The equation given is \( \left(\dfrac{d^2y}{dx^2}\right)^3 + \left(\dfrac{dy}{dx}\right)^2 + y = 0 \). Order is the highest derivative appearing anywhere in the equation, and degree is the power of that highest-order derivative once the equation is written as a polynomial in derivatives, with no fractional or negative powers, and no derivative inside another function like a sine or square root.
- 3, 2: This would be correct only if the highest derivative present were a third-order derivative, but the equation contains only \(\dfrac{d^2y}{dx^2}\) and \(\dfrac{dy}{dx}\), the highest being second-order, not third, so the order given here is wrong.
- 2, 3: The highest derivative present is \(\dfrac{d^2y}{dx^2}\), a second-order derivative, giving order 2. That same term is raised to the power 3 in the equation, giving degree 3. Both values line up with what's actually written.
- 2, 2: The order value of 2 here is correct, since \(\dfrac{d^2y}{dx^2}\) is indeed the highest derivative, but the degree is misread as 2 instead of the actual power of 3 that the highest-order term is raised to.
- 3, 1: Neither value matches, the order is not 3 since no third derivative appears, and even if it were, a degree of 1 doesn't correspond to the cube on the second-derivative term.
Reading the powers directly off the equation, the highest derivative is second order, and it appears cubed.
Therefore, the correct answer is 2, 3.