The differential equation whose solution is ( | vedclass

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Question

(B) Given the equation of a circle: $(x - h)^2 + (y - k)^2 = a^2$ $(i)$Differentiating with respect to $x$: $2(x - h) + 2(y - k) \frac{dy}{dx} = 0$$\R

Vedclass · Accepted Answer

$[1 + (dy/dx)^2]^3 = a^2 (d^2y/dx^2)^2$

Vedclass · Answer

(B) Given the equation of a circle: $(x - h)^2 + (y - k)^2 = a^2$ $(i)$Differentiating with respect to $x$: $2(x - h) + 2(y - k) \frac{dy}{dx} = 0$$\Rightarrow (x - h) + (y - k) \frac{dy}{dx} = 0 \Rightarrow (x - h) = -(y - k) \frac{dy}{dx}$ (ii)Differentiating again with respect to $x$: $1 + \left(\frac{dy}{dx}\right)^2 + (y - k) \frac{d^2y}{dx^2} = 0$$\Rightarrow (y - k) = -\frac{1 + (dy/dx)^2}{d^2y/dx^2}$ (iii)Substituting (iii) into (ii): $(x - h) = \frac{[1 + (dy/dx)^2] \frac{dy}{dx}}{d^2y/dx^2}$ (iv)Substituting (iii) and (iv) into $(i)$:$\left( \frac{[1 + (dy/dx)^2] \frac{dy}{dx}}{d^2y/dx^2} \right)^2 + \left( -\frac{1 + (dy/dx)^2}{d^2y/dx^2} \right)^2 = a^2$$\Rightarrow \frac{[1 + (dy/dx)^2]^2}{(d^2y/dx^2)^2} \left[ (dy/dx)^2 + 1 \right] = a^2$$\Rightarrow [1 + (dy/dx)^2]^3 = a^2 (d^2y/dx^2)^2$

The differential equation whose solution is $(x - h)^2 + (y - k)^2 = a^2$ is ($a$ is a constant).

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