Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| Differential equation | Integrating factor | ||
| (A) | \( \displaystyle\frac{dy}{dx}+3y=e^{-2x} \) | (I) | \( \displaystyle\frac{1}{x} \) |
| (B) | \( \displaystyle\frac{xdy}{dx}+y=3x^2 \) | (II) | \( e^{-x} \) |
| (C) | \( \displaystyle\frac{xdy}{dx}-y=3x^2 \) | (III) | \( x \) |
| (D) | \( \displaystyle\frac{dy}{dx}-y=x \) | (IV) | \( e^{3x} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
In the form \( \frac{dy}{dx}+Py=Q \), the integrating factor is \( e^{\int P\,dx} \).
(A) \( \frac{dy}{dx}+3y=e^{-2x} \)
This matches (IV) \( e^{3x} \).
\( \displaystyle P=3 \)
\( \displaystyle \text{I.F.}=e^{\int 3\,dx}=e^{3x} \)
(B) \( \frac{xdy}{dx}+y=3x^2 \)
This matches (III) x.
Dividing by x,
\( \displaystyle \frac{dy}{dx}+\frac{1}{x}y=3x \)
\( \displaystyle \text{I.F.}=e^{\int\frac{1}{x}dx}=e^{\log x}=x \)
(C) \( \frac{xdy}{dx}-y=3x^2 \)
This matches (I) \( \frac{1}{x} \).
\( \displaystyle \frac{dy}{dx}-\frac{1}{x}y=3x \)
\( \displaystyle \text{I.F.}=e^{-\int\frac{1}{x}dx}=e^{-\log x}=\frac{1}{x} \)
(D) \( \frac{dy}{dx}-y=x \)
This matches (II) \( e^{-x} \).
\( \displaystyle P=-1 \)
\( \displaystyle \text{I.F.}=e^{-x} \)
So the matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II).
Tip: Always divide by the coefficient of dy/dx before reading P; in (B) and (C) the P is \( \pm\frac{1}{x} \), not \( \pm 1 \).