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
Answer & explanation
Correct answer: option 1
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
$\frac{dy}{dx} = \left( x^{1/5} + \frac{d^2y}{dx^2} \right)^4$
$\text{order} = 2, \text{degree} = 4$