Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Differential Equations

Question:

Match List-I with List-II

List-I Differential Equation

List-II General solution

(A) $\frac{dy}{dx}=\frac{y}{x}; x≠0$

(I) $y = cx$: c is an arbitray constant

(B) $xdx – ydy= 0; y ≠ 0, x ≠ 0$

(II) $x^2-y^2 = c$: c is an arbitray constant

(C) $\frac{(x^2-1)}{y^2+1}\frac{dy}{dx}= 1$

(III) $2x+3y= c$: c is an arbitray constant

(D) $2dx+3dy = 0$

(IV) $(x^3-y^3) = c + 3(x + y)$: c is an arbitray constant

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (4) → (A)-(I), (B)-(II), (C)-(IV), (D)-(III)

List-I Differential Equation

List-II General solution

(A) $\frac{dy}{dx}=\frac{y}{x}; x≠0$

(I) $y = cx$: c is an arbitray constant

(B) $xdx – ydy= 0; y ≠ 0, x ≠ 0$

(II) $x^2-y^2 = c$: c is an arbitray constant

(C) $\frac{(x^2-1)}{y^2+1}\frac{dy}{dx}= 1$

(IV) $(x^3-y^3) = c + 3(x + y)$: c is an arbitray constant

(D) $2dx+3dy = 0$

(III) $2x+3y= c$: c is an arbitray constant

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

Integrating:

$\frac{dy}{y}=\frac{dx}{x}$

$\ln y=\ln x + c$

$y=cx$

Matches (I)

(B) $x\,dx - y\,dy = 0$

Integrating:

$\int x\,dx = \int y\,dy$

$\frac{x^2}{2} = \frac{y^2}{2} + c$

$x^2 - y^2 = c$

Matches (II)

(C) $\frac{(x^2-1)}{y^2+1}\frac{dy}{dx}=1$

Rewriting:

$(x^2-1)\,dx = (y^2+1)\,dy$

Integrating:

$\int (x^2-1)\,dx = \int (y^2+1)\,dy$

$\frac{x^3}{3} - x = \frac{y^3}{3} + y + c$

$x^3 - y^3 = c + 3(x + y)$

Matches (IV)

(D) $2\,dx + 3\,dy = 0$

Integrating:

$2x + 3y = c$

Matches (III)

Final Matching:

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