The order and degree of the differential equation $\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^{1/4} + x^{1/5} = 0$ respectively, are

2 and 4

The correct answer is Option (1) → 2 and 4 ##

Given that, $\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^{1/4} = -x^{1/5}$

$\Rightarrow \quad \left( \frac{dy}{dx} \right)^{1/4} = -\left( x^{1/5} + \frac{d^2y}{dx^2} \right)$

On squaring both sides, we get

$\left( \frac{dy}{dx} \right)^{1/2} = \left( x^{1/5} + \frac{d^2y}{dx^2} \right)^2$

Again, on squaring both sides, we have

$\frac{dy}{dx} = \left( x^{1/5} + \frac{d^2y}{dx^2} \right)^4$

$\text{order} = 2, \text{degree} = 4$