Question

The differential form of the equation $y^2+(x-b{)}^2=c$:

• (A) $y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2-1=0$
• (B) $y\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+1=0$
• (C) $y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2+1=0$
• (D) $\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2-1=0$

Answer 1

The Correct Option is C

Explanation:
Given,$y^2+(x-b{)}^2=c$There are two constants $b$ and $c$ so differentiate two timesDifferentiating w.r.t $x$We get,$2y\frac{dy}{dx}+2(x-b)=0$$\Rightarrow y\frac{dy}{dx}=b-x$Differentiating again w.r.t $x$We get,${\left(\frac{dy}{dx}\right)}^2+y\frac{d^{2}y}{d{x}^2}=-1$$\Rightarrow y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2+1=0$Hence, the correct option is (C).