Question

The differential equation of the family of circles touching y – axis at the origin is:

A) $\left( {{x}^{2}}+{{y}^{2}} \right)\dfrac{dy}{dx}-2xy=0$

B) ${{x}^{2}}-{{y}^{2}}+2xy\dfrac{dy}{dx}=0$

C) $\left( {{x}^{2}}-{{y}^{2}} \right)\dfrac{dy}{dx}-2xy=0$

D) $\left( {{x}^{2}}+{{y}^{2}} \right)\dfrac{dy}{dx}+2xy=0$

Answer 1

Draw a rough figure, find the center and radius of the circle and substitute it in the general equation of the circle. Differentiate the expression and find the radius. Substitute radius in the equation of circle.

Complete step-by-step answer:

We know the general equation of circle is ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}-(1)$.

Where (a, b) is the center of the circle and radius of the circle is r.

If the circle touches the y – axis at origin, then the center will be at x –axis. Thus the center of the circle becomes (a, 0) and radius of the circle is a.

Thus with center (a, 0) and radius a the equation of circle becomes i.e. equation (1) becomes,

${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{a}^{2}}$

Let us open the bracket and simplify the above expression.

${{\left( x-a \right)}^{2}}+{{y}^{2}}={{a}^{2}}$

${{x}^{2}}-2ax+{{a}^{2}}+{{y}^{2}}={{a}^{2}}$

$\Rightarrow {{x}^{2}}-2ax+{{y}^{2}}=0$

$\therefore 2ax={{x}^{2}}+{{y}^{2}}-(2)$

Now let us differentiate both sides with respect to ‘x’.

$\dfrac{d}{dx}\left( 2ax \right)=\dfrac{d}{dx}\left( {{x}^{2}}+{{y}^{2}} \right)$

$2a=2a+2y.\dfrac{dy}{dx}$

Divide the entire expression by 2.

$\Rightarrow a=x+y.\dfrac{dy}{dx}-(3)$

Now let us substitute the value of ‘a’ in equation (2).

$2xa={{x}^{2}}+{{y}^{2}}$

$2x\left( x+y.\dfrac{dy}{dx} \right)={{x}^{2}}+{{y}^{2}}$

$2{{x}^{2}}+2xy.\dfrac{dy}{dx}={{x}^{2}}+{{y}^{2}}$

Let us rearrange the expression, we get

$2{{x}^{2}}-{{x}^{2}}-{{y}^{2}}+2xy.\dfrac{dy}{dx}=0$

${{x}^{2}}-{{y}^{2}}+2xy.\dfrac{dy}{dx}=0$

Thus we got the required differential equation.

$\therefore$ Option (B) is the correct answer.

Note: Here we found the first order differential equation. All the linear equations are in the form of derivative of first order. It has only the first derivative such as $\dfrac{dy}{dx}$, where x and y are the two variables and represented as : -
$\dfrac{dy}{dx}=f\left( x,y \right)=y'$.