If U(x) = x + √(1 + x^2), then solution of the differential equation dy/dx + √((1 + y^2)/(1 + x^2)) = 0, is : Option 1: U(x) U(y) = U(c) Option 2: (U(x))/(U(y)) = U(c) Option 3: U(x) - U(y) = U(c) Option 4: U(x) + U(y) = U(c) (where c is arbitrary constant)

Answer: U(x) U(y) = U(c). Solution: Separate variables: dy/(√(1+y^2)) + dx/(√(1+x^2)) = 0. Integrating: (y + √(1+y^2)) + (x + √(1+x^2)) = k. So (U(y)) + (U(x)) = U(c). Hence U(x) * U(y) = U(c).

Calculus

Differential Equations

Medium

If $U(x) = x + \sqrt{1 + x^2}$ , then solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1 + y^2}{1 + x^2}} = 0$ , is :

$U(x) U(y) = U(c)$

$\frac{U(x)}{U(y)} = U(c)$

$U(x) - U(y) = U(c)$

$U(x) + U(y) = U(c)$

(where c is arbitrary constant)

✅ Correct Option: 1

Separate variables: $\frac{dy}{\sqrt{1+y^2}} + \frac{dx}{\sqrt{1+x^2}} = 0$ .

Integrating: $\ln(y + \sqrt{1+y^2}) + \ln(x + \sqrt{1+x^2}) = \ln k$ .

So $\ln(U(y)) + \ln(U(x)) = \ln U(c)$ .

Hence $U(x) \cdot U(y) = U(c)$ .

2026: 29 May Shift 1

Match List-I with List-II

List-I (Differential Equation)	List-II (Degree)
(A) $\left(\dfrac{dy}{dx}\right)^3+\left(\dfrac{d^2y}{dx^2}\right)^2=0$	(I) Not defined
(B) $\left(\dfrac{d^2y}{dx^2}\right)^3+\left(\dfrac{dy}{dx}\right)^2+1=0$	(II) 2
(C) $\sqrt{1+\dfrac{d^2y}{dx^2}}=\dfrac{dy}{dx}+x$	(III) 1
(D) $\dfrac{d^2y}{dx^2}+e^{\frac{dy}{dx}}=0$	(IV) 3

Choose the correct answer from the options given below:

2025: 26 May Shift 2

Curd is at 80° F, five minutes later it came down at 60°F. After another 5 minutes, its temperature became 50° F. Given that the rate of change of temperature is proportional to (T - S), where S is temperature of the surroundings and T is temperature of the curd at any time t. Then the temperature of the surroundings is :

2026: 14 May Shift 2

The general solution of the differintial equation : $x\frac{dy}{dx}-ay=1(a\neq0)$ is

2026: 21 May Shift 2

Match the differential equation (List-I) with its order and degree (List-II)

LIST-I		LIST-II
A. $\dfrac{d^3y}{dx^3} + x^2\left(\dfrac{d^2y}{dx^2}\right) = 0$		I. order =3; degree is not defined
B. $\dfrac{d^3y}{dx^3} + y^2 + e^{\frac{dy}{dx}} = 0$		II. order =3; degree=1
C. $1 - \left(\dfrac{dy}{dx}\right)^2 = \left(a\dfrac{d^2y}{dx^2}\right)^{1/3}$		III. order =2; degree=1
D. $y = x\left(\dfrac{dy}{dx}\right)^3 + \left(\dfrac{d^2y}{dx^2}\right)^4$		IV. order =2; degree=4

Choose the correct answer from the options given below:

2023: 22 May Shift 3

Integerating factor of $(x \log_e x) \frac{dy}{dx} + y = 2 \log_e x$ is :

2026: 30 May Shift 1

The general solution of the differential equation $\frac{dy}{dx}-\frac{y}{x}+\text{cosec}\left(\frac{y}{x}\right)=0$ is

(Consider $\log x=\log_e x$ )

If U(x)=x+1+x2U(x) = x + \sqrt{1 + x^2}U(x)=x+1+x2​, then solution of the differential equation dydx+1+y21+x2=0\frac{dy}{dx} + \sqrt{\frac{1 + y^2}{1 + x^2}} = 0dxdy​+1+x21+y2​​=0, is :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $U(x) = x + \sqrt{1 + x^2}$ , then solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1 + y^2}{1 + x^2}} = 0$ , is :