Q55 / 150 · MHT CET 2024 (03 May Shift 1) · Mathematics

The differential equation $\left[\frac{1+\left(\frac{d y}{d x}\right)^2}{\left(\frac{d^2 y}{d x^2}\right)^{\frac{3}{2}}}\right]^2=\mathrm{k} x$ is of

  • Correct answer
    A
    order $=2$, degree $=3$
  • B
    order $=3$, degree $=2$
  • C
    order $=2$, degree $=2$
  • D
    order $=3$, degree $=3$

Solution

$\begin{aligned} & {\left[\frac{1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2}{\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)}\right]^{\frac{3}{2}}=\mathrm{k} x} \\ & \therefore:\left[\frac{1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2}{\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)}\right]^3=(\mathrm{kx})^2 \\ & \therefore \quad\left[1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2\right]^3=(\mathrm{k} x)^2\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)^3 \\ & \therefore \quad \text { Order }=2, \text { Degree }=3 \end{aligned}$

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

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