If y² = ax² + bx + c, then $y^{3} \frac{d^{2} y}{d x^{2}}$ is

(a) a constant
(b) a function of x only
(c) a function of y only
(d) a function of x and y

Open in App

Solution

(a) a constant

Here,

$y^{2} = a x^{2} + b x + c Now, 2 y \frac{d y}{d x} = 2 a x + b \Rightarrow 2 y \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{2} = 2 a \Rightarrow y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = a \Rightarrow y \frac{d^{2} y}{d x^{2}} + {(\frac{2 a x + b}{2 y})}^{2} = a [∵ 2 y \frac{d y}{d x} = 2 a x + b] \Rightarrow 4 y^{3} \frac{d^{2} y}{d x^{2}} + {(2 a x + b)}^{2} = 4 a y^{2} \Rightarrow y^{3} \frac{d^{2} y}{d x^{2}} = \frac{4 a y^{2} - {(2 a x + b)}^{2}}{4} \Rightarrow y^{3} \frac{d^{2} y}{d x^{2}} = \frac{4 a (a x^{2} + b x + c) - {(2 a x + b)}^{2}}{4} [∵ y^{2} = a x^{2} + b x + c] \Rightarrow y^{3} \frac{d^{2} y}{d x^{2}} = \frac{4 a^{2} x^{2} + 4 a b x + 4 a c - 4 a^{2} x^{2} - b^{2} - 4 a x b}{4} \Rightarrow y^{3} \frac{d^{2} y}{d x^{2}} = \frac{4 a c - b^{2}}{4} = a constant$

Suggest Corrections

Similar questions

If $y^{2}$ = $a x^{2} + b x + c$ , then $y^{3} . \frac{d^{2} y}{d x^{2}}$ is

y^{2} = a x^{2} + b x + c

y^{3} . \frac{d^{2} y}{d x^{2}}

(a x^{2} + b x + c) y + a^{'} x^{2} + b^{'} x + c^{'} = 0

x

y

(a x^{2} + b x + c) y + a^{'} x^{2} + b^{'} x + c^{'} = 0

y = 4 a

y = a x^{2} - b x + c, (a = 0)

y = x

y = 3 x^{2} - 1

Join BYJU'S Learning Program