The differential equation of all parabolas whose axis is y-axis, is

• A

  $\frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$
• B

  $\frac{d^2 y}{d x^2}+\frac{d y}{d x}=0$
• Correct answer
  C

  $x \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$
• D

  $\frac{d^2 y}{d x^2}-y=0$

Solution

Differential Equation of parabolas whose axis is $\mathrm{Y}$ axis, is Let the vertex be $(0, \mathrm{k})$
$\begin{aligned} & \therefore x^2=4 b y-4 b k \\ & \therefore 2 x=4 b \frac{d y}{d x}-0 \Rightarrow b=\left(\frac{x}{2}\right) \times \frac{1}{\left(\frac{d y}{d x}\right)} \\ & 2 x=4 b \frac{d y}{d x}-0 \Rightarrow b=\left(\frac{x}{2}\right) \times \frac{1}{\left(\frac{d y}{d x}\right)} \end{aligned}$ Substituting value of $b$ in eq. (1), we get $\begin{aligned} & \therefore \quad x^2=4\left(\frac{x}{2}\right) \times \frac{1}{\left(\frac{d y}{d x}\right)}(y-k) \\ & \therefore \quad \mathrm{x}^2\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=2 \mathrm{x}(\mathrm{y}-\mathrm{k}) \\ & \therefore \quad x^2\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d y}{d x}\right)(2 x)=2 x \frac{d y}{d x}+2(y-k) \\ & \therefore \quad \mathrm{y}-\mathrm{k}=\frac{\mathrm{x}^2\left(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}}\right)}{2} \\ & \end{aligned}$ Substituting value of $(\mathrm{y}-\mathrm{k})$ in eq. (1), we get $\begin{aligned} & \left.x^2=4\left(\frac{x}{2}\right)\left[\frac{1}{\left(\frac{d y}{d x}\right)}\right]^{\left.\frac{x^2\left(\frac{d^2 y}{d^2}\right)}{2}\right]}\right] \\ & \therefore \quad x^2\left(\frac{d y}{d x}\right)=x^3\left(\frac{d^2 y}{d x^2}\right) \Rightarrow x\left(\frac{d^2 y}{d x^2}\right)-\left(\frac{d y}{d x}\right)=0 \end{aligned}$

Hence, the correct answer is option: 'C'.