Question

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

A
(dydx)34xydydx+8y2=0
B
(dydx)3+4xydydx+8y2=0
C
(dydx)32xydydx+8y2=0
D
(dydx)3+2xydydx+8y2=0

Solution

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

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