The differential equation corresponding to th | vedclass

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(A) The equation of a parabola with axis parallel to the $y$-axis and axis of symmetry $x=1$ is given by $(x-1)^2 = 4a(y-k)$,where $a$ and $k$ are arb

Vedclass · Accepted Answer

$(x-1) \frac{d^2 y}{d x^2} - \frac{d y}{d x} = 0$

Vedclass · Answer

(A) The equation of a parabola with axis parallel to the $y$-axis and axis of symmetry $x=1$ is given by $(x-1)^2 = 4a(y-k)$,where $a$ and $k$ are arbitrary constants.Alternatively,we can write this as $y = A(x-1)^2 + B$,where $A$ and $B$ are arbitrary constants.Differentiating with respect to $x$ once: $\frac{dy}{dx} = 2A(x-1)$.Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = 2A$.From the first derivative,$A = \frac{1}{2(x-1)} \frac{dy}{dx}$.Substituting this into the second derivative: $\frac{d^2y}{dx^2} = 2 \left( \frac{1}{2(x-1)} \frac{dy}{dx} \right) = \frac{1}{x-1} \frac{dy}{dx}$.Rearranging gives: $(x-1) \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$.

The differential equation corresponding to the family of parabolas whose axis is along $x=1$ is

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