).
Therefore, the solution of the given differential equation, satisfying \(y(1) = 0\), is a circle.