Question

If$y=a+b{x}^2$:$a,b$are arbitrary constants, then

• A. $\frac{d^{2}y}{d{x}^2}=2xy$
• B. $\frac{x{d}^{2}y}{d{x}^2}=\frac{dy}{dx}$
• C. $x\frac{d^{2}y}{d{x}^2}–\frac{dy}{dx}+y=0$
• D. $\frac{x{d}^{2}y}{d{x}^2}=2xy$

Answer 1

The correct option is B

$\frac{x{d}^{2}y}{d{x}^2}=\frac{dy}{dx}$

Explanation for the correct option:

Finding the relation:

Given that,

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

Differentiate the given equation with respect to $x$.

$$\begin{gathered} \frac{dy}{dx}=0+2bx\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}\mathbf{\left(}{\mathbf{x}}^{\mathbf{n}}\mathbf{\right)}}{\mathbf{dx}}\mathbf{=}{\mathbf{nx}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{]} \\ =2bx...\left(1\right) \end{gathered}$$

Differentiate again the above equation with respect to $x$.

$\frac{d^{2}y}{d{x}^2}=2b...\left(2\right)$

Multiply by $x$ in equation $\left(2\right)$.

$$\begin{gathered} x\frac{d^{2}y}{d{x}^2}=2bx \\ \Rightarrow x\frac{d^{2}y}{d{x}^2}=\frac{dy}{dx}\left[\mathrm{from}\left(1\right)\right] \end{gathered}$$

Hence, the correct option is B.