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

Answer: ^(-1)y/x = _e(√(x^2+y^2)) + C. Solution: The equation is homogeneous: dy/dx = (x+y)/(x-y). Putting y=vx gives (1-v)/(1+v^2)dv = dx/x. Integrating, ^(-1)v - 12(1+v^2) = x + C, which simplifies to ^(-1)y/x = √(x^2+y^2) + C.

Calculus

Differential Equations

Hard

core

The general solution of the differential equation $(y-x)dy + (x+y)dx = 0$ is: (where C is arbitrary constant)

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

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

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

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

✅ Correct Option: 1

The equation is homogeneous: $\frac{dy}{dx} = \frac{x+y}{x-y}$ . Putting $y=vx$ gives $\frac{1-v}{1+v^2}dv = \frac{dx}{x}$ .

Integrating, $\tan^{-1}v - \frac12\log(1+v^2) = \log x + C$ , which simplifies to $\tan^{-1}\frac{y}{x} = \log\sqrt{x^2+y^2} + C$ .

2025: 19 May Shift 1

Match List-I with List-II

List-I	List-II
Differential Equation	Order and degree of differential equation
(A) $\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 = e^{\frac{dy}{dx}} + 1$	(I) Order = 1, Degree = 2
(B) $\left(\frac{d^2y}{dx^2}\right)^2 + 4\left(\frac{dy}{dx}\right)^3 = e^y - 1$	(II) Order = 2, Degree = 1
(C) $3\left(\frac{dy}{dx}\right) + 4y + e^y = \frac{dx}{dy}$	(III) Order = 2, Degree = 2
(D) $\frac{d^2y}{dx^2} + 3\left(\frac{dy}{dx}\right) = \left(e^y + \frac{dy}{dx}\right)^2$	(IV) Order = 2, Degree = Not defined

Choose the correct answer from the options given below:

2026: 30 May Shift 1

Let $y(x)$ be the solution curve of the differential equation

$\frac{dy}{dx} + 2y^2 = 0, y(1) = 1$ , then $y\left(\frac{3}{4}\right)$ is equal to:

2025: 22 May Shift 1

Which of the following statements is/are true?

(A) $(\tan^{-1} y - x)dy = (1 + y^2)dx$ is a differential equation where variables are separable.

(B) $(1 + x^2)dy + 2xydy = \cot x \ dx (x \neq 0)$ is a first order linear differential equation.

(D) $(xy)dx - (x + y^2)dy = 0$ is a homogeneous differential equation.

Choose the correct answer from the options given below:

2026: 29 May Shift 1

The general solution of the differential equation $(x+1)\dfrac{dy}{dx}=2e^{-y}-1$ is:

(where C is an arbitrary constant)

2022: 23 Aug Shift 1

Match List - I with List - II

List - I (Differential Equation)	List - II (Degree)
A. $\left[1 + (y')^2\right]^2 = y''$	I. $2$
B. $\left[1 + (y'')^3\right]^{\frac{1}{2}} = (y')^3$	II. $4$
C. $(y''')^2 + y'' + 3y' + 5y = e^x$	III. $1$
D. $\left[1 + (y')^3\right]^{\frac{1}{2}} = (y'')^2$	IV. $3$

Choose the correct answer from the option given below:

2025: 3 June Shift 1

If m and n are respectively the order and degree of the differential equation $(\frac{d^2y}{dx^2})^{2} + (\frac{dy}{dx})^3 + y= 4x$ , then the value of $m + n$ is:

The general solution of the differential equation (y−x)dy+(x+y)dx=0(y-x)dy + (x+y)dx = 0(y−x)dy+(x+y)dx=0 is: (where C is arbitrary constant)

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CUET Mathematics Past Year Questions (Topic-Wise)

The general solution of the differential equation $(y-x)dy + (x+y)dx = 0$ is: (where C is arbitrary constant)