Which of the following differential equations has \( y=x \) as one of its particular solution?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{d^{2}y}{dx^{2}}-x^{2}\displaystyle\frac{dy}{dx}+xy=0 \)
A function is a solution if it satisfies the equation identically. For y = x:
\( \displaystyle \frac{dy}{dx}=1,\qquad\frac{d^{2}y}{dx^{2}}=0 \)
Substituting in Option (2):
\( \displaystyle \frac{d^{2}y}{dx^{2}}-x^{2}\frac{dy}{dx}+xy=0-x^{2}(1)+x\cdot x \)
\( \displaystyle =0 \)
The left side is 0 for every x, so y = x is a solution of Option (2).
Why the other options are incorrect
Option (1) \( \displaystyle\frac{d^{2}y}{dx^{2}}-x^{2}\displaystyle\frac{dy}{dx}+xy=x \) — Incorrect. The left side is \( 0-x^{2}+x^{2}=0 \), but the right side is x, and 0 = x fails for x ≠ 0.
Option (3) \( \displaystyle\frac{d^{2}y}{dx^{2}}+x^{2}\displaystyle\frac{dy}{dx}+xy=0 \) — Incorrect. The left side is \( 0+x^{2}+x^{2}=2x^{2} \), which is not 0.
Option (4) \( \displaystyle\frac{d^{2}y}{dx^{2}}+x^{2}\displaystyle\frac{dy}{dx}+2xy=y \) — Incorrect. The left side is \( 0+x^{2}+2x^{2}=3x^{2} \), but the right side is y = x.