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 :
Answer & explanation
Correct answer: option 4
$\frac{\left[1+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2}}{\frac{d^2 y}{d x^2}}=k$
$\left[1+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2}=k\left(\frac{d z y}{d x^2}\right)$
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