The order and degree of a differential equati | vedclass

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Question

(A) Given the family of curves: $y^2 = 2C(x + \sqrt{C}) \quad \dots (i)$Differentiating with respect to $x$:$2y \frac{dy}{dx} = 2C$$\Rightarrow C = y

Vedclass · Accepted Answer

$1, 3$

Vedclass · Answer

(A) Given the family of curves: $y^2 = 2C(x + \sqrt{C}) \quad \dots (i)$Differentiating with respect to $x$:$2y \frac{dy}{dx} = 2C$$\Rightarrow C = y \frac{dy}{dx} \quad \dots (ii)$Substitute the value of $C$ from $(ii)$ into $(i)$:$y^2 = 2 \left( y \frac{dy}{dx} \right) \left( x + \sqrt{y \frac{dy}{dx}} \right)$$y = 2 \frac{dy}{dx} \left( x + \sqrt{y \frac{dy}{dx}} \right)$$y - 2x \frac{dy}{dx} = 2 \frac{dy}{dx} \sqrt{y \frac{dy}{dx}}$Squaring both sides:$(y - 2x \frac{dy}{dx})^2 = 4 \left( \frac{dy}{dx} \right)^2 \left( y \frac{dy}{dx} \right)$$(y - 2x \frac{dy}{dx})^2 = 4y \left( \frac{dy}{dx} \right)^3$The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$.The power of the highest order derivative is $3$,so the degree is $3$.Thus,the order and degree are $1$ and $3$ respectively.

The order and degree of a differential equation obtained by eliminating the arbitrary constant $C$ from the family of curves $y^2 = 2C(x + \sqrt{C})$ are respectively:

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