The differential equation whose solution is ( | vedclass

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

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$\left[1+\left(\frac{dy}{dx}\right)^{2}\right]^{3}=a^{2}\left(\frac{d^{2}y}{dx^{2}}\right)^{2}$

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

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

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