Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

Question:

Match List-I with List-II

List-I

List-II

(A) Integrating factor of $xdy-(y + x^2)dx = 0$

(I) $x^2$

(B) Integrating factor of $xdy + (2y + x^2)dx = 0$

(II) $x^3$

(C) Integrating factor of $(3y-x^2)dx + xdy = 0$

(III) $x$

(D) Integrating factor of $(y + 3x^2)dx + xdy = 0$

(IV) $\frac{1}{x}$

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (1) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

List-I

List-II

(A) Integrating factor of $xdy-(y + x^2)dx = 0$

(IV) $\frac{1}{x}$

(B) Integrating factor of $xdy + (2y + x^2)dx = 0$

(I) $x^2$

(C) Integrating factor of $(3y-x^2)dx + xdy = 0$

(II) $x^3$

(D) Integrating factor of $(y + 3x^2)dx + xdy = 0$

(III) $x$

$\text{(A) } x\,dy - (y + x^2)\,dx = 0$

$\Rightarrow \frac{dy}{dx} = \frac{y + x^2}{x}$

$\text{IF} = e^{\int -\frac{1}{x}\,dx} = \frac{1}{x}$

$\Rightarrow \text{(A)} \rightarrow \text{(IV)}$

$\text{(B) } x\,dy + (2y + x^2)\,dx = 0$

$\Rightarrow \frac{dy}{dx} = -\frac{2y + x^2}{x}$

$\text{IF} = e^{\int \frac{2}{x}\,dx} = x^2$

$\Rightarrow \text{(B)} \rightarrow \text{(I)}$

$\text{(C) } (3y - x^2)\,dx + x\,dy = 0$

$\Rightarrow \frac{dy}{dx} = -\frac{3y - x^2}{x}$

$\text{IF} = e^{\int \frac{3}{x}\,dx} = x^3$

$\Rightarrow \text{(C)} \rightarrow \text{(II)}$

$\text{(D) } (y + 3x^2)\,dx + x\,dy = 0$

$\Rightarrow \frac{dy}{dx} = -\frac{y + 3x^2}{x}$

$\text{IF} = e^{\int \frac{1}{x}\,dx} = x$

$\Rightarrow \text{(D)} \rightarrow \text{(III)}$