Match List-I with List-II. Match the integrating factors: List-I List-II (Differential Equation) (Integrating factor) (A) d y d x + 3 y = e - 2 x (I) 1 x (B) x d y d x + y = 3 x 2 (II) e - x (C) x d y d x - y = 3 x 2 (III) x (D) d y d x -

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Solution

Q.
Match List-I with List-II. Match the integrating factors:

List-I List-II

(Differential Equation) (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:

1 (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

2 (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

3 (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

4 (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Ans.
(1)

Ans. (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Explanation:

$(A)$ $\frac{d y}{d x} + 3 y = e^{- 2 x}$

$\Rightarrow$     $I . F . = e^{\int 3 d x} = e^{3 x}$

$(B)$    $x \frac{d y}{d x} + y = 3 x^{2}$

$\Rightarrow$    $\frac{d y}{d x} + \frac{1}{x} y = 3 x$

   $I . F . = e^{\int \frac{1}{x} d x}$

$= e^{\log x} = x$

$(C)$ $x \frac{d y}{d x} - y = 3 x^{2}$

$\Rightarrow$ $\frac{d y}{d x} - \frac{1}{x} y = 3 x$

    $I . F . = e^{\int \frac{- 1}{x} d x} = e^{- \log x} = \frac{1}{x}$

$(D)$    $\frac{d y}{d x} - y = x$

$\Rightarrow$    $I . F . = e^{\int - 1 d x} = e^{- x}$

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