Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| Differential equation | Integrating factor of the differential equation | ||
| A. | \( \displaystyle\frac{dy}{dx}+2y=xe^{4x} \) | I. | \( \displaystyle\frac{1}{x} \) |
| B. | \( 2x\displaystyle\frac{dy}{dx}+y=6x^3 \) | II. | \( \displaystyle\frac{1}{y} \) |
| C. | \( y\,dx-x\,dy+(\log_e x)\,dx=0 \) | III. | \( e^{2x} \) |
| D. | \( y\,dx-(x+2y^2)\,dy=0 \) | IV. | \( \sqrt{x} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-III, B-IV, C-I, D-II
For \( \frac{dy}{dx}+Py=Q \) the integrating factor is \( e^{\int P\,dx} \); for \( \frac{dx}{dy}+P_1x=Q_1 \) it is \( e^{\int P_1\,dy} \).
(A) \( \frac{dy}{dx}+2y=xe^{4x} \)
This matches (III) \( e^{2x} \).
\( \displaystyle P=2 \)
\( \displaystyle \text{I.F.}=e^{\int 2\,dx}=e^{2x} \)
(B) \( 2x\frac{dy}{dx}+y=6x^3 \)
This matches (IV) \( \sqrt{x} \). Dividing by 2x makes the coefficient of \( \frac{dy}{dx} \) equal to 1:
\( \displaystyle \frac{dy}{dx}+\frac{1}{2x}\,y=3x^2 \)
\( \displaystyle \text{I.F.}=e^{\int\frac{1}{2x}dx} \)
\( \displaystyle =e^{\frac{1}{2}\log x} \)
\( \displaystyle =\sqrt{x} \)
(C) \( y\,dx-x\,dy+(\log_e x)\,dx=0 \)
This matches (I) \( \frac{1}{x} \).
\( \displaystyle x\frac{dy}{dx}=y+\log x \)
\( \displaystyle \frac{dy}{dx}-\frac{1}{x}\,y=\frac{\log x}{x} \)
\( \displaystyle \text{I.F.}=e^{-\int\frac{1}{x}dx} \)
\( \displaystyle =e^{-\log x}=\frac{1}{x} \)
(D) \( y\,dx-(x+2y^2)\,dy=0 \)
This matches (II) \( \frac{1}{y} \). The term \( y^2 \) makes it linear in x, so it is written as an equation for \( \frac{dx}{dy} \):
\( \displaystyle \frac{dx}{dy}=\frac{x+2y^2}{y} \)
\( \displaystyle \frac{dx}{dy}-\frac{1}{y}\,x=2y \)
\( \displaystyle \text{I.F.}=e^{-\int\frac{1}{y}dy} \)
\( \displaystyle =e^{-\log y}=\frac{1}{y} \)
So the matching is A-III, B-IV, C-I, D-II.