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(B) Given the function $y = a(x-a)^{2} \quad ...(1)$ Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2a(x-a) \quad ...(2)$ From $(2)$,$a =

Vedclass · Accepted Answer

$8 y^{3}=\left(\frac{d y}{d x}\right)^{2}\left[2x-\frac{d y}{d x}\right]$

Vedclass · Answer

(B) Given the function $y = a(x-a)^{2} \quad ...(1)$ Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2a(x-a) \quad ...(2)$ From $(2)$,$a = \frac{1}{2} \frac{dy/dx}{x-a}$. Substituting $a$ into $(1)$: $y = \frac{1}{2} \frac{dy/dx}{x-a} (x-a)^2 = \frac{1}{2} (x-a) \frac{dy}{dx}$ Thus,$(x-a) = \frac{2y}{dy/dx}$. Substituting $(x-a)$ back into $(2)$: $\frac{dy}{dx} = 2a \left( \frac{2y}{dy/dx} \right) \implies a = \frac{(dy/dx)^2}{4y}$. Now,substitute $a$ and $(x-a)$ into $y = a(x-a)^2$: $y = \left( \frac{(dy/dx)^2}{4y} \right) \left( \frac{2y}{dy/dx} \right)^2$ $y = \frac{(dy/dx)^2}{4y} \cdot \frac{4y^2}{(dy/dx)^2} = y$. To find the differential equation,we eliminate $a$ from $y = a(x-a)^2$ and $y' = 2a(x-a)$. From $y' = 2a(x-a)$,we have $a = x - \frac{y'}{2a}$ is not correct. Correct approach: $x-a = \frac{y'}{2a}$. Substitute into $y = a(x-a)^2$: $y = a \left( \frac{y'}{2a} \right)^2 = \frac{(y')^2}{4a}$. Since $a = x - (x-a) = x - \frac{y'}{2a}$,we have $2a^2 - 2ax + y' = 0$. Solving for $a$ and substituting leads to the relation $8y^3 = (y')^2 (2x - y')$. Option $B$ is the correct form.

The differential equation obtained from the function $y=a(x-a)^{2}$ is

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