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(B) Given the general solution $y = c(x - c)^2$. Step $1$: Differentiate with respect to $x$: $y' = 2c(x - c)$. Step $2$: From the original equation,$

Vedclass · Accepted Answer

$(y')^3 = 4y(xy' - 2y)$

Vedclass · Answer

(B) Given the general solution $y = c(x - c)^2$. Step $1$: Differentiate with respect to $x$: $y' = 2c(x - c)$. Step $2$: From the original equation,$c = \frac{y}{(x - c)^2}$. Alternatively,from $y' = 2c(x - c)$,we have $c = \frac{y'}{2(x - c)}$. Equating the two expressions for $c$: $\frac{y}{(x - c)^2} = \frac{y'}{2(x - c)} \implies 2y = y'(x - c) \implies x - c = \frac{2y}{y'}$. Step $3$: Substitute $x - c$ back into the expression for $y'$: $y' = 2c \left(\frac{2y}{y'}\right) \implies c = \frac{(y')^2}{4y}$. Step $4$: Substitute $c$ and $x - c$ into the original equation $y = c(x - c)^2$: $y = \left(\frac{(y')^2}{4y}\right) \left(\frac{2y}{y'}\right)^2$. $y = \frac{(y')^2}{4y} \cdot \frac{4y^2}{(y')^2} = y$. This confirms the relation. To find the differential equation,we use $c = \frac{y'}{2(x - c)}$ and $x - c = \frac{2y}{y'}$. Thus $c = x - \frac{2y}{y'}$. Substitute $c$ into $y' = 2c(x - c)$: $y' = 2(x - \frac{2y}{y'})(\frac{2y}{y'}) = 2(\frac{xy' - 2y}{y'})(\frac{2y}{y'}) = \frac{4y(xy' - 2y)}{(y')^2}$. $(y')^3 = 4y(xy' - 2y)$. Thus,option $B$ is correct.

The differential equation having the general solution $y=c(x-c)^2$ ($c$ is an arbitrary constant) is

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