If d/dx≤ft(2(d^2y)/(dx^2))^3=7, then the sum of order and degree of t…

$\frac{d}{dx}\left(2\frac{d^2y}{dx^2}\right)^3=7$

$⇒2×3\left(\frac{d^2y}{dx^2}\right)^2\left(\frac{d^3y}{dx^3}\right)=7$

order = 3, degree = 1

$3 + 1 = 4$