The product of order and degree of the differential equation \(\left(\displaystyle\frac{d^3y}{dx^3}\right)^2 + x^2y\left(\displaystyle\frac{d^2y}{dx^3}\right)^3 = 2x^5\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6
The order is the order of the highest derivative in the equation. The degree is the power of that highest derivative, when the equation is a polynomial in the derivatives.
The highest derivative is \(\frac{d^3y}{dx^3}\), in the first term. The second term is meant to contain the second derivative \(\frac{d^2y}{dx^2}\) (see the note below), which is of lower order, so:
Order = 3
\(\frac{d^3y}{dx^3}\) occurs only to the power 2:
Degree = 2
\(\text{Order} \times \text{Degree} = 3 \times 2 = 6\)
Why the other options are incorrect
Option (1) 3 — Incorrect. 3 is the order alone, not the product.
Option (3) 2 — Incorrect. 2 is the degree alone, not the product.
Option (4) 9 — Incorrect. It takes the power 3 of the second term as the degree. The degree is the power of the highest order derivative only, which is 2.
Note: It seems that there is a mistake in the original NTA paper, which prints the second term as \(x^2y\left(\frac{d^2y}{dx^3}\right)^3\) when it should read \(x^2y\left(\frac{d^2y}{dx^2}\right)^3\). The question can still be solved because the highest derivative, \(\frac{d^3y}{dx^3}\), and its power 2 come from the first term, so the product is 6 either way.