Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Differential Equations | Order and degree | ||
| (A) | \(\displaystyle\frac{dy}{dx}+e^y = 0\) | (I) | order 2, degree not defined |
| (B) | \(\displaystyle\frac{d^2y}{dx^2} =\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}\) | (II) | order 2, degree 1 |
| (C) | \(\displaystyle\left(\frac{d^2y}{dx^2}\right)^2+e^{\left(\frac{dy}{dx}\right)}=0\) | (III) | order 1, degree 1 |
| (D) | \(\displaystyle\frac{d^2y}{dx^2}+x\frac{dy}{dx}-2y= \log x;\ x \gt 0\) | (IV) | order 2, degree 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
The order is the order of the highest derivative. The degree is the power of that highest derivative, defined only when the equation is a polynomial in the derivatives.
(A) \(\displaystyle\frac{dy}{dx}+e^y = 0\)
This matches (III) order 1, degree 1. The highest derivative is \(\frac{dy}{dx}\), to the power 1; \(e^y\) contains y only, not a derivative.
(B) \(\displaystyle\frac{d^2y}{dx^2} =\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}\)
This matches (IV) order 2, degree 2. Squaring both sides removes the fractional power:
\( \displaystyle \left(\frac{d^2y}{dx^2}\right)^2 = \left[1+\left(\frac{dy}{dx}\right)^2\right]^3 \)
Now the equation is a polynomial in the derivatives, and \(\frac{d^2y}{dx^2}\) appears to the power 2.
(C) \(\displaystyle\left(\frac{d^2y}{dx^2}\right)^2+e^{\left(\frac{dy}{dx}\right)}=0\)
This matches (I) order 2, degree not defined. The order is 2, but the term \(e^{\left(\frac{dy}{dx}\right)}\) is not a polynomial in \(\frac{dy}{dx}\), so the degree is not defined.
(D) \(\displaystyle\frac{d^2y}{dx^2}+x\frac{dy}{dx}-2y= \log x;\ x \gt 0\)
This matches (II) order 2, degree 1. The term \(\log x\) contains x only, and \(\frac{d^2y}{dx^2}\) appears to the power 1.
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).
Tip: Remove fractional powers of the derivatives before reading the degree. In (B), reading the degree as 1 straight from the printed form is the usual mistake.