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