Match List-I with List-II
| List-I | List-II | ||
| Linear differential equation | Integrating factor | ||
| (A) | \( x\displaystyle\frac{dy}{dx}-y=2x^3 \) | (I) | \( e^{-x} \) |
| (B) | \( 2x\displaystyle\frac{dy}{dx}+y=3x^2 \) | (II) | \( e^{x} \) |
| (C) | \( \displaystyle\frac{dy}{dx}+y=\cos x-\sin x \) | (III) | \( x^{1/2} \) |
| (D) | \( \displaystyle\frac{dy}{dx}-y=e^{x} \) | (IV) | \( ^{1}/_{x} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
A linear equation in the form \( \frac{dy}{dx}+Py=Q \) has integrating factor \( e^{\int P\,dx} \). The coefficient of \( \frac{dy}{dx} \) must be 1 before P is read.
(A) \( x\frac{dy}{dx}-y=2x^3 \)
This matches (IV) \( ^{1}/_{x} \). Dividing by x,
\( \displaystyle \frac{dy}{dx}-\frac{1}{x}y=2x^2,\quad P=-\frac{1}{x} \)
\( \displaystyle \text{I.F.}=e^{-\int\frac{1}{x}dx}=e^{-\log x}=\frac{1}{x} \)
(B) \( 2x\frac{dy}{dx}+y=3x^2 \)
This matches (III) \( x^{1/2} \). Dividing by 2x,
\( \displaystyle \frac{dy}{dx}+\frac{1}{2x}y=\frac{3x}{2},\quad P=\frac{1}{2x} \)
\( \displaystyle \text{I.F.}=e^{\int\frac{1}{2x}dx}=e^{\frac{1}{2}\log x}=x^{1/2} \)
(C) \( \frac{dy}{dx}+y=\cos x-\sin x \)
This matches (II) \( e^{x} \). Here P = 1, so \( \text{I.F.}=e^{\int 1\,dx}=e^{x} \).
(D) \( \frac{dy}{dx}-y=e^{x} \)
This matches (I) \( e^{-x} \). Here P = −1, so \( \text{I.F.}=e^{\int(-1)\,dx}=e^{-x} \).
So the matching is (A) - (IV), (B) - (III), (C) - (II), (D) - (I).