Question

Find the differential equation corresponding to the family of curves $y=k{(x-k)}^2$ where k is an arbitrary constant.

• A. ${\left(\frac{dy}{dx}\right)}^3-4xy\frac{dy}{dx}+8y^2=0$
• B. ${\left(\frac{dy}{dx}\right)}^3+4xy\frac{dy}{dx}+8y^2=0$
• C. ${\left(\frac{dy}{dx}\right)}^3-2xy\frac{dy}{dx}+8y^2=0$
• D. ${\left(\frac{dy}{dx}\right)}^3+2xy\frac{dy}{dx}+8y^2=0$

Answer 1

The correct option is A ${\left(\frac{dy}{dx}\right)}^3-4xy\frac{dy}{dx}+8y^2=0$

Given $y=k{(x-k)}^2\cdots \left(1\right)$
$\therefore y_1=2k(x-k)$
or
$y_1^2=4k^2{(x-k)}^2=4k^2\frac{y}{k}\therefore \frac{y_1^2}{4y}=k.$
Putting the value of k in (2), we get
$y_1=2.\frac{y_1^2}{4y}[x-\frac{y_1^2}{4y}]$ $\therefore 2y=y_{1}x-\frac{y_1^3}{4y}$
or $8y^2-4xy{y}_1-{\left(y_1\right)}^2=0$
or ${\left(\frac{dy}{dx}\right)}^3-4xy\frac{dy}{dx}+8y^2=0$ is the required equation.