Particular solution of the differential equation x(1 + y^2)dx - y(1 + x^2)dy = 0, given y = 0 when x = 1, is Option 1: (1 + x^2) = 2(1 + y^2) Option 2: (1 + y^2) = 2(1 + x^2) Option 3: (1 + x^2) = 4(1 + y^2) Option 4: (1 + y^2) = 4(1 + x^2)

Answer: (1 + x^2) = 2(1 + y^2). Solution: Given differential equation: x(1 + y^2)dx - y(1 + x^2)dy = 0 Initial condition: When x = 1, y = 0 x(1 + y^2)dx = y(1 + x^2)dy x/(1 + x^2)dx = y/(1 + y^2)dy x/(1 + x^2)dx = y/(1 + y^2)dy For the left side, let u = 1 + x^2, then du = 2x dx: x/(1 + x^2)dx = 1/2 1/udu = 1/2, 1 + x^2, Similarly: y/(1 + y^2)dy = 1/2, 1 + y^2, 1/2(1 + x^2) = 1/2(1 + y^2) + C (1 + x^2) = (1 + y^2) + 2C (1 + x^2) - (1 + y^2) = 2C ((1 + x^2)/(1 + y^2)) = 2C (1 + x^2)/(1 + y^2) = e^2C = A (1 + x^2) = A(1 + y^2) When x = 1, y = 0: (1 + 1^2) = A(1 + 0^2) 2 = A (1 + x^2) = 2(1 + y^2)

Calculus

Differential Equations

Medium

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Particular solution of the differential equation $x(1 + y^2)dx - y(1 + x^2)dy = 0$ , given $y = 0$ when $x = 1$ , is

$(1 + x^2) = 2(1 + y^2)$

$(1 + y^2) = 2(1 + x^2)$

$(1 + x^2) = 4(1 + y^2)$

$(1 + y^2) = 4(1 + x^2)$

✅ Correct Option: 1

Given differential equation: $x(1 + y^2)dx - y(1 + x^2)dy = 0$

Initial condition: When $x = 1$ , $y = 0$

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

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

$\int \frac{x}{1 + x^2}dx = \int \frac{y}{1 + y^2}dy$

For the left side, let $u = 1 + x^2$ , then $du = 2x \, dx$ :

$\int \frac{x}{1 + x^2}dx = \frac{1}{2}\int \frac{1}{u}du = \frac{1}{2}\ln|1 + x^2|$

Similarly:

$\int \frac{y}{1 + y^2}dy = \frac{1}{2}\ln|1 + y^2|$

$\frac{1}{2}\ln(1 + x^2) = \frac{1}{2}\ln(1 + y^2) + C$

$\ln(1 + x^2) = \ln(1 + y^2) + 2C$

$\ln(1 + x^2) - \ln(1 + y^2) = 2C$

$\ln\left(\frac{1 + x^2}{1 + y^2}\right) = 2C$

$\frac{1 + x^2}{1 + y^2} = e^{2C} = A$

$(1 + x^2) = A(1 + y^2)$

When $x = 1$ , $y = 0$ :

$(1 + 1^2) = A(1 + 0^2)$

$2 = A$

$(1 + x^2) = 2(1 + y^2)$

2026: 22 May Shift 1

The general solution of the differential equation $\frac{dy}{dx}=e^{2x-y}+xe^{-y}$ is:

2026: 12 May Shift 2

The solution of the differential equation $\dfrac{dy}{dx} = x^2y^2$ with $y = 1$ at $x = 0$ is

2025: 22 May Shift 2

Match List-I with List-II

List-I	List-II
Differential Equation	Order and Degree
(A) $\left(\frac{d^2y}{dx^2}\right)^2 = e^x\left(\frac{dy}{dx}\right)^4 + 1 = 0$	(I) order = 1 and degree = 2
(B) $\left(\frac{dy}{dx}\right)^2 + xy = 0$	(II) order = 2 and degree = 1
(C) $\left(1 + \frac{dy}{dx}\right)^{3/2} = 4\left(\frac{d^2y}{dx^2}\right)^2$	(III) order = 2 and degree = 2
(D) $\sqrt\frac{d^2y}{dx^2} + 1 = \frac{dy}{dx}$	(IV) order = 2 and degree = 4

Choose the correct answer from the options given below:

2024: 16 May Shift 1

For the differential equation $\left(x \log _{e} x\right) d y=\left(\log _{e} x-y\right) d x$

(A) Degree of the given differential equation is $1$ .

(B) It is a homogeneous differential equation.

(C) Solution is $2y \log _{\mathrm{e}} \mathrm{x}+A=\left(\log _{\mathrm{e}} \mathrm{x}\right)^{2}$ , where $A$ is an arbitrary constant

(D) Solution is $2 y \log _{e} x+A=\log _{e}\left(\log _{e} x\right)$ , where $A$ is an arbitrary constant

Choose the correct answer from the options given below :

2025: 27 May Shift 1

Match List-I with List-II

List-I	List-II
(A) The degree of the differential equation $\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2 = x\sin\left(\frac{dy}{dx}\right)$	(I) 4
(B) The degree of differential equation $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^{1/4} + x^{1/5} = 0$	(II) 1
(C) The degree of differential equation $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 + 6y^5 = 0$	(III) Not defined
(D) The degree of differential equation $1 + \left(\frac{dy}{dx}\right)^4 = 7\left(\frac{d^2y}{dx^2}\right)^3$	(IV) 3

Choose the correct answer from the options given below:

2025: 22 May Shift 1

The differential equation of the family of curves $y = Ae^{3x} + Be^{-3 x}$ , where $a$ and $\beta$ are arbitrary constants, is

Particular solution of the differential equation x(1+y2)dx−y(1+x2)dy=0x(1 + y^2)dx - y(1 + x^2)dy = 0x(1+y2)dx−y(1+x2)dy=0, given y=0y = 0y=0 when x=1x = 1x=1, is

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CUET Mathematics Past Year Questions (Topic-Wise)

Particular solution of the differential equation $x(1 + y^2)dx - y(1 + x^2)dy = 0$ , given $y = 0$ when $x = 1$ , is