Question

If $y^2=a{x}^2+bx+c$, 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 $y$

Answer 1

The correct option is A

a constant

Explanation for the correct option.

Step 1. Find the value of $\frac{dy}{dx}$.

Differentiate the equation $y^2=a{x}^2+bx+c$ with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left(y^2\right)=\frac{d}{dx}(a{x}^2+bx+c) \\ \Rightarrow 2y\frac{dy}{dx}=2ax+b+0 \\ \Rightarrow \frac{dy}{dx}=\frac{2ax+b}{2y}...\left(1\right) \end{gathered}$$

Step 2. Find the value of $y^3\frac{d^{2}y}{d{x}^2}$.

Differentiate the equation $2y\frac{dy}{dx}=2ax+b$ with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left(2y\frac{dy}{dx}\right)=\frac{d}{dx}\left(2ax+b\right) \\ \Rightarrow 2{\left(\frac{dy}{dx}\right)}^2+2y\frac{d^{2}y}{d{x}^2}=2a+0 \\ \Rightarrow 2\left[{\left(\frac{dy}{dx}\right)}^2+y\frac{d^{2}y}{d{x}^2}\right]=2a \\ \Rightarrow {\left(\frac{dy}{dx}\right)}^2+y\frac{d^{2}y}{d{x}^2}=a \\ \Rightarrow y\frac{d^{2}y}{d{x}^2}=a-{\left(\frac{dy}{dx}\right)}^2 \end{gathered}$$

Now using equation $1$ substitute $\frac{2ax+b}{2y}$ for $\frac{dy}{dx}$.

$$\begin{gathered} y\frac{d^{2}y}{d{x}^2}=a-{\left(\frac{2ax+b}{2y}\right)}^2 \\ \Rightarrow y\frac{d^{2}y}{d{x}^2}=a-\left(\frac{4a^2{x}^2+4abx+b^2}{4y^2}\right) \end{gathered}$$

Now multiply both sides by $y^2$.

$$\begin{gathered} y\frac{d^{2}y}{d{x}^2}\times y^2=\left[a-\left(\frac{4a^2{x}^2+4abx+b^2}{4y^2}\right)\right]\times y^2 \\ \Rightarrow y^3\frac{d^{2}y}{d{x}^2}=a{y}^2-\frac{4a^2{x}^2+4abx+b^2}{4} \\ \Rightarrow y^3\frac{d^{2}y}{d{x}^2}=a{y}^2-a^2{x}^2-abx-\frac{b^2}{4} \end{gathered}$$

Now using the given equation substitute $a{x}^2+bx+c$ for $y^2$.

$\begin{array}{rcl}{y}^3\frac{d^{2}y}{d{x}^2}& =& a\left(a{x}^2+bx+c\right)-a^2{x}^2-abx-\frac{b^2}{4}\\ & =& a^2{x}^2+abx+ac-a^2{x}^2-abx-\frac{b^2}{4}\\ & =& ac-\frac{b^2}{4}\end{array}$

So the value of the expression $\begin{array}{rcl}& & y^3\frac{d^{2}y}{d{x}^2}\end{array}$ is $ac-\frac{b^2}{4}$.

As $a,b,c$ are constants so the expression $ac-\frac{b^2}{4}$ is also a constant and so $\begin{array}{l}{y}^3\frac{d^{2}y}{d{x}^2}\end{array}$is a constant.

Hence, the correct option is A.