Match List - I with List - II. Match the integrating factors :
| List - I (Differential Equation) |
List - II (Integrating factor) |
| (A) $\frac{d y}{d x}+3 y=e^{-2 x}$ | (I) $\frac{1}{x}$ |
| (B) $x \frac{d y}{d x}+y=3 x^2$ | (II) $e^{-x}$ |
| (C) $x \frac{d y}{d x}-y=3 x^2$ | (III) $x$ |
| (D) $\frac{d y}{d x}-y=x$ | (IV) $e^{3 x}$ |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 1
B. $x \frac{d y}{d x}+y=3 x^2 \Rightarrow \frac{d y}{d x}+\frac{1}{x}y=\frac{3 x^2}{x}$ (dividing eq by leading coefficient of $\frac{dy}{dx}$)
So integrating factor
$=e^{\int \frac{1}{x} d x}=e^{\log x}=x$ → III
C. $x \frac{d y}{d x}-y=3 x^2 \Rightarrow \frac{d y}{d x}-\frac{1}{x} y=\frac{3 x^2}{x}$ (dividing eq by leading coefficient of $\frac{dy}{dx}$)
So integrating factor
$=e^{\int p ~d x}=e^{\int \frac{-1}{x} d x}=e^{-\ln |x|}=e^{\ln (\frac{1}{x})} = \frac{1}{x}$ → I
D. $\frac{d y}{d x} {-y}=x$
so integrating factor
$=e^{\int p d x}=e^{\int-1 d x}=e^{-x}$ → II