The differential equation having the general | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

The differential equation representing the family of circles having their centres on the $Y$-axis is (where $y_1 = \frac{dy}{dx}$ and $y_2 = \frac{d^2y}{dx^2}$):

If $a$ and $b$ are arbitrary constants,then the differential equation corresponding to the family of curves given by $y=x[a \cos (\log x)+b \sin (\log x)]$ is

The differential equation of all parabolas having vertex at the origin and axis along the positive $Y$-axis is

The differential equation of all parabolas whose axes are parallel to the $Y$-axis is:

$y = ke^{\sin ^{-1} x} + 3$ is a solution of which differential equation?

Vedclass Test Series

Exam Paper Generator

Online Exam Module