The general solution of the differential equation (y-x)dy + (x+y)dx = 0 is: (where C is arbitrary constant) Option 1: ^-1y/x = _e(sqrt(x^2+y^2)) + C Option 2: ^-1y = _esqrt(x^2+y^2) + C Option 3: ^-1x = y_e(x^2+y^2) + C Option 4: ^-1x + ^-1y = C

The answer is '^-1y/x = _e(sqrt(x^2+y^2)) + C'. View full solution:

CUET Mathematics - The general solution of the differential equation (y-x)dy + (x+y)dx = 0 is: (where C is arbitrary constant) | PYQs + Solutions

CUET Mathematics 2026 21st May Shift 2

Calculus

Hard

core

$\tan^{-1}\dfrac{y}{x} = \log_e\left(\sqrt{x^2+y^2}\right) + C$

$\tan^{-1}y = \log_e\sqrt{x^2+y^2} + C$

$\tan^{-1}x = y\log_e(x^2+y^2) + C$

$\tan^{-1}x + \tan^{-1}y = C$

✅ Correct Option: 1

Related questions:

15 June Shift 2

The differential equation whose solution is $Ax^2 + By^2 = 1$ where A and B are arbitrary constant is of :

(A) first order and first degree

(B) second order and first degree

(D) second order

Choose the correct answer from the options given below :

29 May Shift 2

Match List-I with List-II:

List-I	List-II
Differential Equations	Degree/Order
(A) Degree of the differential equation $\frac{d^3y}{dx^3} + 2 \log x.y = 0$	(I) 3
(B) Order of the differential equation $\frac{d^4y}{dx^4} + \left(\frac{dy}{dx}\right)^4 + xy = 0$	(II) 2
(C) Degree of the differential equation $\left(\frac{d^4y}{dx^4}\right)^2 + \left(\frac{dy}{dx}\right)^3 + x^2y = 0$	(III) 1
(D) Order of the differential equation $\frac{d^3y}{dx^3} + y\left(\frac{dy}{dx}\right)^3 = 0$	(IV) 4

7 Aug Shift 2

The differential equation representing family of curves $y = e^{-2x}(a \cos x + b \sin x)$ , where a and b are arbitrary constant, is :