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Question
If y2 = ax2 + bx + c, then \[y^3 \frac{d^2 y}{d x^2}\] is
Options
a constant
a function of x only
a function of y only
a function of x and y
Solution
(a) a constant
Here,
\[y^2 = a x^2 + bx + c\]
\[\text { Now,} \]
\[2y\frac{d y}{d x} = 2ax + b\]
\[ \Rightarrow 2y\frac{d^2 y}{d x^2} + 2 \left( \frac{d y}{d x} \right)^2 = 2a \]
\[ \Rightarrow y\frac{d^2 y}{d x^2} + \left( \frac{d y}{d x} \right)^2 = a \]
\[ \Rightarrow y\frac{d^2 y}{d x^2} + \left( \frac{2ax + b}{2y} \right)^2 = a \left[ \because 2y\frac{d y}{d x} = 2ax + b \right]\]
\[ \Rightarrow 4 y^3 \frac{d^2 y}{d x^2} + \left( 2ax + b \right)^2 = 4a y^2 \]
\[ \Rightarrow y^3 \frac{d^2 y}{d x^2} = \frac{4a y^2 - \left( 2ax + b \right)^2}{4}\]
\[ \Rightarrow y^3 \frac{d^2 y}{d x^2} = \frac{4a\left( a x^2 + bx + c \right) - \left( 2ax + b \right)^2}{4} \left[ \because y^2 = a x^2 + bx + c \right]\]
\[ \Rightarrow y^3 \frac{d^2 y}{d x^2} = \frac{4 a^2 x^2 + 4abx + 4ac - 4 a^2 x^2 - b^2 - 4axb}{4}\]
\[ \Rightarrow y^3 \frac{d^2 y}{d x^2} = \frac{4ac - b^2}{4} = \text { a constant }\]
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