Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

Question:

Match List-I with List-II

List-I Differential Equation

List-II Integrating factor

(A) $x\frac{dy}{dx}-y = 2x^2$

(I) $e^{-y}$

(B) $\frac{dy}{dx}+\frac{y}{x}=2x$

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

(C) $x\frac{dy}{dx}+ 2y = x^2\log x$

(III) $x$

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

(IV) $x^2$

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (3) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

List-I Differential Equation

List-II Integrating factor

(A) $x\frac{dy}{dx}-y = 2x^2$

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

(B) $\frac{dy}{dx}+\frac{y}{x}=2x$

(III) $x$

(C) $x\frac{dy}{dx}+ 2y = x^2\log x$

(IV) $x^2$

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

(I) $e^{-y}$

Each equation is reduced to linear form and integrating factor identified.

(A) $x\frac{dy}{dx}-y=2x^2$

$\frac{dy}{dx}-\frac{1}{x}y=2x$

Integrating factor $=e^{\int -\frac{1}{x}dx}=e^{-\log x}=\frac{1}{x}$

(A) → (II)

(B) $\frac{dy}{dx}+\frac{y}{x}=2x$

Integrating factor $=e^{\int \frac{1}{x}dx}=x$

(B) → (III)

(C) $x\frac{dy}{dx}+2y=x^2\log x$

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

Integrating factor $=e^{\int \frac{2}{x}dx}=e^{2\log x}=x^2$

(C) → (IV)

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

$\frac{dx}{dy}-x=y$ is linear in $x$ with variable $y$

Integrating factor $=e^{\int (-1)dy}=e^{-y}$

(D) → (I)

Final Matching: (A)-(II), (B)-(III), (C)-(IV), (D)-(I).