Question

The differential equation for the family of curves $y = c(x-c)^2$, where c is an arbitrary constant is:

• A. $y = \left(x+2y\frac{dx}{dy}\right)\left(2y\frac{dx}{dy}\right)^2$
• B. $y = \left(x-2y\frac{dx}{dy}\right)\left(2y\frac{dx}{dy}\right)$
• C. $y = \left(x-2y\frac{dx}{dy}\right)\left(2y\frac{dy}{dx}\right)^2$
• D. $y = \left(x-2y\frac{dx}{dy}\right)\left(2y\frac{dx}{dy}\right)^2$

Answer 1

Answer: (D)

(D)

Answer 2

To find the differential equation for the family of curves $y = c(x-c)^2$, we need to eliminate the arbitrary constant 'c'.

Given equation: $y = c(x-c)^2 \quad (1)$

Differentiate equation (1) with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx} [c(x-c)^2]$ Since 'c' is a constant, we use the chain rule for $(x-c)^2$: $\frac{dy}{dx} = c \cdot 2(x-c) \cdot \frac{d}{dx}(x-c)$ $\frac{dy}{dx} = c \cdot 2(x-c) \cdot 1$ $\frac{dy}{dx} = 2c(x-c) \quad (2)$

Now we have two equations:

1. $y = c(x-c)^2$
2. $\frac{dy}{dx} = 2c(x-c)$

We can eliminate 'c' and $(x-c)$ terms. Divide equation (1) by equation (2): $\frac{y}{\frac{dy}{dx}} = \frac{c(x-c)^2}{2c(x-c)}$ $\frac{y}{\frac{dy}{dx}} = \frac{x-c}{2}$

From this, we can express $(x-c)$ in terms of $y$ and $\frac{dy}{dx}$: $x-c = \frac{2y}{\frac{dy}{dx}}$ Since $\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}$, we can write: $x-c = 2y \frac{dx}{dy} \quad (3)$

Now, we need to find 'c'. From equation (3): $c = x - 2y \frac{dx}{dy} \quad (4)$

Substitute the expressions for 'c' from (4) and $(x-c)$ from (3) back into the original equation (1): $y = c(x-c)^2$ $y = \left(x - 2y \frac{dx}{dy}\right) \left(2y \frac{dx}{dy}\right)^2$

This is the differential equation for the given family of curves.