Practicing Success
The differential equation whose solution is $(x-h)^2+(y-k)^2=a^2$ (a is given constant) is: |
$\begin{Bmatrix}1+(\frac{dy}{dx})^2\end{Bmatrix}^3=a^2\frac{d^2y}{dx^2}$ $\begin{Bmatrix}1+(\frac{dy}{dx})^2\end{Bmatrix}^3=a^2(\frac{d^2y}{dx^2})^2$ $\begin{Bmatrix}1+(\frac{dy}{dx})\end{Bmatrix}^3=a^2(\frac{d^2y}{dx^2})^2$ None of these |
$\begin{Bmatrix}1+(\frac{dy}{dx})^2\end{Bmatrix}^3=a^2(\frac{d^2y}{dx^2})^2$ |
$2(x-h)+2(y-k)\frac{dy}{dx}=0;\frac{dy}{dx}=\frac{-(x-h)}{y-k}$ . . . . (i) Differentiate again w.r.t. x; $1+(\frac{dy}{dx})^2+(y-x)\frac{d^2y}{dx^2}=0;(y-x)=\frac{-[1+(\frac{dy}{dx})^2]}{\frac{d^2y}{dx^2}}$ . . . . (ii) Substitute the value of (i), (ii) in original equation $\begin{Bmatrix}1+(\frac{dy}{dx})^2\end{Bmatrix}^3=a^2(\frac{d^2y}{dx^2})^2$ |