Match List-I with List-II.
| List-I | List-II | ||
| Differential equation | Integrating Factor (I.F) | ||
| (A) | $\frac{dy}{dx}-\frac{y}{x}=2x$ | (I) | $x^2-1$ |
| (B) | $\frac{dy}{dx}+\left(\frac{2x}{x^2-1}\right)y=\frac{2}{(x^2-1)^2}$ | (II) | $\sqrt{1-x^2}$ |
| (C) | $\frac{dy}{dx}-\left(\frac{x}{1-x^2}\right)y=\frac{1}{1-x^2}$ | (III) | $\frac{1}{x}$ |
| (D) | $\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{cotx}{1+x^2}$ | (IV) | $1+x^2$ |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
(A) $I.F.=e^{\int -\frac{1}{x}dx}=e^{-\log x}=\frac{1}{x}$ (III)
(B) $I.F.=e^{\int\frac{2x}{x^2-1}dx}=e^{\log|x^2-1|}=x^2-1$ (I)
(C) $I.F.=e^{\int\frac{-x}{1-x^2}dx}=e^{\frac{1}{2}\log|1-x^2|}=\sqrt{1-x^2}$ (II)
(D) $I.F.=e^{\int\frac{2x}{1+x^2}dx}=e^{\log|1+x^2|}=1+x^2$ (IV)