If a and b are the order and degree of differential equation $\frac{\left[1+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2}}{\frac{d^2 y}{d x^2}}=K$ respectively, then the value of $a+2 b$ is :

6

$\frac{\left[1+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2}}{\frac{d^2 y}{d x^2}}=k$

$\left[1+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2}=k\left(\frac{d z y}{d x^2}\right)$

squaring both sides

$\left[1+\left(\frac{d y}{d x}\right)^2\right]^3=k^2\left(\frac{d^2 y}{d x^2}\right)^2$

a = order = 2 (highest order derivative)

b = degree = 2 (power of highest order derivatives)

a + 2b

= 2 + 2 × 2

= 2 + 4

= 6